Question

Ninety five cities provided information on vacancy rates (in percent) in local apartments in the following frequency distribution. The sample mean and the sample standard deviation are 10% and 4.0%, respectively. (You may find it useful to reference the appropriate table: chi-square table or F table)

Vacancy Rate   Frequency
Less than 6      11     
6 up to 9      24     
9 up to 12      38     
12 or more      22     

a. Apply the goodness-of-fit test for normality at the 10% significance level. Do the sample data suggest that vacancy rates do not follow the normal distribution? First, select the appropriate null and alternative hypotheses.

H0: Vacancy rates are not normally distributed with a mean of 10% and a standard deviation of 4%.; HA: Vacancy rates are normally distributed with a mean of 10% and a standard deviation of 4%.
H0: Vacancy rates are normally distributed with a mean of 10% and a standard deviation of 4%.; HA: Vacancy rates are not normally distributed with a mean of 10% and a standard deviation of 4%.

b. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

c. Find the p-value.

p-value < 0.01
p-value 0.10
0.05 p-value < 0.10
0.025 p-value < 0.05
0.01 p-value < 0.025

d. What is the conclusion?

Do not reject H0; there is enough evidence to support the claim that the vacancy rates are not normally distributed
Do not reject H0; there is not enough evidence to support the claim that the vacancy rates are not normally distributed
Reject H0; there is not enough evidence to support the claim that the vacancy rates are not normally distributed
Reject H0; there is enough evidence to support the claim that the vacancy rates are not normally distributed

Answer 1

a)

H0: Vacancy rates are normally distributed with a mean of 10% and a standard deviation of 4%.; HA: Vacancy rates are not normally distributed with a mean of 10% and a standard deviation of 4%.

b)

Class Limits				observed	Normal	Normal	Expected	χ2=(O-E)2/E
lower		Upper	Bin	frequency(O)	probabilty	probabilty(p)	frequency(E=p*ΣO)
	-	6	-6	11	P(X<6)	0.1587	15.08	1.102
6	-	9	6-9	24	P(6<X<9)	0.2426	23.05	0.039
9	-	12	9-12	38	P(9<X<12)	0.2902	27.57	3.947
12	-		12-	22	P(12<X<)	0.3085	29.31	1.822
			Total	95		1.0000	95	6.9104

value of the test statistic =6.910

0.05 p-value < 0.10

Reject H0; there is enough evidence to support the claim that the vacancy rates are not normally distributed