Question

Chpt 9 Assignment i Saved Help Save & Ex Exercise 9-34 A real estate sales agency specializes in selling vacation property. Its records indicate that the mean selling time of vacation property is 90 days. However, it believes that because of the recent hurricanes, the mean selling time is now greater than 90 days. A survey of 100 properties sold recently revealed that the mean selling time was 94 days, with a standard deviation of 24 days. points At the 0.10 significance level, has there been an increase in selling time? a. What is the decision rule? (Round the final answer to 3 decimal places.) eBook Ask Reject Ho: 5 90 and accept Hip> 90 when the test statistic is (Click to select) V D . Print b. What is the value of the test statistic? (Round the final answer to 2 decimal places.) Value of the test statistic = c. What is your decision regarding Ho? (Click to select) Ho.

Answer 1

8(a)

Here we have =94, s=24, n =100

Level of significance: a =0.1 Test is one tailed (right tailed). Degree of freedom: df =n-1=99

Critical value: Rejection Region: 1.29 Ift > 1.29, Reject HO

Excel function for critical value: ETINV(0.2,99)

(b)

Test Statistics: _1-1 F = (94-90)/(24/sqrt(100)) = 1.67

(c)

Decision: Since test statistics lies in rejection region so we reject the null hypothesis

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

	X	(X-mean)^2
	4.56	0.025921
	4.37	0.000841
	4.41	0.000121
	4.37	0.000841
	4.46	0.003721
	4.47	0.005041
	4.38	0.000361
	4.24	0.025281
	4.45	0.002601
	4.28	0.014161
Total	43.99	0.07889

Sample size n =10 Sample sum: x=43.99 Sample mean: F-24 -43.99/10 =4.399 Sample standard deviation: *(x-7) S=1 V Esqrt(0.07889/(10-1)) n-1 =0.0936

(a)

Hypotheses are: H:us 4.36 u > 4.36 H Level of significance: a =0.01 Test is one tailed (right tailed). Degree of freedom: df =n-1=9 Test Statistics: F-u t=- = (4.399-4.36)/(0.0936/sqrt(10)) = 1.318 Critical value: Rejection Region: 2.821 Ift > 2.821, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis

Fail to rejejct. There is no evidence .....

(b)

P-value: P-value=0.11

Answer: 0.1100

Excel function for critical value: Excel function for p-value: ETINV(0.02,9) =TDIST(1.318,9,1)

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