Question

In: Statistics and Probability

3. You have data for a sample of 100 exams taken by 16 year old students....

3. You have data for a sample of 100 exams taken by 16 year old students. You are told that the grades follow a normal distribution, with a variance of 25. You want to use this data to test the null hypothesis that μ=10, vs Ha μ>10. a- What is the power of the test if the true population mean is μ=10.5 and you want a 5% significance level? b- What is the power of the test if the true population mean is μ=11 and you want a 5% significance level? c- what would the sample size need to be to be achieve a power of 80% when testing that μ>10 and μ=10.5 and you want to have a probability of Type I error of 5%

Solutions

Expert Solution

Question 3

Here sample size = n = 100

Here hypothesis are

H0 : μ=10, vs

Ha : μ>10

Here variance = = 25

standard error = sqrt (/n) = sqrt(25/100) = 0.5

Now, significance level = 0.05

Critical value = NORMSINV(0.05) = 1.645

We would reject the null hypothesis if > 10 + 1.645 * 0.5

> 10.8224

so here as we come to know,  true population mean is μ=10.5

so here as we know Power is probability of rejecting the null hypothesis when it is false; in other words, it is the probability of avoiding a type II error.

P( > 10.8224) where ~ N(10.5; 0.5)

z = (10.8224 - 10.5)/0.5 = 0.645

P( > 10.8224) = 1 - P(Z < 0.645) = 1- 0.7405 = 0.2595

(b)

True population mean is μ =11

so here as we know Power is probability of rejecting the null hypothesis when it is false; in other words, it is the probability of avoiding a type II error.

P( > 10.8224) where ~ N(11; 0.5)

z = (10.8224 - 11)/0.5 = -0.355

P( > 10.8224) = 1 - P(Z < -0.355) = 1- 0.3612= 0.6388

(c) Here let say the sample size n

standard error = sqrt(25/n) = 5/

Now, significance level = 0.05

Critical value = NORMSINV(0.05) = 1.645

We would reject the null hypothesis if > 10 + 1.645 * 5/

so here as we come to know,  true population mean is μ=10.5

Now power is 80%

0.80 = P( > 10 + 1.645 * 5/) where ~ N(10.5; 0.5)

P( <  10 + 1.645 * 5/) = 1- 0.80 = 0.20

Z value = NORMSINV(0.20) = -0.8416

-0.8416 = ( 10 + 1.645 * 5/ - 10.5)/(5/)

-0.8416 * 5/ = 1.645 * 5/ - 0.5

n = 618.26 or 619


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