Software can generate samples from (almost) exactly Normal distributions. Here i

Question

Software can generate samples from (almost) exactly Normal distributions.
Here is a random sample of size 5 from the Normal distribution with mean 10 and standard deviation 2:

6.47 7.51 10.10 13.63 9.91

These data match the conditions for a z test better than real data will: the population is very close to Normal and has known standard deviation s = 2, and the population mean is µ = 10.
Test the hypotheses

H0 : µ = 8
Ha : µ ? 8
In Exercise 15.41 (given above), a sample from a Normal population with mean µ = 10 and standard deviation s = 2 failed to reject the null hypothesis H0 : µ = 8 at the a = 0.05 significance level.
Enter the information from this example into the Power of a Test applet.
(Don't forget that the alternative hypothesis is two-sided.)
What is the power of the test against the alternative µ = 10?

Explanation / Answer

Number of cases 5 To find the mean, add all of the observations and divide by 5 Mean 9.524 H0: µ = 8 HA: µ ? 8 Sample mean = 9.524 Standard deviation = 2 Standard error of mean = s / v n Standard error of mean = 2 / v 5 SE = 2/2.2361 Standard error of mean 0.8944 z = (xbar- µ ) / SE z = (9.524-8) / 0.8944 z = 1.7039 alpha =0.05 P( |z| > 1.7039) = 0.0892 (2-tailed test p-value) Conclusion : Do not reject the null hypothesis. To find the power of the test, let us first find the probability of type II error and use the relation: Power = 1-P(type ii error) The significance level is 0.05 and the test is two-tailed. We reject H0 when (xbar-8) / s / v n > 1.96 or (xbar-8) / s / v n > 1.96 < -1.96 s / v n = 0.8944 Reject H0 when (xbar-8)/0.8944 > 1.96 or < -1.96 xbar >8+ (1.96)(0.8944) or xbar < 8 - (1.96)(0.8944) Reject when xbar > 9.753 or xbar < 6.247 P(type II error) = P( do not reject H0/ µ = 10) We do not reject H0 when 6.247

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