Question

The two samples A and B detailed below were taken from normal population of standard deviation 2.5.

A 16 18 23 26 19 24 25 23 21 22 20
B 20 21 23 25 27 24 26 24 28 20 30
Compute a 90% confidence interval for real difference and decide whether the difference of sample means is significant (that the population means are different or same). Also test the hypothsis that the two populations have equal means.

Answer 1

A 16 18 23 26 19 24 25 23 21 22 20

$$\bar{x}_1=\dfrac{237}{11}\approx 21.545$$

B 20 21 23 25 27 24 26 24 28 20 30

$$\bar{x}_2=\dfrac{268}{11}\approx 24.364$$

We need to construct the 90% confidence interval for the difference between the population means $\mu_1-\mu_2.$

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.645.$  

Also, the provided population standard deviations are: $\sigma_1=2.5=\sigma_2$ and the sample sizes are $n_1=11$ and $n_2=11.$

The corresponding confidence interval is computed as shown below:

$$CI=\big(\bar{x}_1-\bar{x}_2-z_c\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}},\bar{x}_1-\bar{x}_2+z_c\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}\big)$$

$$=\big(21.545-24.364-1.645\sqrt{\dfrac{2.5^2}{11}+\dfrac{2.5^2}{11}},$$

$$21.545-24.364+1.645\sqrt{\dfrac{2.5^2}{11}+\dfrac{2.5^2}{11}}\big)$$

$$=(-4.572, -1.065)$$

Therefore, based on the data provided, the 90% confidence interval for the difference between the population means $\mu_1-\mu_2$ is $-4.572<\mu_1-\mu_2<-1.065,$ which indicates that we are 90% confident that the true difference between population means is contained by the interval  $(-4.572, -1.065).$

The following null and alternative hypotheses need to be tested:

$H_0: \mu_1=\mu_2$

$H_1: \mu_1\not=\mu_2$

​This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha=0.1,$ and the critical value for a two-tailed test is $z_c=1.645.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.645\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}$$

$$=\dfrac{21.545-24.364}{\sqrt{\dfrac{2.5^2}{11}+\dfrac{2.5^2}{11}}}\approx-2.644$$

Since it is observed that $|z|=2.644>1.645=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(z<-2.644)=0.0082,$ and since $p=0.0082<0.1=\alpha,$ it is concluded that the null hypothesis is rejected.