8.4.1 Null Hypothesis

In the process of statistical test the hypothesis concerning a population is rejected or accepted based on the sample drawn from the population, the statistician test the hypothesis through observations and makes a probability statement. The hypothesis (i.e., statement) we have assumed is called null hypothesis. It is defined as follows:

Definition 8.2: The hypothesis formulated for the purpose of its rejection under the assumption that it is true is called the null hypothesis.

Null hypothesis is denoted by H₀.

8.4.2 Alternative Hypothesis

The negation of null hypothesis is called alternative hypothesis. It is denoted by H₁. If the null hypothesis is rejected, then the alternative hypothesis is accepted, the acceptance or rejection of null hypothesis is based on sample study. Suppose we want to test the hypothesis that the population mean is equal to the hypothesized mean 60. Symbolically they can be expressed as

$H_0:\mu ={\mu }_0=60$

$\mathrm{where}{\mu }_0={\mu }_{H_0}$

The possible alternative hypothesis can be stated in one of the following form:

$H_1:\mu \ne {\mu }_0,\mathrm{i}\mathrm{.e}.,\mu >{\mu }_0\mathrm{or}\mu <{\mu }_0(\text{two}-\text{tailed}\mathrm{test})$

$H_1:\mu >{\mu }_0(\text{one}-\text{tailed}\mathrm{test})$

$H_1:\mu <{\mu }_0(\text{one}-\text{tailed}\mathrm{test})$

In hypothesis testing, we usually proceed on the basis of null hypothesis, the probability of rejecting null hypothesis when it is true or the level of significance which is very small.

8.8.1 One-Tailed Test

A test of any statistical hypothesis where alternative hypothesis is one sided such as is called one-tailed test.

$\begin{array}{cccc}{H}_0:\mu ={\mu }_0& H_0:\mu ={\mu }_0& H_0:\mu \ge {\mu }_0& H_0:\mu \le {\mu }_0\\ \mathrm{or}& \mathrm{or}& \mathrm{or}& \mathrm{or}\\ H_1:\mu >{\mu }_0& H_1:\mu <{\mu }_0& H_1:\mu >{\mu }_0& H_1:\mu <{\mu }_0\end{array}$

where μ is the parameter to be tested and H₀ is the hypothesized value. When alternative hypothesis is one sided, i.e., one tailed, we apply one-tailed test. There are two types of one-tailed tests. When it is desired to see whether the population mean is as large as some specified value of the mean, we apply the right tailed test. In the right tailed test, the region of rejection (critical region) lies entirely on the right tail on the normal curve (Fig. 8.1).

When it is desired to see whether the population mean is at least as small as some specified value of the mean, we apply the left-tailed test.

In the left-tailed test the region of rejection (critical region) lies entirely on the left tail on the normal curve Fig. 8.2.

8.8.2 Two-Tailed Test

A test of any statistical hypothesis where the alternative hypothesis is two sided such as is called a two-tailed test.

$H_0:\mu ={\mu }_0$

$H_1:\mu \ne {\mu }_0$

where μ is the parameter to be tested and μ₀ is the hypothesized value. In a two-tailed test, there are two rejection regions, one an each tail of the curve (Fig. 8.3). If the significance level is 5% and the test applied is a two-tailed test, the probability of the region area will be 0.05. It is equally splitted on both sides of the curve as 0.025. The region of acceptance in this case is 0.95.

Decision of rejecting Null hypothesis or not rejecting Null hypothesis: The statistical inference made on the basis of the sample mean is called a decision. If the sample mean lies in the region of rejection, null hypothesis is rejected while the alternative hypothesis is accepted, i.e., H₀ is rejected and H₁ is accepted. If the sample mean lies in the region of acceptance, H₀ is accepted and H₁ is rejected. In this case the area in the tails is equal to the level of significance α. In the case of one-tailed test α appears is one tail and for two-tailed test α/2 appears in each tail of the curve. Instead of calculating the region of acceptance and rejection, we can directly calculate the value of the test statistic using the value of the sample mean calculated from the sample data.

