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Chapter 10

Test of Significance—Small Samples

Abstract

This chapter is devoted for the study of t-test and F-test that are known as small tests. The test of hypothesis about the variance of two populations is discussed in this chapter. The students’ t-test for difference of two means, paired t-test are discussed in this chapter.

Keywords

Moments about mean; assumptions for t-test; uses of t-distribution; types of t-test; significance of values of t

10.1 Introduction

In this chapter we discuss tests of significance for small samples. The important tests for small samples are

1. Chi-square test

2. t-test

3. F-test.

χ2 $χ^{2}$ Distribution was already introduced in the previous chapter. We now introduce t and F distributions in this chapter.

t-Distribution

When population standard deviation (SD) is not known and size of the sample is less than or equal to 30, we use t-test. The parameter t was introduced by W.S. Gosset in the year 1908. But t-distribution was established by Fisher in the year 1920. t-Distribution is also known as students t-distribution. Let x₁, x₂, …, x_n be the members of a random sample drawn from a normal population with mean μ and variance σ². We define t-test statistic as

t=x¯–μSn√ $t = \frac{\bar{x} - μ}{\frac{S}{\sqrt{n}}}$

where x¯ $\bar{x}$ is mean of the sample given by ∑xi/n $\sum x_{i} / n$ .

If we take (n−1)S2=∑ni=1(xi–x¯)2=nS2, $S^{2} = \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = n S^{2},$ then we get

t2ν=t2n–1=(x¯–μ)2n(n–1)S2=(x¯–μ)2/(σ2/n)nS2/σ2(v=n−1isdegreesoffreedom) $\frac{t^{2}}{ν} = \frac{t^{2}}{n - 1} = \frac{{(\bar{x} - μ)}^{2} n}{(n - 1) S^{2}} = \frac{{(\bar{x} - μ)}^{2} / (σ^{2} / n)}{n S^{2} / σ^{2}} (v = n - 1 is degrees of freedom)$

Since x_i, i=1, 2, 3, …, n is a random sample from the normal population with mean μ $μ$ and variance σ2 $σ^{2}$ , σ2x¯–μσ/n√~N(0,1) $σ^{2} \frac{\bar{x} - μ}{σ / \sqrt{n}} ~ N (0, 1)$ .

(x¯–μ)2n/σ2, ${(\bar{x} - μ)}^{2} n / σ^{2},$ is the square of a standard normal variate and it is distributed as chi-square variate, with one degrees of freedom. nS2/σ2 $n S^{2} / σ^{2}$ is also distributed as a χ2 $χ^{2}$ -variate with n−1 degrees of freedom. Hence t2/v $t^{2} / v$ is the ratio of two independent χ2 $χ^{2}$ -variates distributed with 1 and ν $ν$ degrees of freedom, respectively. Therefore t2/v $t^{2} / v$ is a β2(12,ν2) $β_{2} (\frac{1}{2}, \frac{ν}{2})$ variate. For different samples, it was shown by Fisher that its distribution is

y=y0(1+t2ν)ν+12,–∞<x<∞ $y = \frac{y_{0}}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}}, - \infty < x < \infty$ (10.1)

(10.1)

The constant y₀ is chosen in such a way that

∫∞–∞ydt=1 $\int_{- \infty}^{\infty} y d t = 1$

i.e., area of the curve given by Eq. (10.1) is unity

i.e.,

∫∞–∞y0(1+t2ν)ν+12dt=1 $\int_{- \infty}^{\infty} \frac{y_{0}}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}} d t = 1$

i.e.,

y0∫∞0(t2ν)−1/2(1+t2ν)12(ν+12)d(t2ν)=1 $y_{0} \int_{0}^{\infty} \frac{{(\frac{t^{2}}{ν})}^{- 1 / 2}}{{(1 + \frac{t^{2}}{ν})}^{\frac{1}{2} (\frac{ν + 1}{2})}} d (\frac{t^{2}}{ν}) = 1$

y0ν1/2β(12,ν2)=1 $y_{0} ν^{1 / 2} β (\frac{1}{2}, \frac{ν}{2}) = 1$

y0=1ν√β(12,ν2) $y_{0} = \frac{1}{\sqrt{ν} β (\frac{1}{2}, \frac{ν}{2})}$

y=f(t)=1ν√β(12,ν2)(1+t2ν)ν+12 $y = f (t) = \frac{1}{\sqrt{ν} β (\frac{1}{2}, \frac{ν}{2}) {(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}}$ (10.2)

(10.2)

Eq. (10.2) is known as t-distribution.

It is called the equation of t-probability curve.

The probability that we get a value of t between the limits t₁ and t₂ is given by

P=∫t2t1y0(1+t2ν)ν+12dt $P = \int_{t_{1}}^{t_{2}} \frac{y_{0}}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}} d t$

where

y0=1ν√β(12,ν2) $y_{0} = \frac{1}{\sqrt{ν} β (\frac{1}{2}, \frac{ν}{2})}$

10.2 Moments About Mean

The origin is mean for a t-distribution. All the moment of odd order about the origin vanish. The moment of even order about the origin is

μ'2r=2∫∞0t2rν√β(ν2,12)⋅dt(1+t2ν)ν+12=2∫∞0t2r(t2ν)−12β(ν2,12)⋅d(t2ν)(1+t2ν)ν+12=ν2β(ν2,12)∫∞0(t2ν)r+12−1(1+t2ν)ν+12d(t2ν) $\begin{matrix} {μ'}_{2 r} & = 2 \int_{0}^{\infty} \frac{t^{2 r}}{\sqrt{ν} β (\frac{ν}{2}, \frac{1}{2})} \cdot \frac{d t}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}} \\ = 2 \int_{0}^{\infty} \frac{t^{2 r} {(\frac{t^{2}}{ν})}^{\frac{- 1}{2}}}{β (\frac{ν}{2}, \frac{1}{2})} \cdot \frac{d (\frac{t^{2}}{ν})}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}} \\ = \frac{ν^{2}}{β (\frac{ν}{2}, \frac{1}{2})} \int_{0}^{\infty} \frac{{(\frac{t^{2}}{ν})}^{r + \frac{1}{2} - 1}}{{(1 + \frac{t^{2}}{ν})}^{\frac{ν + 1}{2}}} d (\frac{t^{2}}{ν}) \end{matrix}$

Substituting,

1+t2ν=y, $1 + \frac{t^{2}}{ν} = y,$

we get

t2ν=1−yy $\frac{t^{2}}{ν} = \frac{1 - y}{y}$

and

μ'2r=νβ(ν2.12)β(r+12,ν2−r),r<ν2 ${μ'}_{2 r} = \frac{ν}{β (\frac{ν}{2} . \frac{1}{2})} β (r + \frac{1}{2}, \frac{ν}{2} - r), r < \frac{ν}{2}$

Therefore

μ2r=(2r−1)(2r−3)…3.1(ν−2)(ν−4)…(ν−2r)⋅νr,n<ν2 $μ_{2 r} = \frac{(2 r - 1) (2 r - 3) \dots 3.1}{(ν - 2) (ν - 4) \dots (ν - 2 r)} \cdot ν^{r}, n < \frac{ν}{2}$

In particular

β1=μ23μ32β2=μ4μ3=0=3(ν−2)(ν−4)=3(1−2ν)(1−4ν) $\begin{array}{l} β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} & = 0 \\ β_{2} = \frac{μ_{4}}{μ_{3}} & = \frac{3 (ν - 2)}{(ν - 4)} \\ = \frac{3 (1 - \frac{2}{ν})}{(1 - \frac{4}{ν})} \end{array}$

As ν→∞ $ν \to \infty$ , we get

β2=Ltν→∞3(1−2ν)(1−4ν)=3 $β_{2} = \underset{ν \to \infty}{Lt} \frac{3 (1 - \frac{2}{ν})}{(1 - \frac{4}{ν})} = 3$

10.3 Properties of Probability Curve

1. Equation of t-probability contains only even powers of t, therefore the curve is symmetrical about the line t=0.

2. t-Distribution extends to infinity on either side and is asymptotic to t-axis at each end.

3. t-Distribution has a greater spread than the normal distribution and its graph is similar to that of normal distribution.

4. Sampling distribution of t is independent of the population parameters μ $μ$ and σ2 $σ^{2}$ , but it depends only on the size of the sample.

5. The curve has a maximum ordinate at t=0, the mean and mode coincide at t=0.

6. As ν→∞ $ν \to \infty$ , y→e−12tν $y \to e^{- \frac{1}{2} t ν}$ hence t is distributed normally.

10.4 Assumptions for t-Test

The t-test is applied under the following assumptions:

1. The samples are drawn from normal population and are random.

2. The population SD may not be known.

3. For testing the equality of two population means, the population variances are regarded as equal.

10.5 Uses of t-Distribution

The t-distribution is used to test the significance of:

1. a mean for small sample when the population variance is not known.

2. the difference between two means for small samples.

The values of t for different degrees of freedom are given at the end of this book.

