3. A simple random sample of a daily lift ticket price from 40 different ski resorts is obtained. The sample mean was found to be $108.50 and the sample standard deviation was found to be $17.90. Use a = 0.05 to test the claim that the mean lift ticket price is different than $105. (Does not equte to) Step 1-3) Claim, Opposite, Ho, H₁ 117105 H.: 47105 M² 105 Ho; M = 105 Step 4) a= D. 05 Critical Value: 796 Step 5) Test Statistic: Step 6) Choose one method: Critical value or P-value test (P-value=_ Circle One: reject Ho or fail to reject Ho Step 7) Conclusion: There's mean lift ticket price is different than $105. 2 test evidence to the claim that the

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3.
A simple random sample of a daily lift ticket price from 40 different ski resorts is obtained. The sample mean was
found to be $108.50 and the sample standard deviation was found to be $17.90. Use a = 0.05 to test the claim that
the mean lift ticket price is different than $105.
(Does not equte to)
Step 1-3) Claim, Opposite, Ho, H₁
17105 H.: 417105
M² 105 Ho ; M=105
Step 4) a = D. 05
Critical Value: 796
Step 5) Test Statistic:
Step 6) Choose one method: Critical value or P-value test (P-value=_
Circle One: reject Ho or fail to reject Ho
Step 7) Conclusion: There's
mean lift ticket price is different than $105.
2 test
evidence to
the claim that the
Transcribed Image Text:3. A simple random sample of a daily lift ticket price from 40 different ski resorts is obtained. The sample mean was found to be $108.50 and the sample standard deviation was found to be $17.90. Use a = 0.05 to test the claim that the mean lift ticket price is different than $105. (Does not equte to) Step 1-3) Claim, Opposite, Ho, H₁ 17105 H.: 417105 M² 105 Ho ; M=105 Step 4) a = D. 05 Critical Value: 796 Step 5) Test Statistic: Step 6) Choose one method: Critical value or P-value test (P-value=_ Circle One: reject Ho or fail to reject Ho Step 7) Conclusion: There's mean lift ticket price is different than $105. 2 test evidence to the claim that the
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