Thus we have the following decision rules:

Remarks

1. When the size of the sample is n≥30, i.e., the sample is large, we make use of the unbiased estimate of σ.
If the population standard deviation, σ, is known and n denotes the size of the sample the test statistic

$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}(\mathrm{normal})$

2. When σ is unknown, and n<30 (i.e., small sample) the statistic is $t=(\overline{x}-\mu )/(\hat{\sigma }/\sqrt{n})$, ($\hat{\sigma }$ is the unbiased estimate of σ).

8.13.1 Test of Significance for Single Mean

Let x₁, x₂, …, x_n be a random sample of size n drawn from a large population with mean μ and variance σ². Let $\overline{x}$ denote the mean of sample and S² denote the variance of the sample.

We know that $\overline{x}~N\left(\mu ,\frac{{\sigma }^2}{n}\right)$

The standard normal variate corresponding to $\overline{x}$

$Z=\frac{\overline{x}-\mu }{SE(\overline{x})}$

where

$SE(\overline{x})\frac{\sigma }{\sqrt{n}}$

We set up the null hypothesis that there is no difference between the sample mean and the population mean. The test statistic is

$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$ (σi known and $SE(\overline{x})=\sigma /\sqrt{n}$).

If σ is not known

$Z=\frac{\overline{x}-\mu }{S/\sqrt{n}}$

where S is the SD of the sample.

8.13.1.1 Solved Examples

Test of significance for single means:

Example 8.1: The heights of college students in a city are normally distributed with SD 6 cm; a sample of 100 students have a mean height of 158 cm. Test the hypothesis that the mean height of college students in the college is 160 cm.

Solution: We have

$\overline{x}$=158 (Mean of the sample)

μ=160 (Mean of the population)

σ=6 (SD)

n=100 (Size of the sample)

Level of significance: 5% and 1%

$H_0:\mu =160$, i.e., difference is not significant

$H_1:\mu \ne 160$

We apply two-tailed test

Test statistic is

$\begin{array}{ll}Z\hfill & =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{158-160}{6/\sqrt{100}}=\frac{-2}{6/10}=–3.333\hfill \\ \therefore \left|Z\right|\hfill & =3.333\hfill \end{array}$

Table value of Z at 5% level of significance=1.96

Since the calculated value of Z at 5% level of significance is greater than the table value of Z, we reject H₀ at 5% level of significance.

Table value of Z at 1% level of significance is 2.58.

Calculated value at 1% level of significance is greater than the table value of Z, hence we reject the null hypothesis H₀ at 1%, i.e., H₀ is rejected both at 1% and 5% level of significance.

Example 8.2: A sample of 400 items is taken from a population whose SD is 10. The mean of the sample is 40. Test whether the sample has come from the population with mean 38. Also calculate 95% confidence interval for the population.

Solution: Here we have

$\begin{array}{ll}\overline{x}\hfill & =40(\text{Mean}\mathrm{of}\mathrm{the}\text{sample})\hfill \\ \mu \hfill & =38(\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{population})\hfill \\ \sigma \hfill & =10(\mathrm{S}\mathrm{D})\hfill \\ n\hfill & =400\hfill \end{array}$

Null hypothesis: H₀: μ=38

i.e., the sample is from the population whose mean=38

Alternative hypothesis: H₁: μ ≠ 38

Level of significance: α=0.05

Table value of Z=Z_α=1.96

Test statistic:

$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{40-38}{10/\sqrt{400}}=\frac{2}{10/20}=\frac{40}{10}=4$

Calculated value of Z=4 is greater than the table value of Z at 0.05 level of significance.

Hence, null hypothesis is rejected, i.e., confidence interval for the population whose mean is 38.

95% confidence interval for the population is given by

$\overline{x}\pm 1.96\cdot \frac{\sigma }{\sqrt{n}}=40\pm 1.96\cdot \frac{10}{\sqrt{400}}=40\pm 0.98$

or

$[40–0.98,40+0.98]=[39.02,40.98]$

Example 8.3: The mean and SD of a population are 11,795 and 14,054, respectively. If n=50 find 95% confidence interval for the mean?

