Question

(3 points) Suppose that you're going to do a test of significance with null hypothesis 30 and alternative hypothesis 30 You take a sample of size 88 and compute the sample mean I assume that the population standard deviation of the variable is o 7.1 and that the sampling distribution of the sample mean is normal a if you carry a hypothesis test with significance level 05, what values of will cause us to reject the null hypothesis in favor of the alternative? null hypothesis is rejected when is smaller than or bigger than What is the power of the best to detect the particular case of the alternative hypothesis at level = 30.52 (Enter your answer as a decimal, not a percentage.) power

Answer 1

As this is a two tailed test, for 0.05 level of significance, we get from the standard normal tables,
P( -1.96 < Z < 1.96) = 0.95

Therefore the critical values here are obtained as:

$= \mu - 1.96*\frac{\sigma}{\sqrt{n}} = 30 - 1.96*\frac{7.1 }{\sqrt{88}}= 28.5165$

$= \mu + 1.96*\frac{\sigma}{\sqrt{n}} = 30 + 1.96*\frac{7.1}{\sqrt{88}} = 31.4835$

Therefore the rejection region here is given as:

Reject null hypothesis if Sample mean is less than 28.5165 or more than 31.4835

For a true mean of 30.5, the power of the test is computed as the probability of rejecting the null hypothesis given that those are the mean values of rejection above.

$P(\bar X < 28.5165) + P(\bar X > 31.4835)$

Converting it to a standard normal variable, we get here:

$P(Z < \frac{28.5165 - 30.5}{\frac{7.1}{\sqrt{88}}}) + P(\bar X >\frac{ 31.4835 - 30.5}{\frac{7.1}{\sqrt{88}}})$

$P(Z <-2.6207 ) + P(\bar X > 1.2994)$

Getting these from the standard normal tables, we get here:

Therefore 10.13% or 0.1013 is the required power of the test here.