10.6 Interval Estimate of Population Mean

Let x¯ $\bar{x}$ denote the sample mean and n denote the size of the sample. Then the interval estimate of the population mean μ $μ$ is given by x¯±t0.05(S/n−−√) $\bar{x} \pm t_{0.05} (S / \sqrt{n})$

x¯±tαSn−1√(whereS2n=S2n−1)S=∑(xi−x¯)2n−−−−−−−√andSn−1isthestandarderrormean. $\begin{array}{l} \bar{x} \pm t_{α} \frac{S}{\sqrt{n - 1}} (where \frac{S^{2}}{n} = \frac{S^{2}}{n - 1}) \\ S = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{n}} and \frac{S}{n - 1} is the standard error mean . \end{array}$

10.7 Types of t-Test

There are two types of t-tests:

1. Unpaired t-test.

2. Paired t-test.

10.8 Significant Values of t

The significant value of t at level of significance (Fig. 10.1) and degrees of freedom ν, $ν,$ for two-tailed test is given by

P{|t|>tν(α)}=αP{|t|≤tν(α)}=1−α $\begin{array}{l} P {| t | > t_{ν} (α)} = α \\ P {| t | \leq t_{ν} (α)} = 1 - α \end{array}$

The significant value of t at level of significance for a one-tailed test can be obtained from those of two-tailed test by referring to the values Zα $Z_{α}$ . The significant values are given α $α$ in the table known as the t-table.

If the calculated value of t exceeds the table value at 5% level of significance then the null hypothesis is rejected at 5%.

Similarly for 1% level of significance, we can reject the null hypothesis at 1% if the calculated value of t exceeds the table value.

10.9 Test of Significance of a Single Mean

To test the significance of a mean of a small sample, the test statistic is

t=x¯−μ(Sn−1√) $t = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})}$

where x¯=∑xi/n $\bar{x} = \sum x_{i} / n$ ; S/(n−1) $S / (n - 1)$ is the standard error of mean; n=size of the sample (n≤30); S=standard deviation =∑(xi−x¯)2n−−−−−−−√ $= \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{n}}$ .

If x¯ $\bar{x}$ is a fraction, we compute S by using the formulae x¯=A+(∑d/n),d=x−A $\bar{x} = A + (\sum d / n), d = x - A$

S2=1n−1[∑d2−∑(d)2n] $S^{2} = \frac{1}{n - 1} [\sum d^{2} - \frac{\sum {(d)}^{2}}{n}]$

where A is the assumed mean.

If the calculated value of |t| $| t |$ is less than the table value of tα $t_{α}$ (α=level of significance), we accept null hypothesis for the degrees of freedom ν $ν$ .

If the calculated value of |t| $| t |$ is greater than the table value tα $t_{α}$ of at the level of significance α, $α,$ for the degrees of freedom ν, $ν,$ we reject the null hypothesis and accept the alternative hypothesis.

10.9.1 Solved Examples

Example 10.1: A random sample of size 7 from a normal population gave a mean of 977.51 and a SD of 4.42. Find 95% confidence interval for the population mean?

Solution: Here we have n=7,x¯=977.51,df=n−1=6andS=4.42 $n = 7, \bar{x} = 977.51, d f = n - 1 = 6 and S = 4.42$

SE=Sn−1√=4.427−1√=4.426√=4.422.4494=1.8045 $S E = \frac{S}{\sqrt{n - 1}} = \frac{4.42}{\sqrt{7 - 1}} = \frac{4.42}{\sqrt{6}} = \frac{4.42}{2.4494} = 1.8045$

95% confidence limits for μ $μ$ are

x¯−t0.05Sn−1√<μ<x¯+t0.05Sn−1√ $\bar{x} - t_{0.05} \frac{S}{\sqrt{n - 1}} < μ < \bar{x} + t_{0.05} \frac{S}{\sqrt{n - 1}}$

i.e., 977.51–(2.447) (1.8045)<μ<977.51+(2.447) (1.8045)

i.e., 977.51–4.4156<μ<977.51+4.4156

i.e., 973.094<μ<981.9256

The required confidence interval is (973.094, 981.9256).

Example 10.2: A sample of 10 camshafts intended for use in gasoline engine has an average eccentricity of 0.044 in. The data may be treated as a random sample from a normal population. Determine a 95% confidence interval for actual mean eccentricity of the camshaft.

Solution: It is given that

n=size of the sample=10

S=Standard deviation=0.044 in.

x¯ $\bar{x}$ =1.02 in.

Degrees of freedom ν $ν$ =n–1=10–1=9

Sn√=0.04410√=0.0443.1622=0.0139 $\frac{S}{\sqrt{n}} = \frac{0.044}{\sqrt{10}} = \frac{0.044}{3.1622} = 0.0139$

The confidence interval is

x¯±tα/2Sn√=1.02±2.262×0.0139=1.02±0.0314 $\begin{array}{l} \bar{x} \pm t_{α / 2} \frac{S}{\sqrt{n}} & = 1.02 \pm 2.262 \times 0.0139 \\ = 1.02 \pm 0.0314 \end{array}$

i.e.,

(1.02−0.0314,1.02+0.0314)=(0.9886,1.0514) $\begin{array}{r} (1.02 - 0.0314, 1.02 + 0.0314) = (0.9886, 1.0514) \end{array}$

Example 10.3: Find the students t for following variable values in a sample of eight.

−4, −2, −2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.

Solution: Table for computing mean x¯ $\bar{x}$ and S

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$
−4	−4.25	18.0625
−2	−2.25	5.0625
−2	−2.25	5.0625
0	−0.25	0.0625
2	1.75	3.0625
2	1.75	3.0625
3	2.75	7.5625
3	2.75	7.5625
∑xi=2 $\sum x_{i} = 2$		49.5080

x¯S=∑xin=28=14=0.25=∑(xi−x¯)2n−−−−−−−√=49.50008−1−−−−−√=7.071428−−−−−−−√=2.6592 $\begin{array}{l} \bar{x} & = \frac{\sum x_{i}}{n} = \frac{2}{8} = \frac{1}{4} = 0.25 \\ S & = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{n}} = \sqrt{\frac{49.5000}{8 - 1}} = \sqrt{7.071428} \\ = 2.6592 \end{array}$

Test statistic:

t=x¯−μ(Sn√)=0.25−02.6592×8–√=(0.25)(2.8284)2.6592=0.2659 $\begin{array}{l} t & = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n}})} = \frac{0.25 - 0}{2.6592} \times \sqrt{8} \\ = \frac{(0.25) (2.8284)}{2.6592} \\ = 0.2659 \end{array}$

Example 10.4: A certain stimulus administered to each of 12 patients resulted in the following increases of blood pressures:

5, 2, 8, −1, 3, 0, 6, −2, 1, 5, 0, 4.

Can it be calculated that the stimulus will be in general accomplished by an increase in blood pressure given that for 11 degrees of freedom the value of is 2.201?

Solution: For 11 df=2.201 (given)

Meanofthesamplex¯=∑xin=5+2+8−1+3+0+6−2+1+5+0+412=3112=2.581 $\begin{array}{l} Mean of the sample \bar{x} = \frac{\sum x_{i}}{n} & = \frac{5 + 2 + 8 - 1 + 3 + 0 + 6 - 2 + 1 + 5 + 0 + 4}{12} \\ = \frac{31}{12} = 2.581 \end{array}$

Calculation of S

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$
5	2.42	5.8564
2	−0.58	0.3364
8	5.42	29.3764
−1	−3.58	12.8164
3	0.42	0.1764
0	−2.58	6.6564
6	3.42	11.6964
−2	−4.58	20.9764
1	−1.58	2.4964
5	2.42	5.8564
0	−2.58	6.6564
4	1.42	2.0614
		104.9268

S2=∑(xi−x¯)2n=104.926812−1=9.5389 $S^{2} = \frac{\sum {(x_{i} - \bar{x})}^{2}}{n} = \frac{104.9268}{12 - 1} = 9.5389$

∴S=9.5389−−−−−√=3.0885 $∴ S = \sqrt{9.5389} = 3.0885$

Assuming that mean of the universe is zero, we get

t=x¯−μ(Sn√)=2.58−03.0885×12−−√=(2.58)(3.464)3.0885=2.8936 $\begin{array}{l} t & = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n}})} = \frac{2.58 - 0}{3.0885} \times \sqrt{12} \\ = \frac{(2.58) (3.464)}{3.0885} = 2.8936 \end{array}$

Since the calculated value of t for 11 df at 5% level of significance is greater than the table value of t, we reject null hypothesis and conclude that stimulus will be in general accompanied by an increase in blood pressure.

Example 10.5: A random blood sample to test fasting sugar for 10 taxi drivers gave the following data (in mg/dL):

70, 120, 110, 101, 88, 83, 95, 107, 100, 98.

Do this support, the assumption of population mean of 100 mg/dL. Find a reasonable range in which the most of the mean fasting sugar test of 10 taxi drivers lie?