Solution: It is given that $\overline{x}=11,795$, σ=14,054, n=50

Therefore

$\frac{\sigma }{\sqrt{n}}=\frac{14054}{\sqrt{50}}=\frac{14054}{7.071}=1987.55$

and

$1.96\cdot \frac{\sigma }{\sqrt{n}}=(1.96)(1987.55)=3895.607$

Hence the 95% confidence interval is

$\begin{array}{ll}\overline{x}\pm 1.96\cdot \frac{\sigma }{\sqrt{n}}\hfill & =[11795–3895.607,1175+3895.607]\hfill \\ \hfill & =[7899.39,15690.60]\hfill \end{array}$

Example 8.4: It is claimed that a random sample of 49 types has a mean life of 15,200 km. This sample has drawn from a population whose mean is 15,150 km and a SD of 1200 km. Test the significance at 0.05 levels.

Solution: We have $\overline{x}=15,200$, μ=15,150, σ=1200, and n=49

Null hypothesis: H₀: μ=15,150

Alternative hypothesis: H₁: μ ≠ 15,150 (two-tailed test)

Level of significance=α=0.05

Table value of Z=1.96

Test statistic is

$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{15200-15150}{1200/\sqrt{49}}=\frac{50}{171.42}=0.29168$

Since the calculated value of Z is less than the table value at 5% level, we accept null hypothesis.

Example 8.5: A sample size of 400 was drawn and the sample mean was found to be 98. Test whether this sample could have come from a normal population with mean 100 and SD 8 at 5% level of significance.

Solution: Here we have $\overline{x}=98$, μ=100, σ=8, and n=400

Null hypothesis: H₀: μ=100

i.e., The sample comes from a normal population with mean 100.

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

Level of significance=α=0.05

Table value of Z at 5% level of significance=1.96

Test statistic:

$\begin{array}{ll}Z\hfill & =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{98-100}{8/\sqrt{400}}=\frac{-2}{8/20}=\frac{-10}{2}=-5\hfill \\ \left|Z\right|\hfill & =|-5|=5>1.96\hfill \end{array}$

Since the calculated value of Z is greater than the table value of Z at 5% level of significance, we reject H₀, i.e., sample was not drawn from the normal population with mean 100 and SD 8.

Example 8.6: The average marks in mathematics of a sample of 100 students was 51 with a SD of 6 marks, could this have been a random sample from a population with average marks of 50?

Solution: Here we have $\overline{x}=51$, S=6, μ=50, n=100

Null hypothesis: H₀: 50, The sample is from a population with an average of 50 marks.

i.e.,

$\mu =50$

Alternate hypothesis: $H_1:\mu \ne 50$

Let the level of significance be taken as 5%

Table value of Z for 5% level of significance is |Z|=1.96

$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{51-50}{6/100}=\frac{1}{6/10}=\frac{10}{6}=1.666$

Since the calculated value of Z is less than the table value of Z at 5% level of significance, we accept H₀, i.e., sample is drawn from a population with a mean of 50 marks. The difference is not significant.

Example 8.7: The mean of certain production process is known to be 50 with a SD of 2.5. The production manager may welcome any change in mean value toward higher side but would like to safe guard against decreasing values of mean. He takes a sample of 12 items that gives a mean value of 46.5 what inference should the manager take for production process on the basis of sample results. Use 5% level of significance for the purpose.

Solution: $\overline{x}=46.5$, μ=50, σ=2.5

Null hypothesis $H_0:\mu =50$

Alternative hypothesis $H_1:\mu <50$ (one-sided test)

Level of significance α=5%

Table value of Z=1.645

$\begin{array}{ll}Z\hfill & =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{46.5-50}{2.5/\sqrt{12}}=\frac{-3.5}{2.5/3.464}=\frac{-3.5}{0.721}=-4.854\hfill \\ \left|Z\right|\hfill & =|-4.854|=4.854\hfill \end{array}$

i.e., calculated value of Z is greater than the table value of Z.

Hence H₀ is rejected.

Therefore the production process shows a mean that is significantly less than the population mean and this calls for taking same corrective measures concerning the production process.

