o Aplia: Student Question x + HOD a coursesapliacom/af/servlet/quizzquiz, action

Question

o Aplia: Student Question x + HOD a coursesapliacom/af/servlet/quizzquiz, action-takeQuiz&quizprobGuid;=QNAPCOA201o1oooooo3bd7500a0000&ctedelia.martin; eQ Scorch A 1 in 5 7. The probability of Type II errors, the power curve, and sample size Aa Aa Suppose an architectural firm specializing in the structural restoration and renovation of historic homes and early barns is deciding whether to open a branch of the company in Wilmington, Vermont. Market research commissioned by the firm indicates that the Vermont location will be profitable only if the mean age of houses and barns located within a 100-mile radius of Wilmington is greater than 65 years. The architectural firm conducts a hypothesis test to determine whether 1, the mean age of structures located within a 100-mile radius of Wilmington, is greater than 65 years. The test is conducted at a = .05 level of significance using a random sample of n = 144 houses and bams located in the specified area. The population standard deviation of the age of structures is assumed to be known with a value of a = 16.8 years. The firm will open a Vermont branch only if it rejects the null hypothesis that the mean age of structures in the specified area is less than or equal to 65 years. To summarize this hypothesis test refer to the chart given below. Reject the Null = Open the Branch Fail to Reject = Do Not Open the Branch Use the Distributions tool to help you answer the questions that follow, Momas Distribution Mean = 68.0 Standard Deviation - 3.0 Session 59:17 Timeout OO 61 9:20 PM YEbp 7 x w) Paos 9 O2 9 10/21/2017

Explanation / Answer

Null Hypothesis is rejected when the sample mean is

we will reject the null hypothesis when the sample mean is above the upper bound which is

x = H + Z95% (s/ sqrt(n) = 65 + 1.645 * (16.8/ sqrt(144) = 65 + 1.645 * (16.8/12) = 67.303

{here standard error of the mean = 16.8/ sqrt(n) = 16.8/12 = 1.4}

we will reject the null if x > 67.303

THe firm will not open a branch when x < 67.303 years

Now if true mean = 67 years

Probability of type II error = Pr(x < 67.303 ; 67; 1.4)

Z = (67.303 - 67)/1.4 = 0.2164

so Pr (Type II error) = Pr(Z < 0.5314) = 0.5857

The power of the test is = 1 - 0.5857= 0.4143

Based on the level of significance it has selected for its test, the firm is willing to risk a 0.05 probability of opening an unsucceful branch.

Now, it is willing to fail to reject the null hypothesis is 0.10

so Pr(Type II error) = 0.10 when True = 0.68

let say new sample size = n

so the upper bound of sample mean above which we will reject the null hypothesis = 65 + 1.645 * 16.8/ sqrt(n)

= 65 + 27.636/ sqrt(n)

so there is 0.10 probability that we faile to reject the null hypothesis.

Pr (Type II error ) = 0.10

stanadard error of true mean = 16.8/ (n)

NORM ( 65 +27.636/(n) ; 68 ; 16.8/ (n) ) = 0.10

The Z value = -1.28 as the value will be on left side of true mean

(65 + 27.636(n) -68)/ 16.8/ (n) = -1.28

-3 + 27.636/(n) = -21.504/(n)

49.14/(n) = 3

(n) = 16.38

n = 269

so sample size must be increased to 269. Option B is correct

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.