Question

1. A local sports organization has started to use e-tickets for events, which saves them resources and time. Going to this format has resulted in customer complaints. Data collected by the organization is listed below and represents the number of complaints per month.

14	14	18	12	19	14	13
12	15	15	14	13	13	12

At the 0.025 level of significance, can the organization conclude the mean number of complaints per month is less than 15 per month? (Use unrounded values for both Mean and Sample Standard Deviation when doing this question)

Answer 1

The sample size is n = 14 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	14	196
	14	196
	18	324
	12	144
	19	361
	14	196
	13	169
	12	144
	15	225
	15	225
	14	196
	13	169
	13	169
	12	144
Sum =	198	2858

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 15

Ha: μ < 15

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.025, and the critical value for a left-tailed test is t_c = -2.16

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = -1.522 > t_c = -2.16, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.076, and since p = 0.076 > 0.025, 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 less than 15, at the 0.025 significance level.