Example 8.8: A random sample of 100 students gave a mean height of 58 kg, with a standard deviation (SD) 4 kg. Test the hypothesis that the mean weight in the population is 60 kg.

Solution: Here $\overline{x}=58$, S=4, μ=60, n=100

$\begin{array}{c}{H}_0:\mu =60\mathrm{kg}\\ H_1:\mu \ne 60\mathrm{kg}\end{array}$

Level of significance α=5%

Table value of Z_α=1.96

$\begin{array}{ll}Z\hfill & =\frac{\overline{x}-\mu }{S/\sqrt{n}}=\frac{58-60}{4/\sqrt{100}}=-5\hfill \\ \left|Z\right|\hfill & =|-5|=5\hfill \end{array}$

Since the calculated value of Z is greater than the table value at 5% level of significance, H₀ is rejected.

$\therefore \mu \ne 60$

8.13.2 Test of Significance for Difference of Means of Two Large Samples

Let ${\overline{x}}_1$ be the mean of independent random sample of size n₁ from a population with mean μ₁ and variance ${\sigma }_1^2$ and let ${\overline{x}}_2$ be the mean of independent random sample of size n₂ from a population with mean μ₂ and variance ${\sigma }_2^2$, where n₁ and n₂ are large.

Clearly,

${\overline{x}}_1~N\left({\mu }_1,\frac{{\sigma }_1^2}{n_1}\right)\mathrm{and}{\overline{x}}_2~N\left({\mu }_2,\frac{{\sigma }_2^2}{n_2}\right)$

and

$Z=\frac{({\overline{x}}_1-{\overline{x}}_2)-E({\overline{X}}_1-{\overline{X}}_2)}{S.E.({\overline{x}}_1-{\overline{x}}_2)}~N(0,1)$

i.e., Z is the standard normal variate.

Setting up the null hypothesis, we have

$\begin{array}{ll}E({\overline{x}}_1-{\overline{x}}_2)\hfill & =E({\overline{x}}_1)–E({\overline{x}}_2)\hfill \\ \hfill & ={\mu }_1-{\mu }_2=0\hfill \end{array}$

Since ${\overline{x}}_1$ and ${\overline{x}}_2$ are independent, the covariance terms vanish. We have

$\begin{array}{ll}\mathrm{Var}({\overline{x}}_1–{\overline{x}}_2)\hfill & =\mathrm{Var}({\overline{x}}_1)+\mathrm{Var}({\overline{x}}_2)\hfill \\ \hfill & =\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}\hfill \end{array}$

Hence under the null hypothesis H₀: μ₁=μ₂, the test statistic is

$Z=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$ (8.1)

(8.1)

If ${\sigma }_1={\sigma }_2=\sigma$, then the test statistic is

$Z=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{{\sigma }^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sigma \sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$ (8.2)

(8.2)

If σ₁ and σ₂ are not known then the test statistic is

$Z=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}$

where

${\sigma }_1\ne {\sigma }_2$

If ${\sigma }_1={\sigma }_2$ and σ is not known, we compute σ² by using the formula

${\sigma }^2=\frac{n_1{S}_1^2+n_2{S}_2^2}{n_1+n_2}$

In this case the test statistic is

$Z=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{n_1{S}_1^2+n_2{S}_2^2}{n_1+n_2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

substitute S₁ for and S₂ for provided n₁ and n₂ are large.

i.e.,

$Z=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{n_1{S}_1^2+n_2{S}_2^2}{n_1{n}_2}}}$ (8.3)

(8.3)

If we want to test the null hypothesis ${\mu }_1-{\mu }_2=\delta$ where δ is a specified constant against one of the alternatives

$\begin{array}{c}{\mu }_1-{\mu }_2\ne \delta \\ {\mu }_1-{\mu }_2<\delta \end{array}$

or

${\mu }_1-{\mu }_2>\delta$

We apply likelihood ratio technique and at the test based on ${\overline{x}}_1-{\overline{x}}_2$. The corresponding critical regions for the test can be written as

$\left|Z\right|\ge Z_{\alpha /2},Z\ge Z_{\alpha }$

and

$Z\le Z_{\alpha }$

where

$Z=\frac{{\overline{X}}_1-{\overline{X}}_2-\delta }{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

When we deal with independent random samples drawn from populations with unknown variances, that may not be normal. We substitute S₁ for σ₁ and S₂ for σ₂ provided n₁ and n₂ are large.

