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saw5 [17]
3 years ago
6

A company produces refrigerators for beer kegs. The refrigerators are supposed to maintain a mean temperature of 43°F, ideal for

a certain type of German pilsner. The owner of the brewery does not agree with the company and claims that the mean temperature is incorrect. Identify the type II error for the test.
Mathematics
1 answer:
Deffense [45]3 years ago
3 0

Answer:

C) Fail to reject the claim that the mean temperature is equal to 43°F when it is actually different from 43°F.

Options:

A) Fail to reject the claim that the mean temperature is equal to 43°F when it is actually 43°F.

B) Reject the claim that the mean temperature is equal to 43°F when it is actually 43°F.

C) Fail to reject the claim that the mean temperature is equal to 43°F when it is actually different from 43°F.

D) Reject the claim that the mean temperature is equal to 43°F when it is actually different from 43°F.

Explanation:

The null hypothesis H0: µ=43°F (a true mean temperature maintained by refrigerator is equal to 43°)

The alternative hypothesis Ha: µ<>43 (a true mean temperature maintained by refrigerator is not equal to 43).

A type II error does not reject null hypothesis H0 when it is false.  Therefore, the type II error for the test fails to reject the claim that the mean temperature is equal to 43°F when it is actually different from 43°F.

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Answer:

z(s) is in the rejection region we reject H₀     μ₁  = μ₂  and support the claim that at CI 95 % the means of the two groups differs

Step-by-step explanation:

Sample 1:

Sise sample   n₁  =  115

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Sample 2:

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We will choose CI = 95 %   then significance level  α  = 5 %

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z(c) for 0,025 is from z-table   z(c) = 1,96

Test Hypothesis:

Null Hypothesis                                  H₀           μ₁  = μ₂

Alternative Hypothesis                      Hₐ            μ₁  ≠ μ₂

The alternative hypothesis tells us that the test is a two-tail test.

z(s)  =  ( μ₁  - μ₂ ) / √ σ₁²/n₁  + σ₂²/n₂

z(s)  = ( 169,9  -163,3 ) / √ (24,8)² /115   +  ( 25,8)²/235

z(s)  =  6,6  / √5,35 + 2,83

z(s)  =  6,6  / 2,86

z(s) = 2,30

Comparing  |z(c)|    and  |z(s)|

z(s) > z(c)

z(s) is in the rejection region we reject H₀     μ₁  = μ₂  and support the claim that at CI 95 % the means of the two groups differs

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