1. If a random sample of 28 homes south of Center Street in Provo has a mean sel
ID: 3155134 • Letter: 1
Question
1.
If a random sample of 28 homes south of Center Street in Provo has a mean selling price of $145,050 and a standard deviation of $4850, and a random sample of 15 homes north of Center Street has a mean selling price of $148,350 and a standard deviation of $5675, 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.
2.
Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is = 19 inches. However, a survey reported that of a random sample of 23 fish caught, the mean length was x = 18.0 inches, with estimated standard deviation s = 2.6 inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than = 19 inches? Use = 0.05. (Use 3 decimal places for the test statistic and the critical value. Use 4 decimal places for the P-value.)
Conclusion
Reject the null hypothesis, there is sufficient evidence that the average fish length is less than 19 inches.
Reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.
Fail to reject the null hypothesis, there is sufficient evidence that the average fish length is less than 19 inches.
Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.
test statistic critical value P-valueExplanation / Answer
Q1.
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=145050
Standard Deviation(s.d1)=4850 ; Number(n1)=28
Y(Mean)=148350
Standard Deviation(s.d2)=5675; Number(n2)=15
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =145050-148350/Sqrt((23522500/28)+(32205625/15))
to =-1.909
| to | =1.909
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 14 d.f is 2.145
We got |to| = 1.90936 & | t | = 2.145
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -1.9094 ) = 0.077
Hence Value of P0.05 < 0.077,Here We Do not Reject Ho
[ANSWERS]
t = -1.909
P-value : 0.077
Fail to reject the null hypothesis, there is not significant evidence of a difference in means.
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