Solution: We have n=10

∑xix¯=∑xin=70+120+110+101+88+83+95+107+100+91=972=97210=97.2 $\begin{array}{l} \sum x_{i} & = 70 + 120 + 110 + 101 + 88 + 83 + 95 + 107 + 100 + 91 \\ = 972 \\ \bar{x} = \frac{\sum x_{i}}{n} & = \frac{972}{10} = 97.2 \end{array}$

Null Hypothesis H₀: μ=100

Alternative Hypothesis H₁: μ ≠ 100 (two-tailed test)

Degrees of freedom ν=n−1=10−1=9

Level of significance α=0.05

Table value of t for 9 df at 5% level of significance=2.262

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$
70	–27.2	739.84
120	22.8	519.84
110	12.8	163.84
101	3.8	14.44
88	–9.2	84.64
83	–14.2	201.64
95	–2.2	4.84
107	9.8	96.04
100	2.8	7.84
98	0.8	0.64
		∑(xi–x¯)2=1833.60 $\sum {(x_{i} - \bar{x})}^{2} = 1833.60$

S2=∑(xi−x¯)2n=1833.6010−1=1833.609=203.74 $S^{2} = \frac{\sum {(x_{i} - \bar{x})}^{2}}{n} = \frac{1833.60}{10 - 1} = \frac{1833.60}{9} = 203.74$

Standarderrorofmean=Sn√=203.74√10√=20.374−−−−−√ $Standard error of mean = \frac{S}{\sqrt{n}} = \frac{\sqrt{203.74}}{\sqrt{10}} = \sqrt{20.374}$

Test statistic:

t=x¯−μ(Sn√)=−2.820.374√=−2.84.5137=−0.6203 $t = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n}})} = \frac{- 2.8}{\sqrt{20.374}} = \frac{- 2.8}{4.5137} = - 0.6203$

∴|t|=0.6203<2.262 $∴ | t | = 0.6203 < 2.262$

i.e., calculated value of t at 5% level of significance for 9 df is less than the table value.

Hence we accept null hypothesis.

The 95% confidence limits for μ are

x¯±t0.05Sn√=97.2±2.262(4.51) $\bar{x} \pm t_{0.05} \frac{S}{\sqrt{n}} = 97.2 \pm 2.262 (4.51)$

i.e., [97.2–10.20, 97.2+10.20]

i.e., [87.0, 107.40]

i.e., The 95% confidence interval for is [87.0, 107.40].

Example 10.6: A sample of 26 bulbs gives a mean life of 990 hours with a SD of 20 hours. The manufacturer claims that the mean life of the bulbs is 1000 hours. Is the sample not up to the standard?

Solution: We have

n=26x¯=990μ=1000SD=20df=n–1=26–1=25 $\begin{array}{l} n = 26 \\ \bar{x} = 990 \\ μ = 1000 \\ S D = 20 \\ d f = n - 1 = 26 - 1 = 25 \end{array}$

Null Hypothesis H₀: The sample is up to standard

Alternative Hypothesis H₁: The sample is not up to the standard

t=x¯−μ(Sn−1√)=990−10002026−1√=−10(205)=−5020=−2.5 $\begin{array}{l} t & = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})} = \frac{990 - 1000}{\frac{20}{\sqrt{26 - 1}}} = \frac{- 10}{(\frac{20}{5})} \\ = \frac{- 50}{20} = - 2.5 \end{array}$

∴|t|=2.5 $∴ | t | = 2.5$

Table value of t for 25 df at 5% level is 1.708. Since the calculated value of t is greater than the table value of t, we reject null hypothesis.

Therefore the sample is not up to the standard.

Example 10.7: The average breaking strength of the steel rods is specified to be 18.5 thousand pounds. To test this, a sample of 14 rods was tested. The mean and SDs obtained were 17.85 and 1.955, respectively. Is the result of the experiment significant?

Solution: Here we have n=14, x¯=17.85, $\bar{x} = 17.85,$ μ=18.5 $μ = 18.5$

S=SD=1.955 $S = S D = 1.955$

Degrees of freedom=n–1=14–1=13

Table value of t for 13 df at 5% level of significance=2–16 (i.e., critical value)

Null Hypothesis H₀: The result of the experiment is not significant.

Alternative Hypothesis H₁: μ≠18.5 $μ \neq 18.5$

The test statistic

t=x¯−μ(Sn−1√)=17.85−18.51.95513√=−0.650.542=−1.199 $t = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})} = \frac{17.85 - 18.5}{\frac{1.955}{\sqrt{13}}} = \frac{- 0.65}{0.542} = - 1.199$

Therefore

|t|=1.199<2.166 $| t | = 1.199 < 2.166$

Calculated value of t is less than the table value of t. Hence we accept null hypothesis, i.e., the result of the experiment is not significant.

Example 10.8: A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025 cm. A random sample of 10 washers was found to have a thickness of 0.024 cm, with SD of 0.002 cm. Test the significance of the deviation (value of t for 9 degrees of freedom at 5% level of significance is 2.262).

Solution: Size of the given sample=10 (small sample)

Sample mean=x¯=0.024cm $\bar{x} = 0.024 cm$

Population mean μ=0.025cm $μ = 0.025 cm$

S=Standard deviation=0.002 cm

Degrees of freedom=n−1=10−1=9

Table value of t for 9 df at 5% level of significance=2.262

Null hypothesis H₀: The difference between x¯ $\bar{x}$ and μ is not significant.

Alternative hypothesis H₁: μ≠0.025 $μ \neq 0.025$

The test statistic

t=x¯−μ(Sn−1√)=0.024−0.0250.00210−1√=−0.010.002×3=−1.5 $\begin{array}{l} t & = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})} = \frac{0.024 - 0.025}{\frac{0.002}{\sqrt{10 - 1}}} \\ = \frac{- 0.01}{0.002} \times 3 = - 1.5 \end{array}$

∴|t|=1.5<2.262 $∴ | t | = 1.5 < 2.262$

Calculated value of t is less than the table value of t at 5% level of significance and 9 df. We accept null hypothesis and conclude that the difference between x¯ $\bar{x}$ and μ $μ$ is not significant.

Example 10.9: A sample of 20 items has a mean 42 units and SD 5 units. Test the hypothesis that it is a random sample from a normal population with mean 46 units.

Solution: We have n=20 (small sample), x¯=46, $\bar{x} = 46,$ μ=46, $μ = 46,$ S=5 $S = 5$

Null Hypothesis H₀: μ=46 $μ = 46$

Alternative Hypothesis H₁: μ≠46 $μ \neq 46$

Degrees of freedom=n−1=20−1=19

Table value of t for 19 df at 5% level of significance=2.09

The test statistic

t=x¯−μ(Sn−1√)=42−46519√=−419√5=−4×4.35885=−3.4871 $\begin{array}{l} t = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})} & = \frac{42 - 46}{\frac{5}{\sqrt{19}}} \\ = \frac{- 4 \sqrt{19}}{5} = \frac{- 4 \times 4.3588}{5} \\ = - 3.4871 \end{array}$

Therefore

|t|=3.4871>2.09 $| t | = 3.4871 > 2.09$

Calculated value of t is greater than the table value at 5% level for 19 df.

Hence we reject null hypothesis H₀: μ=46 $μ = 46$

i.e., the difference between sample mean and population mean is significant.

Example 10.10: A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025 cm. A random sample of 10 washers was found to have an average thickness of 0.024 cm with a SD of 0.002 cm. Test the significance of the deviation of mean (Table value of t for df at 5% level is 2.2627).

Solution: We have n=10 (small sample), x¯=0.024, $\bar{x} = 0.024,$ μ=0.025, $μ = 0.025,$ S=0.002 $S = 0.002$

Null Hypothesis H₀: μ=0.025cm $μ = 0.025 cm$

Alternative Hypothesis H₁: μ≠0.025 $μ \neq 0.025$

Degrees of freedom=n−1=10−1=9

Table value of t for 9 df at 5% level of significance=2.262

The test statistic

t=x¯−μ(Sn−1√)=0.024−0.0250.0023=−(0.001)×30.002=−1.5 $\begin{array}{l} t & = \frac{\bar{x} - μ}{(\frac{S}{\sqrt{n - 1}})} = \frac{0.024 - 0.025}{\frac{0.002}{3}} \\ = \frac{- (0.001) \times 3}{0.002} = - 1.5 \end{array}$

Therefore

|t|=1.5<2.262 $| t | = 1.5 < 2.262$

Calculated value of t is less than the table value at 5% level for 9df.

Hence we accept null hypothesis. We conclude that there is no significant deviation between the sample mean and population mean.

10.10 Student’s t-Test for Difference of Means

Suppose we want to test if two independent samples x1,x2,x3,…,xn1 $x_{1}, x_{2}, x_{3}, \dots, x_{n_{1}}$ and y1,y2,y3,…,yn2 $y_{1}, y_{2}, y_{3}, \dots, y_{n_{2}}$ of sizes n₁ and n₂ have been drawn from the normal population with means μ1 $μ_{1}$ and μ2 $μ_{2}$ , respectively.

