Answer to Question #195958 in Statistics and Probability for jack

Question

Take a sample of 60 grades and test the following:

1) Test for the hypothesis that the average grade is not 75. 

2) Test for the hypothesis that the average grade below is higher than 75.

3) Calculate the confidence limits for the population mean at 95 %

Answer 1

The sample of 60 grades taken were:

The following information was deduced:

Hypothesized mean $\left(\mu \right)$ $=75$

Sample standard deviation $\left(s\right)$ $=1$

Sample size $\left(n\right)\:$ $=60$

Sample mean $\left(\overline{X}\right)\:$ $=30.5$

Significance level $\left(\alpha \right)$ $=0.05$

1). Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.671.

The rejection region for this right-tailed test is $R=\left\{t:t>1.671\right\}$

The t-statistic is computed as follows:

$t=\frac{\overline{X}-\mu _o}{\frac{s}{\sqrt{n}}}$

$t=\frac{30.5-75}{\frac{1}{\sqrt{60}}}$

$=-344.696$

Since it is observed that $t=-344.696\le t_c=1.671$ , it is then concluded that the null hypothesis is not rejected.

2). Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.671.

The rejection region for this right-tailed test is $R=\left\{t:t<1.671\right\}$

The t-statistic is computed as follows:

$t=\frac{30.5-75}{\frac{1}{\sqrt{60}}}$

$t=\frac{30.5-75}{\frac{1}{\sqrt{60}}}$

$=-344.696$

Since it is observed that $t=-344.696< t_c=1.671$ , it is then concluded that the null hypothesis is rejected.

3). $t=-344.696< t_c=1.671$

The 95% confidence interval is $30.242<\mu <30.758$ .