Answer to Question #234845 in Statistics and Probability for liche

Question

A random sample of 20 business statistics II students gave a summary of the information

on how they spend their time studying in hours, in the week before the busines statistics

II nal examinations: sample mean = 40 hours, sample standard deviation is 15. It is

assumed that the study time follows an approximate normal distribution.

(a) Test at the 5% signicance level the null hypothesis that the population mean is

not more than 50 hours.

(b) Test at the 2:5% signicance level the null hypothesis that the varaince is atleast

110 hours.

(c) Construct 95% CI for  and 2

Answer 1

(a) The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq50$

$H_1:\mu>50$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test $df=n-1=20-1=19$ degrees of freedom is $t_c=1.729133.$

The rejection region for this right-tailed test is $R=\{t:t>1.729133\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{40-50}{15/\sqrt{20}}\approx-2.981424$$

Since it is observed that $t=-2.981424<1.729133=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=19, t=-2.981424$ is

$p=0.996165,$ and since $p=0.996165>0.05=\alpha,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is not more than 50 hours, at the $\alpha=0.05$ significance level.

(b) The following null and alternative hypotheses need to be tested:

$H_0: \sigma^2\geq110$

$H_1:\sigma^2<110$

This corresponds to a left-tailed test test, for which a Chi-Square test for one population variance will be used. Based on the information provided, the significance level is $\alpha=0.025,$  $df=n-1=20-1=19$ degrees of freedom is $\chi_c^2=8.9065.$

The rejection region for this left-tailed test is $R=\{\chi^2: \chi^2<8.9065\}.$  

The Chi-Squared statistic is computed as follows:

$$\chi^2=\dfrac{(n-1)s^2}{\sigma^2}=\dfrac{(20-1)(15)^2}{110}\approx38.8636$$

Since it is observed that $\chi^2=38.8636>8.9065=\chi_c^2,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is enough evidence to claim that the population variance $\sigma^2$  is at least

$110 (hours)^2,$ at the 0.025 significance level.

(c) The critical value for $\alpha=0.05$ and $df=n-1=20-1=19$ degrees of freedom is $t_c=z_{1-\alpha/2;n-1}=2.093024.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(40-2.093024\times\dfrac{15}{\sqrt{20}}, 40+2.093024\times\dfrac{15}{\sqrt{20}})$$

$$=(32.9798, 47.0202)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $32.9798<\mu<47.0202,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(32.9798, 47.0202).$

The critical values for $\alpha=0.05$ and $df=19$ degrees of freedom are $\chi_L^2=\chi_{1-\alpha/2, n-1}=8.9065$ and $\chi_U^2=\chi_{\alpha/2, n-1}=32.8523.$ The corresponding confidence interval is computed as shown below:

$$CI(Variance)=(\dfrac{(n-1)s^2}{\chi_{\alpha/2, n-1}}, \dfrac{(n-1)s^2}{\chi_{1-\alpha/2, n-1}})$$

$$=(\dfrac{19(15)^2}{32.9798}, \dfrac{19(15)^2}{8.9065})\approx(129.6248,479.9865)$$

Therefore, based on the data provided, the 95% confidence interval for the population variance is $(129.6248,479.9865).$