Answer to Question #345824 in Statistics and Probability for Ally

Question

A school teacher suspects the claim that the mean number of students that use library materials in a certain school is at most 450. To check the claim, the professor checks a random sample of 100 library records and obtain that the mean number of students using library materials is 458 with a standard deviation of 9. What would be the teacher's conclusion using 0.05 level of significance?

a. State the null and alternative hypothesis in symbols.

b. Choose the test statistic applied where 𝛼 = 0.05

c. Determine the critical points

d. Computation of the test statistic

e. Decision

f. Conclusion

Answer 1

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

$H_0:\mu\le450$

$H_1:\mu>450$

b.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 $\alpha = 0.05,$ $df=n-1=99.$ 

c. The critical value for a right-tailed test,  $\alpha = 0.05,$ $df=99$ is $t_c =1.660391.$

The rejection region for this right-tailed test is $R = \{t:t>1.660391\}.$

d. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{458-450}{9/\sqrt{100}}=8.8889$$

e. Since it is observed that $t=8.8889>1.660391=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=99$ degrees of freedom, $t=8.8889$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

f. Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 250, at the $\alpha = 0.05$ significance level.