Assume that the population variances are equal

The statistic t

t=(x¯−y¯)−(μ1−μ2)S1n1+1n2√ $t = \frac{(\bar{x} - \bar{y}) - (μ_{1} - μ_{2})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

where

x¯=∑xin1,y¯=∑yin2S2=1n1+n2−2[∑n1(xi−x¯)2+∑n1(yi−y¯)2] $\begin{array}{l} \begin{matrix} \bar{x} = \frac{\sum x_{i}}{n_{1}}, & \bar{y} = \frac{\sum y_{i}}{n_{2}} \end{matrix} \\ S^{2} = \frac{1}{n_{1} + n_{2} - 2} [\sum_{1}^{n} {(x_{i} - \bar{x})}^{2} + \sum_{1}^{n} {(y_{i} - \bar{y})}^{2}] \end{array}$

is an unbiased estimate of the population variance σ2 $σ^{2}$ .

t follows t-distribution with degrees of freedom n1+n2−2 $n_{1} + n_{2} - 2$

Under the null hypothesis H₀, that

1. Samples have been drawn from the populations with the same means, i.e., μ1=μ2 $μ_{1} = μ_{2}$

2. Sample means x¯ $\bar{x}$ and y¯ $\bar{y}$ do not differ significantly.

We compute the test statistic

t=(x¯−y¯)S1n1+1n2√ $t = \frac{(\bar{x} - \bar{y})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

The degrees of freedom is n1+n2−2 $n_{1} + n_{2} - 2$

If |t| $| t |$ is less than the table value of tα $t_{α}$ for a given level of significance α, $α,$ and df=n1+n2−2 $df = n_{1} + n_{2} - 2$

We accept null hypothesis H₀, otherwise reject the null hypothesis.

The confidence interval for μ1−μ2 $μ_{1} - μ_{2}$ is a (100, 1−α $1 - α$ )% interval given by

x¯−y¯±tα/2∑(xi−x¯)2+∑(yi−y¯)2n1+n2−2−−−−−−−−−−−−−−√[1n1+1n2] $\bar{x} - \bar{y} \pm t_{α / 2} \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2} + \sum {(y_{i} - \bar{y})}^{2}}{n_{1} + n_{2} - 2}} [\frac{1}{n_{1}} + \frac{1}{n_{2}}]$

i.e.,

x¯−y¯±tα/2S1n1+1n2−−−−−−−√ $\bar{x} - \bar{y} \pm t_{α / 2} S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}$

where the degrees of freedom is ν=n1+n2−2 $ν = n_{1} + n_{2} - 2$ .

Example 10.11: The nicotine content in milligrams of the samples of tobacco were found as follows:

Sample A	24	27	26	21	25
Sample B	27	30	28	31	22	36

Can it be said that the two samples come from normal populations with the sample mean.

Solution: We have

x¯=24+27+26+21+255=1235=24.6y¯=27+30+28+31+22+366=1746=29 $\begin{array}{l} \bar{x} = \frac{24 + 27 + 26 + 21 + 25}{5} = \frac{123}{5} = 24.6 \\ \bar{y} = \frac{27 + 30 + 28 + 31 + 22 + 36}{6} = \frac{174}{6} = 29 \end{array}$

Calculation of sample SD’s

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
24	–0.6	0.36	27	–2	4
27	2.4	5.76	30	1	1
26	1.4	1.96	28	–1	1
21	–3.6	12.96	31	2	4
25	0.4	0.16	22	–7	49
−	−	−	36	7	49
123	0	21.2	174	0	108

∴∑(xi−x¯)2=21.2,S2∑(yi−y¯)2=108=1n1+n2−2[∑n1(xi−x¯)2+∑n1(yi−y¯)2]=15+6+1[21.2+108]=14.35 $\begin{array}{l} ∴ \sum {(x_{i} - \bar{x})}^{2} = 21.2, & \sum {(y_{i} - \bar{y})}^{2} = 108 \\ S^{2} & = \frac{1}{n_{1} + n_{2} - 2} [\sum_{1}^{n} {(x_{i} - \bar{x})}^{2} + \sum_{1}^{n} {(y_{i} - \bar{y})}^{2}] \\ = \frac{1}{5 + 6 + 1} [21.2 + 108] = 14.35 \end{array}$

∴S=14.35−−−−√=3.78 $∴ S = \sqrt{14.35} = 3.78$

Null Hypothesis H₀: μ1=μ2 $μ_{1} = μ_{2}$ , i.e., the two samples have been drawn from normal populations with the same mean

Alternative Hypothesis H₁: μ1≠μ2 $μ_{1} \neq μ_{2}$

Level of significance=5%,

Degrees of freedom=n1+n2−2=5+6−2=9 $n_{1} + n_{2} - 2 = 5 + 6 - 2 = 9$

t0.05 $t_{0.05}$ for 9 df=2.262

Test statistic is

t=(x¯−y¯)S1n1+1n2√=24.6−293.7815+16√=−1.92 $t = \frac{(\bar{x} - \bar{y})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{24.6 - 29}{3.78 \sqrt{\frac{1}{5} + \frac{1}{6}}} = - 1.92$

∴|t|=1.92<2.262 $∴ | t | = 1.92 < 2.262$

The calculated value of t is less than the table value of t at 5% level of significance for 9 df. Therefore we accept H₀.

The samples are drawn from normal populations with the same mean.

Example 10.12: A group of 7-week-old chicken reared on a high-protein diet weigh 12, 15, 11, 16, 14, 14, and 16 ounces. A second group of 5 chickens similarly treated except that they receive a low-protein diet weigh 8, 10, 14, 10, and 13 ounces. Test whether there is evidence that additional protein has increased the weight of the chicken (The table value of t for ν $ν$ =10 at 5% level of significance is 2.33).

Solution: Let the weights of chicken reared on high-protein diet be denoted by x_i and the weights of chicken reared on low-protein diet be denoted by y_i

We have

x¯=∑xin1y¯=∑yin2=12+15+11+16+14+14+167=987=14,=8+10+14+10+135=555=11 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} & = \frac{12 + 15 + 11 + 16 + 14 + 14 + 16}{7} = \frac{98}{7} = 14, \\ \bar{y} = \frac{\sum y_{i}}{n_{2}} & = \frac{8 + 10 + 14 + 10 + 13}{5} = \frac{55}{5} = 11 \end{array}$

Null Hypothesis H₀: μ1=μ2 $μ_{1} = μ_{2}$

i.e., additional protein has not increased the weight of the chickens

Alternative Hypothesis H₁: μ1>μ2 $μ_{1} > μ_{2}$ (one-tailed test)

Computation of SD.

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
12	−2	4
15	1	1	8	−3	9
11	−3	9	10	−1	1
16	2	4	14	3	9
14	0	0	10	−1	1
14	0	0	13	2	4
16	2	4
98		22	55		24

S=∑(xi−x¯)2+∑(yi−y¯)2n1+n2−2−−−−−−−−−−−−−−√=22+247+5−2−−−−−√=4610−−√=4.6−−−√=2.1447 $S = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2} + \sum {(y_{i} - \bar{y})}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{22 + 24}{7 + 5 - 2}} = \sqrt{\frac{46}{10}} = \sqrt{4.6} = 2.1447$

Level of significance=5% (α=0.05)

Critical value, i.e., table value of t for 10 df at 5% level=1.81

Test statistic is

t=(x¯−y¯)S1n1+1n2√=14−112.144717+15√=32.1447×25√5+7√=3×5.9162.1447×3.4641=17.7487.4294=2.3888 $\begin{array}{l} t & = \frac{(\bar{x} - \bar{y})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{14 - 11}{2.1447 \sqrt{\frac{1}{7} + \frac{1}{5}}} \\ = \frac{3}{2.1447} \times \frac{\sqrt{25}}{\sqrt{5 + 7}} = \frac{3 \times 5.916}{2.1447 \times 3.4641} \\ = \frac{17.748}{7.4294} = 2.3888 \end{array}$

Since the calculated value of t is greater than the table value of t, at 5% level of significance for 10 degrees of freedom. We reject H₀ and conclude that additional protein has increased the weight of the chicken.

Example 10.13: Two salesmen A and B working in a certain district from a sampling survey conducted by the head office. The following results were obtained. State whether there is any significant difference in the average sales between the two salesmen?

