Question

According to a recent survey of the 23​,000 funeral homes of a certain​ nation, funeral homes collected an average of ​$6,500 per​ full-service funeral last year. A random sample of 36 funeral homes reported revenue data for the current year. Among other​ measures, each reported its average fee for a​ full-service funeral. These data​ (in thousands of​ dollars) are shown in the accompanying table. Complete parts a through c below.

b. Conduct the test at α=0.05

Do the sample data provide sufficient evidence to conclude that the average fee this year is higher than last​ year?

Find the test statistic.

z=
c. In conducting the​ test, was it necessary to assume that the population of average​ full-service fees was normally​ distributed?

6.9
9.5
4.8
8.5
7.5
6.7
5.9
8.5
6.6
11.6
6.3
5.9
7.1
6.1
5.2
6.6
6.1
7.8
7.3
6.4
6.6
5.4
7.1
5.5
9.5
7.1
7.5
5.8
6.1
5.5
7.3
7.5
6.3
5.3
7.8
6.5

Answer 1

From the data: = 6863.9, s = 1422.5

____________________

The Hypothesis

H0: = 6500

Ha: > 6500

This is a one tailed test

____________________

The Test Statistic: The test statistic is given by the equation:

_____________________

The p Value:    The p value (Right Tail) for Z = 1.53, is; p value = 0.063

The Critical Value:   The critical value (Right Tail) at = 0.05, Z critical = -1.645

The Decision Rule:   If Z observed is > Z critical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since Z observed (1.53) is < Z critical (1.645), We Fail to Reject H0.

Also since P value (0.063) is > (0.05) , We Fail to Reject H0.

The Conclusion: There isn’t sufficient evidence at the 95% significance level to conclude that the average fees this year is higher than last year.

__________________________

(c) Yes, Since we were using the z distribution and population standard deviation was unknown, we had to assume that the population from which the sample was randomly picked was normally distributed. This assumption was necessary so that we have a sample which was normally distributed and hence we could use the z distribution.

__________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#	X	Mean	(x - mean)2
1	6900	6863.900	1303.210
2	9500	6863.900	6949023.210
3	4800	6863.900	4259683.210
4	8500	6863.900	2676823.210
5	7500	6863.900	404623.210
6	6700	6863.900	26863.210
7	5900	6863.900	929103.210
8	8500	6863.900	2676823.210
9	6600	6863.900	69643.210
10	11600	6863.900	22430643.210
11	6300	6863.900	317983.210
12	5900	6863.900	929103.210
13	7100	6863.900	55743.210
14	6100	6863.900	583543.210
15	4200	6863.900	7096363.210
16	6600	6863.900	69643.210
17	6100	6863.900	583543.21
18	7800	6863.900	876283.21
19	7300	6863.900	190183.21
20	6400	6863.900	215203.21
21	6600	6863.900	69643.21
22	5400	6863.900	2143003.21
23	7100	6863.900	55743.21
24	5500	6863.900	1860223.21
25	9500	6863.900	6949023.21
26	7100	6863.900	55743.210
27	7500	6863.900	404623.210
28	5800	6863.900	1131883.21
29	6100	6863.900	583543.21
30	5500	6863.900	1860223.21
31	7300	6863.900	190183.21
32	7500	6863.900	404623.21
33	6300	6863.900	317983.21
34	5300	6863.900	2445783.21
35	7800	6863.900	876283.21
36	6500	6863.900	132423.21

n	36
Sum	247100
Average	6863.9
SS	70823055.56
Variance = SS/n-1	2023515.873
Std Dev	1422.50