Statistics Female Eating Disorders SRS Normal Assumption

Statistics - Female Eating Disorders - SRS, Normal Assumption - Probability of a Type II Error

As you may have guessed, there is a part a and a part b, which you can find at https://www.chegg.com/homework-help/questions-and-answers/statistics-female-eating-disorders-srs-normal-assumption-hypothesis-testing-p-value-q8951481 if you need to or want to help there if it\'s not already done.

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Solution

Here test of the hypothesis is,

H₀ : µ = 18 Vs H₁ : µ > 18

n is the number of females = 10

= 1.34

We have to find the probability of type II error when µ = 19.

The test is right sided.

alpha = 0.05 (assume)

critical value is 1.645 (since the test is right tail)

Critical region is Z > 1.645

where Z is the standardised normal score.

Z = (Xbar - µ) / ( / sqrt(n))

1.645 > ( Xbar - 18) / (1.34/sqrt(10) )

Xbar - 18 > 0.6971

Xbar > 18.6971

This is the critical region or rejection region.

Xbar   18.6971 is the accceptance region.

Now we can find probability of type II error.

P(type II error) = P(fail ro reject H₀ / µ = 19)

= P(Xbar 18.6971 / µ = 19)

= P(Z (18.6971 - 19) / [1.34 / sqrt(10) ] )

= P(Z -0.3029 / 0.4237)

= P( Z -0.71 )

= 0.2388

This is the probability of type II error.

In the next part we have given that type II error = 0.2

We have to find sample size n.

P(type II error ) = P(Xbar < 18.6971 / µ = 19)

0.2 = P(Z (18.6971 - 19) / [1.34 / sqrt(n)] )

0.2 = P(Z -0.3029 / [ 1.34 / sqrt(n) ] )

0.2 = - 0.3029 / (1.34 / sqrt(n))

sqrt(n) = -0.8848

n = (-0.8848)²

= 0.7828

Approximately =1.