Question

Supposethat z is a normally distributed variable with variance 4. You collect a sample of size n for which you get a sample average value of 3. You are asked to test the null hypothesis that the mean is smaller than 2. Find the sample size n above which you can reject the null hypothesiswith a 95% of confidence.

Answer 1

Answer:

The sample size is 4

Step-by-step explanation:

Null hypothesis: The mean is 2

Alternate hypothesis: The mean is less than 2

Mean = 3

sd = sqrt(variance) = sqrt(4) = 2

At 95% confidence level, t-value is 1.960

Assuming the lower bound of the mean is 1.04

Lower bound = mean - (t×sd/√n)

1.04 = 3 - (1.96×2/√n)

3.92/√n = 3 - 1.04

3.92/√n = 1.96

√n = 3.92/1.96

√n = 2

n = 2^2

n = 4