Question

A developer collects a random sample of the ages of houses from two neighborhoods and finds that the summary statistics for each are as shown. Assume that the data come from a distribution that is Normally distributed. Complete parts a through c below. 1 2 n1=30 n2=35 y1=51.7 y2=43.2 s1=7.29 s2=7.98 Test the null hypothesis at alphaαequals=0.05 using the pooled​ t-test. Identify the null and alternative hypotheses. Choose the correct answer below. A. Upper H 0H0​: mu 1 minus mu 2 equals 0μ1−μ2=0 Upper H Subscript Upper AHA​: mu 1 minus mu 2 greater than 0μ1−μ2>0 B. Upper H 0H0​: mu 1 minus mu 2 equals 0μ1−μ2=0 Upper H Subscript Upper AHA​: mu 1 minus mu 2 less than 0μ1−μ2<0 C. Upper H 0H0​: mu 1 minus mu 2 not equals 0μ1−μ2≠0 Upper H Subscript Upper AHA​: mu 1 minus mu 2 equals 0μ1−μ2=0 D. Upper H 0H0​: mu 1 minus mu 2 equals 0μ1−μ2=0 Upper H Subscript Upper AHA​: mu 1 minus mu 2 not equals 0μ1−μ2≠0 Compute the​ t-statistic. Let the difference of the sample means be y overbar 1 minus y overbar 2y1−y2. tequals=nothing ​(Round to two decimal places as​ needed.) Find the​ P-value. The​ P-value is nothing. ​(Type an integer or decimal rounded to three decimal places as​ needed.) State the conclusion. Recall that alphaαequals=0.05. ▼ Fail to reject Reject Upper H 0H0. There is ▼ insufficient sufficient evidence to reject the claim that the mean age of houses is ▼ different the same in the two neighborhoods. ​b) Find a​ 95% confidence interval using the pooled degrees of freedom. The confidence interval is (__,__) years. ​(Round to two decimal places as​ needed.) ​c) A​ 95% confidence interval for the mean difference in ages of houses in the two neighborhoods was ​(4.72,12.28​). Is this result different than the result of the​ pooled-t confidence​ interval? Explain why or why not. Choose the correct answer below. A.Yes. Since the standard deviations are fairly​ close, the two methods will result in essentially the same confidence intervals and hypothesis tests. B.No. Since the means are fairly​ close, the two methods will result in essentially the same confidence intervals and hypothesis tests. C.Yes. Since the means are fairly​ close, the two methods will result in essentially the same confidence intervals and hypothesis tests. D.No. Since the standard deviations are fairly​ close, the two methods will result in essentially the same confidence intervals and hypothesis tests.

Answer 1

I have used Minitab and the command is as follows

Stat > Basic stats> 2 sample t-test(Assume equal var) > select the data > ok

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Two-Sample T-Test and CI

Method

μ₁: mean of Sample 1

µ₂: mean of Sample 2

Difference: μ₁ - µ₂

Equal variances are assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Sample 1	30	51.70	7.29	1.3
Sample 2	35	43.20	7.98	1.3

Estimation for Difference

Difference	Pooled StDev	95% CI for Difference
8.50	7.67	(4.69, 12.31)

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ ≠ 0

T-Value	DF	P-Value
4.45	63	0.000

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t =4.45

P-value = 0.000

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Reject Ho

sufficient, same

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b) (4.69, 12.31)

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c) D. No. Since the standard deviations are fairly​ close, the two methods will result in essentially the same confidence intervals and hypothesis tests

NOTE - Here result means conclusion