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Excel Project # 2

Model Soln

March 2005

Part A: Estimate the true mean Price with a 95 percent confidence interval.

t - Confidence Interval Estimate

Confidence Level 0. Sample Mean 94900 Sample SD 31030. Sample size 44 Critical value 2. The 0.95 confidence interval estimate: Lower value 85465. Upper value 104334.

1. Critical value = t 43 = 2.

2. Margin of error = (CV)(SE) = 2.01669*31030.9/sqrt(44) = 9434.

3. We are 95 percent confident that the true Price is between 85465.7 and 104334.

Part B: Using a 5 percent level of significance, test the hypotheses that the average Price

is less than 100000 dollars.

Hypothesis test for the Mean using the t -

distribution

Significance level = 0. Ho: μ = (^100000) Sample mean = 94900 Sample SD = 31030. Sample size = 44 Test statistic = - 1.

Two tailed test

Lower critical value - 2. Upper critical value 2.01 67 p - value 0.

Lower tailed test

Critical value - 1. p - value 0.

Upper tailed test

Critical value 1. p - value 0.

1. H 0 : μ >= 100000 vs. H 1 : μ < 100000

2. Test statistic = t = - 1.

3. Critical value = t 43 = - 1.6811 and the p - value = 0.

4. Since the p - value = 0.1409 > α=0.05, we will not reject H 0 and conclude that the

Price is not less than 100000.

Part C: Estimate the true proportion Price with a 95 percent confidence interval.

p - Confidence Interval Estimate

Confidence Level 0.

of success in sample 7

Sample size 44 Sample Proportion 0. Critical value 1. The 0.95 confidence interval estimate: Lower value 0. Upper value 0.

1. The critical value z is 1.

2. The margin of error is (CV)(SE) = (1.96)(.0551)= 0.10 81

3. We are 95 percent confident that the true proportion is between 0.0510 and

0.2672.

Part D: Does the average Price (of houses) differ when Garage = 0 or 1?

t-Test: Two-Sample Assuming Equal Variances Price 0-Stall Price 1-Stall Mean 72900 851 28. Variance 216333333.3 303800659. Observations 7 14 Pooled Variance 276179398. Hypothesized Mean Difference 0 df 19 t Stat - 1. P(T<=t) one-tail 0. t Critical one-tail 1. P(T<=t) two-tail 0. t Critical two-tail 2.

1. The hypotheses are H 0 : μ 0 = μ 1 vs. H 1 : μ 0 ≠ μ 1.

2. The test statistic t = - 1.

3. The critical value t 19 = ±2.

4. Since the p-value = 0.12843 > 0.05, do not reject H 0 ; there is no evidence that the

price is different when a house has no garage or a one-stall garage.