Question

. Suppose a random sample of 25 is taken from a population that follows a normal distribution with unknown mean and a known variance of 144. Provide the null and alternative hypotheses necessary to determine if there is evidence that the mean of the population is greater than 100. Using the sample mean, Y, as the test statistic and a rejection region > k}, find the value of k so that α = 0.15. of the form - Using the sample mean, Y, as the test statistic and the rejection region > k}, find the Type II error rate when the true mean of the population is known to be 110 (α = 0.15 i.e., use the same value of k obtained in the previous part). To decrease the Type II error rate we can increase the size of the sample taken. How large would the sample need to be so that α 0.15 and 0.05 when using the test statistic Y, rejection region {y > kh and true value of the mean to be 110?

Answer 1

Hypotheses are:

Ho : μ 100 . Ha : μ > 100

Test is right tailed. The critical value of sample mean for which we will reject the null hypothesis using excel function "=NORMSINV(1-0.15)" is 1.036.

So critical value of sample mean for which we will reject the null hypothesis is

$z=\frac{\bar{Y}-\mu}{\sigma/\sqrt{n}}$

1.036-Y-100 12/v25 1.036 =

102.4864

The rejection region is :

y> 102.4864)

The z-score for
102.4864
and
μ = 1 10
is

102.4864- 110 =-3.13 12//25

The type II error is

3-P(z <-3.13) = 0.0009

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Here we have σ-12,CL-0.85, β-0.05 Level of significance: α-0.15 Power-0.95 Test is two tailed so critical values are: 1-a1.036 'l-β-1.64 δ-14-61-10 Delta: Required sample size is: n-1.036+1.64) 12)/10)A2 10.31180544 Hence, the required sample size is 11. Excel function for critical value: -NORMSINV(1-(0.15/2)) -NORMSINV(1-0.05) 21-a 21-β