When the sizes of the independent random samples, i.e., n₁ and n₂ are small and σ₁ and σ₂ are unknown the above test cannot be used. For independent random samples from two normal population having the same unknown variable, we use the test

$Z=\frac{{\overline{X}}_1-{\overline{X}}_2-\delta }{S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

where

$S_1=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$

(Test of significance of difference between two means).

Example 8.9: Random samples drawn from two places gave the following data relating to the heights of children:

Test at 5% level that the mean height is the same for the children at two places.

Solution: Here we have

$\begin{array}{ll}{\overline{x}}_1=68.50,\hfill & {\overline{x}}_2=68.58\hfill \\ n_1=1200,\hfill & n_2=1500\hfill \\ {\sigma }_1=2.5,\hfill & {\sigma }_2=3.0\hfill \end{array}$

Level of significance=0.05

$Z_{0.05}=1.96$

$\begin{array}{ll}SE({\overline{x}}_1–{\overline{x}}_2)\hfill & =\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\hfill \\ \hfill & =\sqrt{\frac{{(2.5)}^2}{1200}+\frac{3^2}{1500}}\hfill \\ \hfill & =\sqrt{\frac{6.25}{1200}+\frac{9}{1500}}=0.1058\hfill \end{array}$

Null hypothesis $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis $H_1:{\mu }_1\ne {\mu }_2$ (Two-tailed test)

$\begin{array}{ll}{Z}_{0.05}\hfill & =1.96(\mathrm{table}\mathrm{value})\hfill \\ Z\hfill & =\frac{{\overline{X}}_1-{\overline{X}}_2}{SE(\overline{x_1}-\overline{x_2})}=\frac{68.50-68.58}{0.1058}=-0.756\hfill \end{array}$

|Z|=0.756<1.96, i.e., calculated value of Z is less than the table value. Hence null hypothesis is accepted.

Example 8.10: The mean produce of a sample of 100 fields is 200 lb, per acre with a standard deviation (SD) of 10 lb. Another sample of 150 fields gives the mean of 220 lb, with a SD 12 lb. Can the two samples be considered to have been taken from the same population whose SD is 11 lb. Use 5% level of significance.

Solution: It is given that

${\overline{x}}_1=200\mathrm{lb},{\overline{x}}_2=220\mathrm{lb}$

$n_1=100,n_2=150$

${\sigma }_1=10\mathrm{lb},{\sigma }_2=12\mathrm{lb}$

$\sigma =\mathrm{SD}\mathrm{of}\mathrm{the}\mathrm{population}=11\mathrm{lb}$

$H_0:{\mu }_1={\mu }_2$

$H_1:{\mu }_1\ne {\mu }_2(\text{two}-\text{tailed}\mathrm{test})$

Level of significance=5%

$Z_{\alpha }=1.96$

Test statistic:

$\begin{array}{ll}Z\hfill & =\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{{\sigma }^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\frac{200-220}{\sqrt{{(11)}^2\left(\frac{1}{100}+\frac{1}{150}\right)}}\hfill \\ \hfill & =\frac{-20}{\sqrt{(121)\left(\frac{250}{(100)(150)}\right)}}=\frac{-20}{1.42}=-14.08\hfill \\ \therefore \left|Z\right|\hfill & =|-14.08|=14.08\hfill \end{array}$

The calculated value of Z is greater than the table value at 5% level of significance.

Hence we reject null hypothesis H₀.

Therefore the difference between the means of the two samples is significant and it is not due to sampling fluctuations.

Example 8.11: Two sets of 100 students each were taught to read by two different methods. After the instructions were over, a reading test given to them revealed ${\overline{x}}_1=73.4$, ${\overline{x}}_2=70.3$, S₁=8, and S₂=10. Test the hypothesis that ${\mu }_1={\mu }_2$.

