Question

A radio station wants to control the time allotted to unpaid public service commercials. If there are too many such commercials, the station loses revenue; if there are too few, the station loses points with the Federal Communications Commission. The target figure is 2.2 commercial minutes per hour. Data on the minutes spent by the radio station on unpaid public service commercials are reported in the following table for a randonm sample of 30 hours. 1.8 2.2 1.5 2.5 2.1 1.71.4 2.7 2.3 2.3 1.9 2.6 2.1 1.7 1.4 1.6 2.4 2.2 2.9 1.5 2 2.4 2.81.8 1.9 2.3 2.1 1.5 2 2.4 Using the sample data, please calculate and interpret the 90% and 99% confidence intervals for the mean time spent on unpaid public service commercials by the radio station. Based on your results, do you think the radio station is achieving its target for the time it spends on the unpaid public service commercials? Please justify your answer. What would be the probability that the population mean time spent on unpaid public service commercials could be less than the upper limit of the 99% confidence interval? Please explain which confidence interval is wider and why. What do you think will happen to the margins of error and the confidence intervals if the sample size was to be reduced to 15 observations? Please justify your answer. a. 5 points] b. 2 points)

Answer 1

First we need to find the mean and SD of data set. Following table shows the calculations:

	Time, X	(X-mean)^2
	1.8	0.071289
	2.2	0.017689
	1.5	0.321489
	2.5	0.187489
	2.1	0.001089
	1.7	0.134689
	1.4	0.444889
	2.7	0.400689
	2.3	0.054289
	2.3	0.054289
	1.9	0.027889
	2.6	0.284089
	2.1	0.001089
	1.7	0.134689
	1.4	0.444889
	1.6	0.218089
	2.4	0.110889
	2.2	0.017689
	2.9	0.693889
	1.5	0.321489
	2	0.004489
	2.4	0.110889
	2.8	0.537289
	1.8	0.071289
	1.9	0.027889
	2.3	0.054289
	2.1	0.001089
	1.5	0.321489
	2	0.004489
	2.4	0.110889
Total	62	5.18667

Sample size n=30

The sample mean is

Σ z. _ 62 _ 2.067 30

The sample standard deviation is

2 Σ(x-x) rl 30-1-~ 0.4229

The 90% confidence interval is

Since populaiton SD is unknown so t critical value will be used. Level of significance: a -0.1 : α=0.1 Critical value: 1.699 Degree of freedom: 1-29 The requried confidence interval is: 2.067 (0.4229*1.699/sqrt(30) =2.067 ± 0.131 =(1.936,2.198) Excel function for critical value: TINV(0.1,29) Hence, the required confidence interval is (1.936,2.198).

Since this confidence interval does not contain 2.2 so we cannot conclude that ratio station is achieving its target.

--------------------------------------------------

The 99% confidence interval is

Since populaiton SD is unknown so t critical value will be used. Level of significance: α-001 Critical value: t. 2.756 Degree of freedom: df -n-1-29 The requried confidence interval is: -2.067+ (0.4229*2.756/sqrt(30) 2.067 0.213 (1.854,2.28) Excel function for critical value TINV(0.01,29) Hence, the required confidence interval is (1.854,2.28).

Since this confidence interval contains 2.2 so we can conclude that ratio station is achieving its target.

(b)

The upper limit of 99% confidence interval is 2.28. Out of 30 data values, 19 are less than 30 so required probability is

19 /30 = 0.633

The 99% confidence interval is wider because it has greater critical value.

As the sample reduces, margin of error will increases. So confidence interval become wider.