Question

If a random sample of 41 items produces x Overscript bar EndScripts = 127.4 and s = 21.6, what is the 98% confidence interval for μ? Assume x is normally distributed for the population. What is the point estimate?

Answer 1

Solution :

Given that,

t /2,df = 2.423

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

= 2.423 * (21.6 / 41)

Margin of error = E = 8.2

The 98% confidence interval estimate of the population mean is,

- E < < + E

127.4 - 8.2 < < 127.4 + 8.2

119.2 < < 135.6

(119.2 , 135.6)

Point estimate = sample mean = = 127.4