	A	B
No. of sales	20	18
Average sales (in Rs.)	170	203
Standard deviation (in Rs.)	20	25

Solution: We have

n₁=20,	n₂=18
x¯1=170, ${\bar{x}}_{1} = 170,$	x¯2=205 ${\bar{x}}_{2} = 205$
S₁=20,	S₂=25

Degrees of freedom=n1+n2−2=20+16−2 $n_{1} + n_{2} - 2 = 20 + 16 - 2$

Level of significance α=0.05

Table value of t=1.9 (for 36 df at 5%)

Standarderrorofdifferenceofmeans=SE(x¯1−x¯2)=S1n1+1n2−−−−−−−√ $\begin{array}{l} Standard error of difference of means & = S E ({\bar{x}}_{1} - {\bar{x}}_{2}) \\ = S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} \end{array}$

where

S=n1S21+n2S22n1+n2−2−−−−−−−−√=(20)(20)2+(18)(25)220+18−2−−−−−−−−−−−−√=800+1125036−−−−−−−√=23.1240 $\begin{array}{l} S & = \sqrt{\frac{n_{1} S_{1}^{2} + n_{2} S_{2}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{(20) {(20)}^{2} + (18) {(25)}^{2}}{20 + 18 - 2}} \\ = \sqrt{\frac{800 + 11250}{36}} = 23.1240 \end{array}$

Therefore

SE(x¯1−x¯2)=23.1240120+118−−−−−−√=23.12409+10180−−−−√=23.142019180−−−√=23.1420×0.3248=7.5128 $\begin{array}{l} S E ({\bar{x}}_{1} - {\bar{x}}_{2}) & = 23.1240 \sqrt{\frac{1}{20} + \frac{1}{18}} = 23.1240 \sqrt{\frac{9 + 10}{180}} \\ = 23.1420 \sqrt{\frac{19}{180}} = 23.1420 \times 0.3248 \\ = 7.5128 \end{array}$

Test statistic is

t=(x¯1−x¯2)S1n1+1n2√=170−2057.5128=−357.5128=−4.6587 $t = \frac{({\bar{x}}_{1} - {\bar{x}}_{2})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{170 - 205}{7.5128} = \frac{- 35}{7.5128} = - 4.6587$

∴|t|=4.6587>1.96 $∴ | t | = 4.6587 > 1.96$

i.e., the calculated value of t is greater than the table value of t at 5% level of significance, for 36 df.

Hence we reject null hypothesis and conclude that there is a significant difference in the average sales between two salesmen.

Example 10.14: A group of 5 patients treated with medicine A weigh 42, 39, 48, 60, and 41 kg. A second group of 7 patients from the same hospital with medicine B weigh 38, 42, 56, 64, 68, 69, and 62 kg. Do you agree with the claim that the medicine B increases the weight significantly (The value of t at 5% level of significance for 10 df is 2.228).

Solution: Let x_i denote the weights of patients treated with medicine A, and y_i denote the weights of patients treated with medicine B.

We have n₁=5, n₂=7, df=n₁+n₂–2=5+7–2=10

Null hypothesis H₀: μA=μB $μ_{A} = μ_{B}$

i.e., there is no significant difference between the medicines A and B regards their effect on increase in weight.

Alternative hypothesis H₁: μA≠μB $μ_{A} \neq μ_{B}$ (Two-tailed test)

Table value of t at 5% level of significance for 10 df=2.228

x¯=∑xin1=42+39+48+60+415=2305=46,y¯=∑yin2=38+42+56+64+68+69+627=59 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} = \frac{42 + 39 + 48 + 60 + 41}{5} = \frac{230}{5} = 46, \\ \bar{y} = \frac{\sum y_{i}}{n_{2}} = \frac{38 + 42 + 56 + 64 + 68 + 69 + 62}{7} = 59 \end{array}$

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
42	–4	16	38	–19	361
39	–7	49	42	–15	225
48	2	4	56	–1	1
60	14	196	64	7	49
41	–5	25	68	11	121
–	–	–	69	12	144
–	–	–	62	5	25
230	–	290	399	–	926

S=∑(xi−x¯)2+∑(y1−y¯)2n1+n2−2−−−−−−−−−−−−−−√=290+9265+7−2−−−−−−√=121.610−−−−√=11.027 $\begin{array}{l} S & = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2} + \sum {(y_{1} - \bar{y})}^{2}}{n_{1} + n_{2} - 2}} = \sqrt{\frac{290 + 926}{5 + 7 - 2}} \\ = \sqrt{\frac{121.6}{10}} = 11.027 \end{array}$

Test statistic is

t=(x¯−y¯)S1n1+1n2√=46−5711.02715+17√=−11×35√(11.027)12√=−11×5.916(11.027)×3.464=−65.07638.197=−1.703 $\begin{array}{l} t & = \frac{(\bar{x} - \bar{y})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \\ = \frac{46 - 57}{11.027 \sqrt{\frac{1}{5} + \frac{1}{7}}} = \frac{- 11 \times \sqrt{35}}{(11.027) \sqrt{12}} \\ = \frac{- 11 \times 5.916}{(11.027) \times 3.464} \\ = \frac{- 65.076}{38.197} = - 1.703 \end{array}$

Therefore

|t|=1.703 $| t | = 1.703$

The calculated value of t is less than the table value of t at 5% level of significance, for 10 df. Therefore we accept null hypothesis H₀.

There is no significant difference between medicines A and B as regards the increase in weights.

Example 10.15: The students of two schools were measured for their heights, one school was in east coast another in west coast, where there is slight difference in weather. The sampling results are as follows:

East coast	43	45	48	49	51	51	–	–	–
West coast	47	49	51	53	54	55	55	56	57

Find whether there is any impact of weather on height taking other variable constant. (given t_0.05 for 13 df=2.16)?

Solution: Here we have

n1=6,n2=9 $n_{1} = 6, n_{2} = 9$

Let east coast school be denoted by x and west coast student height be denoted by y. Then

x¯=∑xin1y¯=∑yin2=43+45+48+49+51+526=2886=48=47+49+51+53+54+55+56+579=4779=53 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} & = \frac{43 + 45 + 48 + 49 + 51 + 52}{6} = \frac{288}{6} = 48 \\ \bar{y} = \frac{\sum y_{i}}{n_{2}} & = \frac{47 + 49 + 51 + 53 + 54 + 55 + 56 + 57}{9} = \frac{477}{9} = 53 \end{array}$

Null hypothesis H₀: Weather has no impact on the heights of the students

Alternative hypothesis H₁: Weather has impact on the heights of the students

Degree of freedom=6+9–2=13

Level of significance=5%

Table value of t for 13 df=2.16

S2=1n1+n2−2[∑n1(x1−x¯)2+∑n1(y1−y¯)2]=60+906+9−2=15013=11.5 $\begin{array}{l} S^{2} & = \frac{1}{n_{1} + n_{2} - 2} [\sum_{1}^{n} {(x_{1} - \bar{x})}^{2} + \sum_{1}^{n} {(y_{1} - \bar{y})}^{2}] \\ = \frac{60 + 90}{6 + 9 - 2} = \frac{150}{13} = 11.5 \end{array}$

∴S=11.5−−−−√=3.39 $∴ S = \sqrt{11.5} = 3.39$

Test statistic is

t=(x¯−y¯)S1n1+1n2√=48−533.3916+19√=−53.39518√=−51.76=−2.84 $\begin{array}{l} t & = \frac{(\bar{x} - \bar{y})}{S \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{48 - 53}{3.39 \sqrt{\frac{1}{6} + \frac{1}{9}}} \\ = \frac{- 5}{3.39 \sqrt{\frac{5}{18}}} = \frac{- 5}{1.76} = - 2.84 \end{array}$

∴|t|=2.84 $∴ | t | = 2.84$

The calculated value of t is greater than the table value of t at 5% level of significance for 13 df.

Hence we reject null hypothesis and accept the alternative hypothesis.

Exercise 10.1

1. Define t-distribution and write the properties of t-probability curve.

2. Ten students are selected at random from a college and their heights are found to be 100, 104, 108, 110, 118, 120, 122, 124, 126, and 118 cm. In the height of these data, discuss the suggestion that the mean height of the students of the college is 110 cm. Use 5% level of significance.

Ans: Null hypothesis is accepted, i.e., mean height of the students is 110 cm

3. A machine that produces mica insulating washers having a thickness of 10 miles (1 mile=0.001 in.). A sample of 10 washers has an average thickness of 9.52 miles with a SD of 0.60 miles. Find the value of t?

4. Ten cartons are taken at random from an automatic machine. The mean net weight of the 10 cartons is 11.8 kg and SD is 0.15 kg. Does the sample means differ significantly from the intended height of 12 kg (t_0.05 for 9 df=2.26).

Ans: H₀ rejected. The sample means differ

5. A fertilizer mixing machine is set to give 12 kg of nitrate for every quintal bag of fertilizer. Ten 100 kg bags are examined. The percentages of nitrate are as follows. 11, 14, 13, 12, 13, 12, 1, 3, 14, 11, 12 is there reason to be believe the machine is effective (t_0.05 for 9 df=2.262).

Ans: There is no reason to believe the t the machine is defective

6. Two samples of sizes 6 and 5 gave the following data:

	Sample I	Sample II
Mean	40	50
SD	8	10

Is the difference of the mean significant? The value of t for t degrees of freedom at 5% level of significance is 2.26.

Ans: The difference of means is not significant

7. Samples of two types of sodium vapor were tasted for the length of life and the following data were obtained:

	Type I (h)	Type II (h)
Sample No.	8	7
Sample mean	1234	1036
Sample SD	36	40

Is the difference in the mean sufficient to warrant an inference that type I is superior to type II regarding the length of life (Type I is superior to type II)?

