Suppose that we have a random sample of size n = 100 from a normal distribution

Question

Suppose that we have a random sample of size n = 100 from a normal distribution with mean and variance 2 = 16.

From the random sample, we wish to test the following hypothesis:

H0 : 80 versus Ha : > 80

1. Find the value of c in the rejection region of the form " X > c " when the significant level is = 0.05.

2. Calculate the power of the above test at = 79.
3. Under the significance level = 0.05, if we want to make the power of the above test at = 81 to be 0.9 (i.e. the probability of type 2 error at = 81 is equal to 0.1), what is the minimum sample size to achieve this goal?

Explanation / Answer

Suppose that we have a random sample of size n = 100 from a normal distribution with mean and variance 2 = 16.

From the random sample, we wish to test the following hypothesis:

H0 : 80 versus Ha : > 80

1. Find the value of c in the rejection region of the form " X > c " when the significant level is = 0.05.

Answer : so the test is the one directional test here so rejection region here is the region where we will reject the null hypothesis so

Pr (X >c; H ; /n) = 0.05

where c = critical value to reject the null ; H = HYpothesis population mean n = sample size

so from Z - table z value for the given probability = 1.645

standard error of the sampling se0 = /n = 4/10 = 0.4

so (c-H / ( /n) = 1.645

c = 1.645 * (4/ 100) + 80 = 1.645 * 0.4 + 80 = 80.658

so rejection region c > 80.658

2. Calculate the power of the above test at = 79.

firrst we will calculate that means the probability of type II error that means failure to reject the null hypothesis even if it is wrong.

so Pr(Type II error) = Pr( x < 80.658; 79; 0.4)

Z = (80.658 - 79)/ 0.4 = 4.145

Pr (Type II error) = 0.9999

so Power = 1 - = 0.0001 or say 0

3. Under the significance level = 0.05, if we want to make the power of the above test at = 81 to be 0.9 (i.e. the probability of type 2 error at = 81 is equal to 0.1), what is the minimum sample size to achieve this goal?

Let say sample size will be n

so Pr( Type II error) = Pr (x < H + Zcr * (/n); t , se)

Here H = 80

Zcr = 1.645 for alpha = 0.05 (one tailed) ; = 4

n = we have to calculate and se = 4/n

Here type II error = 0.1

so for this probability to be 0.1 the Z - value = -1.28

so Z value = (80 + Zcr * (/n)  - 81)/ (se) = -1.28

(80 + 1.645 * (4/n) - 81) = -1.28 (4/n)

2.925  (4/n) = 1

n = 2.925 * 4 = 11.7

n = 136.89 equals to 137

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