Answer to Question #212305 in Statistics and Probability for Manasa Reddyrajula

Question

Consider the following information to validate the hypothesis that the mean of the population

is at most 30 at 5% level of significance.

Sample size = 45, mean of the sample = 38, variance = 4

What will be conclusion if the level of significance is taken as 1%? Mention the observations, if

any.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0: \mu\leq 30$

$H_1:\mu>30$

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 $\alpha=0.05,$

$df=n-1=45-1=44$ degrees of freedom, and the critical value for a right-tailed test is $t_c=1.68023.$

The rejection region for this right-tailed test is $R=\{t:t>1.68023\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-30}{4/\sqrt{45}}=13.4164$$

Since it is observed that $t=13.4164>1.68023=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $\alpha=0.05, df=44, t=13.4164,$right-tailed is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

$\mu$  is at most 30, at the $\alpha=0.05$ significance level.

2. The following null and alternative hypotheses need to be tested:

$H_0: \mu\leq 30$

$H_1:\mu>30$

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 $\alpha=0.01,$

$df=n-1=45-1=44$ degrees of freedom, and the critical value for a right-tailed test is $t_c=2.414134.$

The rejection region for this right-tailed test is $R=\{t:t>2.414134\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{38-30}{4/\sqrt{45}}=13.4164$$

Since it is observed that $t=13.4164>2.414134=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $\alpha=0.05, df=44, t=13.4164,$right-tailed is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

$\mu$  is at most 30, at the $\alpha=0.01$ significance level.

Therefore there is not enough evidence to claim that the population mean $\mu$

$\mu$  is at most 30, at $\alpha=0.05$ significance level and at $\alpha=0.01$ significance level.