Question

6.) A random sample of 15 females aged 19-50 was taken to measure their daily iron intake. Their mean daily iron intake was 16.2 mg with a standard deviation of 4.7 mg.
(a) (10 pts) Find a 95% confindence interval for the mean daily iron intake of all females aged 19-50.

(b) (5 pts) Based upon the confidence interval you found in part (a), would you conclude that the mean daily iron intake of all females aged 19-50 is 19mg? Explain your reasoning.

Answer 1

Solution:

Given:

Sample size = n = 15

Sample mean = mg

Sample standard deviation = mg

Part a) Find a 95% confidence interval for the mean daily iron intake of all females aged 19-50.

Formula:

where

t_c is t critical value for c = 95%  confidence level

Thus two tail area = 1 - c = 1 - 0.95 = 0.05

df = n - 1 =  15 - 1 = 14
Look in  t table for df = 14 and two tail area = 0.05 and find t critical value

t_c = 2.145

thus

thus

_{( Round final answer to specified number of decimal places)}

Part b) Based upon the confidence interval you found in part (a), would you conclude that the mean daily iron intake of all females aged 19-50 is 19 mg? Explain your reasoning.

Since both the limits of 95% confidence interval for the mean daily iron intake of all females aged 19-50 are less than claimed mean of 19 mg, we can not conclude that the mean daily iron intake of all females aged 19-50 is 19 mg.