A researcher wants to test if the mean price of houses in an area is greater than $145,000. A random sample of 35 houses selected from the area produces a mean price of $149,700. Assume that  $13,100, and that the test is to be made at the 1% significance level.a. What is the critical value of z?A) 1.88 B) 2.17 C) 1.96 D) 2.58b. What is the value of the test statistic, z, rounded to three decimal places?c. What is the p-value for this hypothesis test, rounded to four decimal places?d. Should you reject or fail to reject the null hypothesis in this test? (State your answer as "reject" or "fail to reject", but don't include the quotation marks.) A researcher wants to test if the mean price of houses in an area is greater than $145,000. A randomsample of 35 houses selected from the area produces a mean price of $149,700. Assume that O =$13,100, andthat the test is to be made at the 1% significance level.а.What is the critical value of z?A) 1.88B) 2.17C) 1.96D) 2.58b.What is the value of the test statistic, z, rounded to three decimal places?c. What is the p-value for this hypothesis test, rounded to four decimal places?d.Should you reject or fail to reject the null hypothesis in this test? (State your answer as "reject" or"fail to reject", but don't include the quotation marks.)

Name: A researcher wants to test if the mean price of houses in an area is g
Uploaded: 2024-03-20T07:07:12.000Z
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Description: A researcher wants to test if the mean price of houses in an area is greater than $145,000. A random sample of 35 houses selected from the area produces a mean price of $149,700. Assume that  $13,100, and that the test is to be made at the 1% signi

a. Computation of critical value: α=0.01α2=0.0051-α2=0.995 The critical value of z-distribution can be obtained using the excel formula “=NORM.S.INV(0.995)”. The critical value is 2.58. Thus, the critical value of z is 2.58.b. Let μ denotes the population mean price of houses in an area. The test is that the mean price of houses in an area is greater than $145,000. The hypothesis is, Null hypothesis: H0:μ≤$145,000 Alternative hypothesis: H1:μ>$145,000 The value of the test statistic is, z=x¯-μσn=149700-1450001310035=47002214.3041=2.123 Thus, the value of the test statistic is 2.123.c. The hypothesis is right tail. The p-value for right tailed test is, p-value=Pz>2.123=1-Pz<2.123 The probability of z less than 2.123 can be obtained using the excel formula “=NORM.S.DIST(2.123,TRUE)”. The probability value is 0.9831. The required p-value is, p-value=1-Pz<2.123=1-0.9831=0.0169 Thus, the p-value is 0.0169. d. Decision rule: If p-value≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The p-value (0.0169) is greater than the level of significance (0.01). Based on the decision rule, fail to reject the null hypothesis. Correct Answer: Fail to reject.