Question

The heights (measured in inches) of men aged 20 to 29 follow approximately the normal distribution with a standard deviation of 2.8 inches. A random sample of 16 American men aged 20 to 29 yields a sample mean of 70.8 inches. Find a 90% confidence interval for the average height of all American men between the ages of 20 and 29. (Please give responses to 3 decimal places)

Answer 1

I 90%(μ)= [69.648;71.951]

Explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqr(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

Z(5%)= is the critical value of the standardized normal distribution. In this case is 1.645

σ= standard deviation. In this case 2.8 inches

mean= 70.8 inches

n= number of observations

Then, the confidence interval (90%):

I 90%(μ)= 70.8+- [1.645*(2.8/sqr(16))

I 90%(μ)= 70.8+- [1.1515)

I 90%(μ)= [70.8-1.1515;70.8+1.1515]

I 90%(μ)= [69.6485;71.9515]

I 90%(μ)= [69.648;71.951]