8. Sandal powder is packed into packets by a machine. A random sample of 12 packets is drawn and their weights are found to be (in kg) 0.49, 0.48, 0.47, 0.48, 0.49, 0.50, 0.51, 0.49, 0.50, 0.51, 0.48. Test if the average packing can be taken as 0.5 kg (t_0.05 for 11 df=2.20)?

Ans: H₀ rejected

9. The following results are obtained from a sample of 10 boxes of biscuits:
Mean weight of the content=490 g
SD of the weights=9 g
Could the sample come from a population having a mean of 500 g (t_0.05 for 9 df=2.26)

Ans: H₀ is rejected

10. The wages of 10 workers taken at random from a factory are given as follows:
Wages (in Rs.): 578, 572, 570, 568, 572, 578, 570, 572, 596, 584
It is possible that the mean height of all workers of this factory could be Rs. 580 (given at 9 df t_0.05=2.26)?

Ans: H₀ is accepted

11. The 9 items of a sample have the following values 45, 47, 50, 52, 48, 47, 49, 53, 51. Does the mean values differ significantly from the assumed mean 47.5 (for 8 df, t_0.05=2.31)?

Ans: H₀ is accepted

12. Write short notes on:

a. Students t-distribution

b. Properties of t-distribution

13. Test whether the sample having values 63, 63, 64, 55, 66, 69, 70, 70, and 71 have been chosen from a population with mean of 65 at 5% level of significance (t_0.05=2.31 for 8 df)?

Ans: The difference is significant

14. A certain medicine was administered to each of 10 patient’s results in the following increases in the blood pressure:
8, 8, 7, 5, 4, 1, 0, 0, –1, –1
Concluded that the medicine was responsible for the increase in blood pressure.

Ans: The difference is significant. The medicine is responsible for the increase in the blood pressure

15. The weight of 10 people of a locality are found to be 70, 67, 62, 68, 61, 68, 70, 64, 66 kg. Is it reasonable to believe that the average weights of the people of locality are greater than 64 kg. Test at 5% level of significance (Given t_0.05=1.83 for 9 df).

Ans: H₀ is rejected. Average weight is greater than 64 kg

16. Following table contains the data resulting from a sample of managers trained under different t programs:

Program sampled	Mean sensitivity of the program	Number of managers observed	Estimate SD for sensitivity after the program
Formal	92%	12	15%
Informal	84%	15	19%

Test at 0.05 level of significance whether the sensitivity achieved by formal program is significantly higher than that achieved under informal program?

Ans: H₀ rejected: The sensitivity achieved by formal program is higher than that achieved under the informal program

17. A random sample of 10 tins of oil filled in by an automatic machine gave the following weights (in kg): 2.05, 2.01, 2.04, 1.91, 1.97, 1.97, 2.04, 2.02. Can we accept at 5% level of significance the claim that the average weight of the tin is 2 kg.

Ans: The average weight of the tin is 2 kg is accepted

18. A random sample of size 16 has 53 as its mean. The sum of the squares of the deviations taken from the mean is 1.50. Can this sample be regarded as taken from the population having 56 as its mean? Obtain 95% and 99% confidence limits of the mean population (for v=15, t_0.01=2.95, t_0.05=2.131).

Ans: H₀ is rejected, [50.316, 54.684], [50.67, 55.33]

19. A sample of 20 items has mean 42 units and SD 5 units. Test the hypothesis that it is a random sample from a normal population with mean 45 units?

Ans: The sample could not have come from the given population

20. The means of two random samples of size 9 and 7 are 196.42 and 198.82, respectively. The sum of the squares of the deviations from the mean is 26.94 and 18.73 respectively. Can the sample be considered that they have been drawn from the same normal population?

Ans: The two samples are not drawn from the same population

21. Two horses A and B were tested according to the time (in seconds) to run a particular track with the following results:

Horse A	28	30	32	33	33`	29	34
Horse B	29	30	30	24	27	27	-

Test whether you can discriminate between two horses you can use the fact that at 5% level of significance value of t for 11 df is 2.2?

Ans: The medicines A and B do not differ significantly as regards their effect on the increase of weight

22. A group of 5 patients treated with medicine A weighed 42, 39, 48, 60 and 41 kg. Second group of 7 patients from the same hospital treated with medicine B weighed 38, 42, 56, 64, 68, 69, and 62 kg. Do you agree with the claim that medicine B increases the weight significantly?

Ans: The medicines A and B do not differ significantly as regards their effect on the increase of weight

23. Memory capacity of 10 students was tested before and after training. State whether the training was effective or not from the following scores:

Before	12	14	11	8	7	10	3	0	5	6
After	15	16	10	7	5	13	10	2	8	8

Ans: Not significant

24. An IQ test was administered on 5 persons before and after they were trained. The results are given below. Test whether there is any change in IQ after the training program?

Ans: No change

25. Measurements of a sample of weights were determined as
8.3, 10.6, 9.7, 8.8, 10.2, 9.4.
Determine unbiased and efficient estimates of

a. The population mean

b. The population variance

Ans: 9.5 kg, 0.736

26. Ten students were given intensive coaching for a month in mathematics. The scores obtained in tests 1 and 5 are given below:

Serial no. of students:	1	2	3	4	5	6	7	8	9	10	11
Marks in first test:	50	52	53	60	65	67	48	69	72	80
Marks on second test:	65	55	65	65	60	67	49	82	74	86

Does the scores from test 1 to test 5 show an improvement? Test at 5% level of significance (t_0.05 for 9 df=1.833 and two-tailed test t_0.05 for 9 df=2.262)?

Ans: For one-tailed test null hypothesis is rejected

27. A certain stimulus administered to each of 12 patients resulted in the following changes in the blood pressure 5, 2, 8, −1, 3, 0, −2, 1, 5, 0, 4, 6. Can it be concluded that the stimulus in general be accompanied by an increase in blood pressure?

Ans: H₀ rejected. The stimulus in general be accompanied by an increase in blood pressure

28. Experience shows that a fixed dose of certain drug causes an average increase of pulse rate by 10 beats per minute with SD of 4. A group of 9 patients given the same dose showed the following increase: 13, 15, 14, 10, 8, 12, 16, 9, 20. Test 5% level of significance whether this group is different in response to the drug?

Ans: H₀ rejected: The given group is different in response to the drug

10.11 Paired t-Test

In this case the size of two samples are equal, i.e., n₁=n₂=n

Let x₁, x₂, …, x_n and y₁, y₂, …, y_n be two samples. Let d₁, d₂, …, d_n be the differences between the corresponding members of the two samples.

d¯=∑dn $\bar{d} = \frac{\sum d}{n}$

and

S=∑(d−d¯)2n−1 $S = \frac{\sum {(d - \bar{d})}^{2}}{n - 1}$

then the t-statistic is defined as

t=d¯−0Sn√ $t = \frac{\bar{d} - 0}{\frac{S}{\sqrt{n}}}$

t-statistic is applied for n−1 degrees of freedom.

10.11.1 Solved Examples

Example 10.16: A certain diet was newly introduced to each of 12 pigs resulted in the following increase in body weight:

6, 3, 8, −2, 3, 0, −1, 1, 6, 0, 5, and 4.
Can you conclude that the diet was effective in increasing the weight of pigs (given t_0.05=2.20 for 11 df)?
Solution: Here we have n=12

d1=6,d2=3,d3=8,d4=–2,d5=3,d6=0,d7=–1,d8=1,d9=6,d10=0,d11=5,d12=4 $\begin{array}{l} d_{1} = 6, d_{2} = 3, d_{3} = 8, d_{4} = - 2, d_{5} = 3, d_{6} = 0, \\ d_{7} = - 1, d_{8} = 1, d_{9} = 6, d_{10} = 0, d_{11} = 5, d_{12} = 4 \end{array}$

where d_i denotes the increase in weights of the pigs.

