Question

A recent survey indicated that the average amount spent for breakfast by business managers was $7.58 with a standard deviation of $0.42. It was felt that breakfasts on the West Coast were higher than $7.58. A sample of 81 business managers on the West Coast had an average breakfast cost of $7.65. Find the P-value for the test.

Answer 1

The provided sample mean is Xˉ=7.65 and the known population standard deviation is σ=0.42, and the sample size is n = 81.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=7.58

Ha: μ>7.58

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is

zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=1.5≤zc​=1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0668, and since p=0.0668≥0.05, 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 7.58, at the 0.05 significance level.

please rate my answer and comment for doubts.