Question

Need a second opinion on this Stats problem. Thanks

How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours, with a standard deviation of .65 hour. Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating or drinking each day.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1.22

sample standard deviation = s = 0.65

sample size = n = 1001

Degrees of freedom = df = n - 1 = 1000

At 95% confidence level the t is ,

t _/2,df = t_0.025,1000 = 1.962

Margin of error = E = t_/2,df * (s /n)

= 1.962 * (0.65 / 1001)

= 0.040

The 95% confidence interval estimate of the population mean is,

- E < < + E

1.22 - 0.040 < < 1.22 - 0.040

1.18 < < 1.26

(1.18 , 1.26)