Question

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? Calculate the p-value.

Answer 1

The provided sample mean is Xˉ=137

and the sample standard deviation is s = 45,

and the sample size is n = 30.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 150

Ha: μ > 150

(2) Rejection Region

Based on the information provided, the significance level is α=0.01,

and the critical value for a right-tailed test is t_c = 2.462

The rejection region for this right-tailed test is

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

t  =−1.582 ≤  tc​ = 2.462,

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.9378,

and since p=0.9378 ≥ 0.01,

it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is greater than 150, at the 0.01 significance level.

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