Null hypothesis H₀: There is no significant difference in the body weights of the pigs before and after introduction of the new diet, i.e., μx=μy $μ_{x} = μ_{y}$

Alternative hypothesis H₁: μx≠μy $μ_{x} \neq μ_{y}$

Degrees of freedom=n−1=12−1=11

Level of significance=5%, t_0.05 for 11 df=2.20

∑dd¯=∑dn∑d2=6+3+8+(–2)+3+0+(–1)+1+6+0+5+4=33=3312=2.75=62+32+82+(–2)2+32+02+(–1)2+12+62+02+52+42=201 $\begin{array}{l} \sum d & = 6 + 3 + 8 + (- 2) + 3 + 0 + (- 1) + 1 + 6 + 0 + 5 + 4 = 33 \\ \bar{d} = \frac{\sum d}{n} & = \frac{33}{12} = 2.75 \\ \sum d^{2} & = 6^{2} + 3^{2} + 8^{2} + {(- 2)}^{2} + 3^{2} + 0^{2} + {(- 1)}^{2} + 1^{2} + 6^{2} + 0^{2} + 5^{2} + 4^{2} = 201 \end{array}$

∴S2S=1n−1[∑d2((∑d)2n)]=112−1[201−(33)212]=111[201−90.75]=10.0227=10.0227−−−−−−√=3.1658 $\begin{array}{l} ∴ S^{2} & = \frac{1}{n - 1} [\sum d^{2} (\frac{{(\sum d)}^{2}}{n})] = \frac{1}{12 - 1} [201 - \frac{{(33)}^{2}}{12}] \\ = \frac{1}{11} [201 - 90.75] = 10.0227 \\ S & = \sqrt{10.0227} = 3.1658 \end{array}$

Test statistic t is

t=d¯−0Sn√=2.753.165812√=2.75×3.4643.1658=3.0091 $t = \frac{\bar{d} - 0}{\frac{S}{\sqrt{n}}} = \frac{2.75}{\frac{3.1658}{\sqrt{12}}} = \frac{2.75 \times 3.464}{3.1658} = 3.0091$

Since the calculated value of t is greater than the table value of t at 5% level of significance for 11 df, we reject null hypothesis. Therefore the new diet is effective in increasing the body weight of the pigs.

10.12 F-Distribution

In this section we introduce F-distribution and another statistic F

F-statistic is defined by

F=S21S22, $F = \frac{S_{1}^{2}}{S_{2}^{2}},$

where

S21>S22 $S_{1}^{2} > S_{2}^{2}$

and

S21S22=1n1−1∑(x−x¯)2=1n2−1∑(y−y¯)2 $\begin{array}{l} S_{1}^{2} & = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} \\ S_{2}^{2} & = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} \end{array}$

For the degrees of freedom ν1=n1−1,ν2=n2−1 $ν_{1} = n_{1} - 1, ν_{2} = n_{2} - 1$

The value of F is always greater than 1.

S21<S22(i.e.,S21<S22) $S_{1}^{2} < S_{2}^{2} (i . e ., S_{1}^{2} < S_{2}^{2})$

F is given by

F=S22S21 $F = \frac{S_{2}^{2}}{S_{1}^{2}}$

So that the value of F is greater than 1.

F-test is based on F-distribution which is defined as the ratio of two independent chi-square variates.

F-distribution with (ν1,ν2) $(ν_{1}, ν_{2})$ degrees of freedom is given by the probability function.

dp=ν12ν11ν12ν22Fν12−2β(ν12,ν22)(ν1F+ν2)12(ν1+ν2)dF,0<F<∞ $d p = \frac{ν_{1}^{\frac{1}{2} ν_{1}} ν_{2}^{\frac{1}{2} ν_{2}} F^{\frac{ν_{1}}{2} - 2}}{β (\frac{ν_{1}}{2}, \frac{ν_{2}}{2}) {(ν_{1} F + ν_{2})}^{\frac{1}{2} (ν_{1} + ν_{2})}} d F, 0 < F < \infty$

From the definition of F, we have

ν1ν2F=(n1−1)S21σ2(n2−1)S22σ2 $\frac{ν_{1}}{ν_{2}} F = \frac{(n_{1} - 1) \frac{S_{1}^{2}}{σ^{2}}}{(n_{2} - 1) \frac{S_{2}^{2}}{σ^{2}}}$

The numerator and the denominator of the second member are independent χ2 $χ^{2}$ variates with ν1 $ν_{1}$ and ν2 $ν_{2}$ degrees of freedom respectively. Hence (ν1/ν2)F $(ν_{1} / ν_{2}) F$ is a β2((ν1/2),(ν2/2)) $β_{2} ((ν_{1} / 2), (ν_{2} / 2))$ variate so that the probability that a random value of F will fall in the interval dF is given by

F-Probability curve is J-shaped. If the curve is bell shaped if ν1>2 $ν_{1} > 2$ .

The distribution of F is independent of the population variance and depends on ν1 $ν_{1}$ and ν2 $ν_{2}$ only.

F-test is used to test:

1. Whether two independent samples have been drawn from the normal populations with same variance.

2. Whether the two independent estimates of the population variances are homogeneous or not.

F-test is based on the following assumptions:

1. The distribution in each group showed to be normally distributed.

2. Error should be independent of each observed value.

3. Variance within each group should be equal for all groups.

When the sample sizes are large, the assumption of normality is not necessary.

10.12.1 Solved Examples

Example 10.17: Two horses A and B were tested according to the time (in seconds) to run on a particular track with the following results:

Horse A	28	30	32	33	33	29	34
Horse B	29	30	30	24	27	29	–

Test whether the two horses have the same running capacity?

Solution: We have n₁=7, n₂=6

Then

x¯=∑xin1=2197=31.28,y¯=∑yin2=1696=28.2 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} = \frac{219}{7} = 31.28, & \bar{y} = \frac{\sum y_{i}}{n_{2}} = \frac{169}{6} = 28.2 \end{array}$

Let the time taken by horse A be denoted by x and the time taken by horse B be denoted by y

Degrees of freedom=n₁–1, n₂–1=6, 5

Critical value, i.e., level of significance=5%

x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
28	−3.28	10.75	28	0.	0.64
30	−1.28	1.64	30	1.8	3.24
32	0.72	0.52	30	1.8	3.24
33	1.72	2.96	24	−4.2	17.64
33	1.72	2.96	27	−1.2	1.44
29	−2.28	5.20	29	0.8	0.64
34	2.72	7.40	–	–	–
219		31.43	169		26.84

S21=1n1−1∑(x−x¯)2=31.436=5.224S22=1n2−1∑(y−y¯)2=26.845=5.368 $\begin{array}{l} S_{1}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{31.43}{6} = 5.224 \\ S_{2}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{26.84}{5} = 5.368 \end{array}$

Null Hypothesis H₀: The two horses have the same running capacity, i.e., σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$

Alternative Hypothesis H₁: σ21≠σ22 $σ_{1}^{2} \neq σ_{2}^{2}$

Table value of F for (5, 6) degrees of freedom at 5% level of significance is 4.39.

Test statistic is F=S22S21=5.3685.224=1.02(sinceS22>S21) $F = \frac{S_{2}^{2}}{S_{1}^{2}} = \frac{5.368}{5.224} = 1.02 (since S_{2}^{2} > S_{1}^{2})$

Since the calculated value of F is less than the table value of F for (5, 6) df at 5% level of significance, we accept H₀.

The two horses have the same running capacity.

Example 10.18: In a sample of 8 observations, the sum of squares of deviations of the sample values from the sample mean was 94.5 and in another sample of 10 observations it was found to be 101.7. Test whether the difference is significant?

Solution: It is given that

∑(x−x¯)2=94.5,n1=8∑(y−y¯)2=101.7,n2=10S21=1n1−1∑(x−x¯)2=94.58−1=13.5S22=1n2−1∑(y−y¯)2=1010.710−1=11.3 $\begin{array}{l} \sum {(x - \bar{x})}^{2} = 94.5, n_{1} = 8 \\ \sum {(y - \bar{y})}^{2} = 101.7, n_{2} = 10 \\ S_{1}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{94.5}{8 - 1} = 13.5 \\ S_{2}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{1010.7}{10 - 1} = 11.3 \end{array}$

Null Hypothesis H₀: σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$ , i.e., The difference is not significant

Alternative Hypothesis H₁: σ21≠σ22 $σ_{1}^{2} \neq σ_{2}^{2}$

Level of significance: 5%

Critical value, i.e., table value of F for (8−1, 10−1)=(7, 9) df is 3.29.

F=S21S22=13.511.3=1.195 $F = \frac{S_{1}^{2}}{S_{2}^{2}} = \frac{13.5}{11.3} = 1.195$

Since the calculated value of F is less than the table value of F at 5% level of significance for (7, 9) df, we accept Null hypothesis and conclude that the difference is not significant.

Example 10.19: In a test given two groups of students drawn from two normal populations the marks are as follows:

Sample A	18	20	36	50	49	36	34	49	41
Sample B	29	28	26	35	30	44	46

Examine 5% level whether the two populations have the same variance.