Solution: It is given that

$\begin{array}{l}{\overline{x}}_1=73.4,{\overline{x}}_2=70.3\hfill \\ S_1=8,S_2=10\hfill \\ n_1=n_2=100\hfill \end{array}$

Null hypothesis $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$

$\begin{array}{ll}Z\hfill & =\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=\frac{73.4-70.3}{\sqrt{\frac{64}{100}+\frac{100}{100}}}\hfill \\ \hfill & =\frac{3.1}{\sqrt{164}}\times \sqrt{100}=\frac{31}{12.806}=2.4207\hfill \end{array}$

|Z|=2.4207>1.96, therefore at 5% level of significance H₀ is rejected.

|Z|=2.4207<2.58, therefore at 1% Level of significance H₀ is accepted.

Example 8.12: Samples of students were drawn from two univariates and from their weights in kilograms and standard deviations (SDs) are calculated. Make a large sample test the significance of the difference between the means.

Solution: We have

$\begin{array}{ll}{\overline{x}}_1=55,\hfill & {\overline{x}}_2=57\hfill \\ {\sigma }_1=10,\hfill & {\sigma }_2=15\hfill \\ n_1=400,\hfill & n_2=100\hfill \end{array}$

Null hypothesis $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$

Level of significance=0.05

Table value of Z=1.96

Test statistic:

$\begin{array}{ll}Z\hfill & =\frac{\overline{X_1}-\overline{X_2}}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}=\frac{55-57}{\sqrt{\frac{100}{400}+\frac{225}{100}}}\hfill \\ \hfill & =\frac{-2}{\sqrt{\frac{100+900}{400}}}=\frac{-4}{\sqrt{10}}=-1.26\hfill \\ \therefore \left|Z\right|\hfill & =1.26<1.96\hfill \end{array}$

i.e., calculated value of Z is less than the table value of Z at 5% level of significance. Hence H₀ is accepted.

Example 8.13: In a certain factory there are two independent processes for manufacturing the same item. The average from one process is formed to be 120 gm with a standard deviation (SD) of 400 items. Is the difference between the mean weights significant at 10% level of significance?

Solution: The given values are

${\overline{x}}_1=120,{\overline{x}}_2=124$

$S_1=12,S_2=14$

$n_1=250,n_2=400$

Level of significance=10%

Null hypothesis $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$ (two-tailed test)

Z-value=1.645 at 10% level

Test statistic is

$\begin{array}{ll}Z\hfill & =\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}=\frac{120-124}{\sqrt{\frac{{12}^2}{250}+\frac{{14}^2}{400}}}\hfill \\ \hfill & =\frac{-4}{1.0325}=-3.87\hfill \\ \left|Z\right|\hfill & =3.87>1.645(\mathrm{i}\mathrm{.e}.,Z\mathrm{at}10\%\mathrm{level})\hfill \end{array}$

Since the calculated value of Z is greater than the table value of Z, we reject null hypothesis. Therefore there is a significance difference between the two sample mean weights.

8.13.3 Test of Significance for the Difference of SDs of Two Large Samples

If S₁ and S₂ are the SD of two independent samples and $H_0:{\sigma }_1={\sigma }_2$ is the null hypothesis, the test statistic is

$Z=\frac{S_1-S_2}{S.E(S_1-S_2)}~N(0,1)$

If ${\sigma }_1,{\sigma }_2$ are known, then for large samples

$Z=\frac{S_1-S_2}{\sqrt{\frac{{\sigma }_1^2}{2n_1}+\frac{{\sigma }_2^2}{2n_2}}}$

If ${\sigma }_1,{\sigma }_2$ are not known, the test statistic is

$Z=\frac{S_1-S_2}{\sqrt{\frac{S_1^2}{2n_1}+\frac{S_2^2}{2n_2}}}$

where

$\sqrt{\frac{S_1^2}{2n_1}+\frac{S_2^2}{2n_2}}\mathrm{is}\mathrm{the}\mathrm{SE}(S_1–S_2)$

Example 8.14: Intelligence test of two groups of boys and girls gave the following results: