Ninety cities provided information on vacancy rates (in percent) in local apartm

Question

Ninety 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 3.6% respectively. Use Table 3 Vacancy Rate (in percent) Less than 6 6 up to 9 9 up to 12 12 or more Frequency 16 20 31 23 a. Choose the appropriate null and alternative hypotheses Ho: Vacancy rates are not normally distributed with a mean of 10% and a standard deviation of 3.6%. ; HA: Vacancy rates are normally distributed with a mean of 10% and a standard deviation of 36% ·H): Vacancy rates are normally distributed with a mean of 10% and a standard deviation of 3.6%. ; HA: Vacancy rates are not normally distributed with a mean of 10% and a standard deviation of 36% b. Calculate the value of the test statistic at the 1% significance level. (Round the z value to 2 decimal places, all other intermediate values to at least 4 decimal places and final answer to 2 decimal places.) Test statistic c. Approximate the p-value p-value

Explanation / Answer

a)

Ho: vacancy rate are normally distributed...........

b)

applying chi square goodness of fit test:

test statistic =2.30 ( please try 2.61 if this does not work due to rounding)

c)

p value >= 0.200

mean = 10 std deviation = 3.6 degree of freedom = k-m-1     = 3 Class Limits observed Normal Normal Expected 2=(O-E)2/E lower Upper Bin frequency(O) probabilty probabilty(p) frequency(E=p*O) - 6 -6 16 P(X<6) 0.1333 11.993 1.338 6 - 9 6-9 20 P(6<X<9) 0.2573 23.160 0.431 9 - 12 9-12 31 P(9<X<12) 0.3202 28.814 0.166 12 - inf 12-inf 23 P(12<X<inf) 0.2893 26.033 0.353 Total 90 1 90 2.289

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