Solution:

Null Hypothesis H₀: σ2A=σ2B $σ_{A}^{2} = σ_{B}^{2}$

Alternative Hypothesis H₁: σ2A≠σ2B $σ_{A}^{2} \neq σ_{B}^{2}$

Level of significance: 5%,

We have

x¯=∑xin1=18+20+36+50+49+36+34+49+419=3339=37y¯=∑yin2=29+28+26+35+30+44+467=2387=34 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} = \frac{18 + 20 + 36 + 50 + 49 + 36 + 34 + 49 + 41}{9} = \frac{333}{9} = 37 \\ \bar{y} = \frac{\sum y_{i}}{n_{2}} = \frac{29 + 28 + 26 + 35 + 30 + 44 + 46}{7} = \frac{238}{7} = 34 \end{array}$

Calculation of variances

Sample A			Sample B
x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
18	−19	361
20	−17	289	29	−5	25
36	−1	1	28	−6	36
50	13	169	26	−8	64
49	12	144	35	−1	1
36	−1	1	30	−4	16
34	−3	9	44	−10	100
49	12	144	46	−12	144
41	4	16
		1134			386

S2A=1n1−1∑(x−x¯)2=11349−1=141.75S2B=1n2−1∑(y−y¯)2=3867−1=64.33 $\begin{array}{l} S_{A}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{1134}{9 - 1} = 141.75 \\ S_{B}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{386}{7 - 1} = 64.33 \end{array}$

We have

S2A>S2BF=S2AS2B=141.7564.33=2.203 $\begin{array}{l} S_{A}^{2} > S_{B}^{2} \\ F = \frac{S_{A}^{2}}{S_{B}^{2}} = \frac{141.75}{64.33} = 2.203 \end{array}$

Degreesoffreedom=(ν1,v2)=(n1−1,n2−1)=(9–1,7–1)=(8,6) $\begin{array}{l} Degrees of freedom & = (ν_{1}, v_{2}) = (n_{1} - 1, n_{2} - 1) \\ = (9 - 1, 7 - 1) \\ = (8, 6) \end{array}$

Table value of F at 5% for (8,6) df=4.15.

Calculated value of F is less than the table value of F at 5% level (8,6) df. Hence we accept H₀ and conclude that the populations from where the samples have been drawn have the same variance.

Example 10.20: Two samples of sizes 9 and 8 gave the sum of the squares of deviations from their respective means equal to 160 and 91, respectively. Can they be regarded as drawn from the same normal population?

Solution: It is given that

∑(x−x¯)2=160,n1=9∑(y−y¯)2=91,n2=8S21=1n1−1∑(x−x¯)2=1609−1=20S22=1n2−1∑(y−y¯)2=918−1=13 $\begin{array}{l} \sum {(x - \bar{x})}^{2} = 160, n_{1} = 9 \\ \sum {(y - \bar{y})}^{2} = 91, n_{2} = 8 \\ S_{1}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{160}{9 - 1} = 20 \\ S_{2}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{91}{8 - 1} = 13 \end{array}$

Null Hypothesis H₀: Both the samples have come from the same normal population.

Alternative Hypothesis H₁: The samples are not from the same normal population.

Since S21>S22 $S_{1}^{2} > S_{2}^{2}$ , we have

Degrees of freedom (n₁−1, n₂−1)=(8, 7)

Level of significance: 5%

Table value of F at 5% level of significance for (8,7) df=3.73

Test statistic:

F=S21S22=2013=1.5385 $F = \frac{S_{1}^{2}}{S_{2}^{2}} = \frac{20}{13} = 1.5385$

Calculated value of F is less than the table value of F at 5% level (8, 6) df. Hence we accept null hypothesis.

The two samples can be regarded as drawn from the same normal population.

Example 10.21: Two samples are drawn from two normal populations from the following data, test whether the two samples have the same variances at 5% level?

Sample I		60	65	71	74	76	82	85	87
Sample II	64	66	67	85	78	88	86	85	63	91

Solution: Null hypothesis H₀: σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$ , i.e., the two samples have same variances.

Alternative hypothesis H₁: σ21≠σ22 $σ_{1}^{2} \neq σ_{2}^{2}$ (Two-tailed test).

x¯=∑xin1=60+65+71+74+76+82+85+878=6008=75,y¯=∑yin2=64+66+67+85+78+88+86+85+63+9110=77010=77 $\begin{array}{l} \bar{x} = \frac{\sum x_{i}}{n_{1}} = \frac{60 + 65 + 71 + 74 + 76 + 82 + 85 + 87}{8} = \frac{600}{8} = 75, \\ \bar{y} = \frac{\sum y_{i}}{n_{2}} = \frac{64 + 66 + 67 + 85 + 78 + 88 + 86 + 85 + 63 + 91}{10} = \frac{770}{10} = 77 \end{array}$

Sample A			Sample B
x	x–x¯ $x - \bar{x}$	(x–x¯)2 ${(x - \bar{x})}^{2}$	y	y–y¯ $y - \bar{y}$	(y–y¯)2 ${(y - \bar{y})}^{2}$
			61	−16	256
60	−15	225	66	−11	121
65	−10	100	67	−10	100
71	−4	16	85	8	64
74	−1	1	78	1	1
76	1	1	88	11	121
82	7	49	86	9	81
85	10	100	85	8	64
87	12	144	63	−14	196
			91	14	196
600		636	770		1200

S21=1n1−1∑(x−x¯)2=6368−1=90.857S22=1n2−1∑(y−y¯)2=120010−1=133.33 $\begin{array}{l} S_{1}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{636}{8 - 1} = 90.857 \\ S_{2}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{1200}{10 - 1} = 133.33 \end{array}$

Since S22>S21 $S_{2}^{2} > S_{1}^{2}$

F=S22S21=133.3390.851=1.467 $F = \frac{S_{2}^{2}}{S_{1}^{2}} = \frac{133.33}{90.851} = 1.467$

Calculated value of F for (n₁−1, n₂−1)=(9, 7) df at 5% level=3.68.

Therefore the calculated value of F is less than table value of F at 5% level of significance for (9, 7) df.

We accept null hypothesis. The samples 1 and 11 have the same variance.

Example 10.22: In a laboratory experiment, two samples of blood gave the following results:

Sample	Size	Sample SD	Sum of squares of deviations from mean
1	10	15	90
2	12	14	108

Test the equality of sample variances at 5% level of significance?

Solution: Here we have n₁=10, n₂=12

x¯=15,y¯=14∑(x−x¯)2=90,∑(y−y¯)2=108S21=1n1−1∑(x−x¯)2=9010−1=10S22=1n2−1∑(y−y¯)2=10812−1=9.82 $\begin{array}{l} \bar{x} = 15, \bar{y} = 14 \\ \begin{array}{l} \sum {(x - \bar{x})}^{2} = 90, & \sum {(y - \bar{y})}^{2} = 108 \end{array} \\ S_{1}^{2} = \frac{1}{n_{1} - 1} \sum {(x - \bar{x})}^{2} = \frac{90}{10 - 1} = 10 \\ S_{2}^{2} = \frac{1}{n_{2} - 1} \sum {(y - \bar{y})}^{2} = \frac{108}{12 - 1} = 9.82 \end{array}$

Null hypothesis H₀: σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$

Alternative hypothesis H₁: σ21≠σ22 $σ_{1}^{2} \neq σ_{2}^{2}$

Test statistic:

F=S21S22=109.82=1.018 $F = \frac{S_{1}^{2}}{S_{2}^{2}} = \frac{10}{9.82} = 1.018$

Table value of F at 5% level of significance for (10−1, 12−1)=(9, 11) df is 2.90.

Therefore the calculated value of F is less than the table value. Hence we accept null hypothesis. The two programs have the same variance.

Exercise 10.2

1. In one sample of 10 observations, the sum of squares of the deviations of the sample values from the sample mean was 120 and in another sample of 12 observations it was 314. Test whether the difference is significant at 5% level of significance?

Ans: H₀ accepted, i.e., σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$

2. Two random samples gave the following results:

Sample	Size	Sample	Sum of squares of deviations from mean
1	10	15	90
2	12	14	108

Test whether the samples came from the same population?

Ans: The samples came from the populations with equal variance. σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$ , i.e., H₀ accepted

3. Write short notes on F-distribution and F-test.

4. Mention the applications of F-test.

5. Two independent samples are of sizes 8 and 10. The sum of squares of deviations of sample values from their respective sample means were 84.4 and 102.6. Test whether the difference of variances of the populations is significant or not?

Ans: σ21=σ22 $σ_{1}^{2} = σ_{2}^{2}$ , i.e, Accept H₀

6. Two random samples of sizes 8 and 7 had the following values of variances:

Sample A	9	11	13	11	15	9	12	14
Sample B	10	12	10	14	9	8	10

Do these eliminates of population variance differ significantly?

Ans: Not significant

7. Two independent samples of sizes 9 and 7 from a normal population has the following values of the variables:

Sample I	18	13	12	15	12	14	16	14	15
Sample II	16	19	13	16	18	13	15

Do these eliminates of population variance differ significantly?

Ans: H₀ accepted. The difference is not significant

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Table of Contents for Chapter 10. Test of Significance—Small Samples

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Test of Significance—Small Samples

Abstract

Keywords

10.1 Introduction

10.2 Moments About Mean

10.3 Properties of Probability Curve

10.4 Assumptions for t-Test

10.5 Uses of t-Distribution

10.6 Interval Estimate of Population Mean

10.7 Types of t-Test

10.8 Significant Values of t

10.9 Test of Significance of a Single Mean

10.9.1 Solved Examples

10.10 Student’s t-Test for Difference of Means

10.11 Paired t-Test

10.11.1 Solved Examples

10.12 F-Distribution

10.12.1 Solved Examples

Table of Contents for
Chapter 10. Test of Significance—Small Samples