If a random sample of 25 homes south of Center Street in Provo has a mean sellin

Question

If a random sample of 25 homes south of Center Street in Provo has a mean selling price of $145,225 and a standard deviation of $4550, and a random sample of 21 homes north of Center Street has a mean selling price of $148,575 and a standard deviation of $5800, can you conclude that there is a significant difference between the selling price of homes in these two areas of Provo at the 0.05 level? Assume normality.

(a) Find t. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion.

-Reject the null hypothesis, there is not significant evidence of a difference in means.

-Reject the null hypothesis, there is significant evidence of a difference in means.

-Fail to reject the null hypothesis, there is significant evidence of a difference in means.

-Fail to reject the null hypothesis, there is not significant evidence of a difference in means.

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0  
At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    145225          
X2 =    148575          
              
Calculating the standard deviations of each group,              
              
s1 =    4550          
s2 =    5800          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    25          
n2 = sample size of group 2 =    21          
Thus, df = n1 + n2 - 2 =    44          
Also, sD =    1558.847254          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -2.149023896   [ANSWER, TEST STATISTIC[

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ii.      
              
where uD = hypothesized difference =    0          
              
Also, using p values, as this is two tailed,              
              
p =    0.037173264   [ANSWER]

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b)

As P < 0.05, we

OPTION B: -Reject the null hypothesis, there is significant evidence of a difference in means. [ANSWER]      
              

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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