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Semenov [28]
3 years ago
12

At many golf​ clubs, a teaching professional provides a free​ 10-minute lesson to new customers. A golf magazine reports that go

lf facilities that provide these free lessons​ gain, on​ average, ​$2 comma 100 in green​ fees, lessons, or equipment expenditures. A teaching professional believes that the average gain exceeds ​$2 comma 100. Complete parts a through c below. a. In order to support the claim made by the teaching​ professional, what null and alternative hypotheses should you​ test? Upper H 0​: mu equals ​$2 comma 100 Upper H Subscript a​: mu greater than ​$2 comma 100 b. Suppose you select alphaequals0.01. Interpret this value in the words of the problem. The probablility that the null hypothesis is ▼ when the average gain ▼ is is not is less than exceeds ​$2 comma 100 is 0.01. c. For alphaequals0.01​, specify the rejection region of a​ large-sample test. Choose the correct answer below. A. z less than minus 2.33 B. z less than minus 2.575 or z greater than 2.575 C. zless thanminus1.28 D. zless thanminus1.96 or zgreater than1.96 E. minus2.575less thanzless than2.575 F. z greater than 2.33 G. minus1.96less thanzless than1.96 H. zgreater than1.28
Mathematics
1 answer:
Airida [17]3 years ago
7 0

Answer:

Step-by-step explanation:

a) The objective of the study is test the claim that the average gain in the green fees , lessons or equipment expenditure for participating golf facilities is less than $2,100 under the claim the null and alternative hypothesis are,

H₀ : μ = $2,100

H₀ : μ < $2,100

B) Suppose you selects α = 0.01

The probability that the null hypothesis is rejected when the average gain is $2,100 is 0.01

C) For α = 0.01

specify the rejection region of a large sample test

At the given level of significance 0.01 and the test is left-tailed then rejection level of a large-sample = < - 1.28

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Answer:

1)

Null hypothesis            H₀ : μ = 25,809

Alternative hypothesis H₁ : μ < 25,809

2) Test Statistic = -1.44

3) Conclusion:

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Step-by-step explanation:

Given the data in the question;

Sample mean x" = 24,638

sample size n = 46

standard deviation σ = 5531

level of significance ∝ = 0.10

NULL and ALTERNATIVE HYPOTHESIS

Null hypothesis            H₀ : μ = 25,809

Alternative hypothesis H₁ : μ < 25,809

TEST STATISTICS

Z = (x"-μ) / σ√n

we substitute

Z = (24,638 - 25,809) / (5531/√46)

Z = -1171 / 815.5

Z = -1.44

Test Statistic = -1.44

Now, from normal z-table;

P-value = P( Z < -1.44 ) = 0.0749

P-value = 0.0749

Since P-value ( 0.0749 ) is less than level of significance ( 0.10 ), we reject H₀.

Conclusion:

The result is significant, there is sufficient evidence to support the bride’s hope at the 0.10 level of significance.

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Step-by-step explanation:

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Answer:

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