Question

(please make answer clear than you! :)

A sample of size =n66 is drawn from a normal population whose standard deviation is =σ8.9. The sample mean is =x50.35.

Part 1 of 2 (a) Construct a 98% confidence interval for μ. Round the answer to at least two decimal places. A 98% confidence interval for the mean is <μ <

(b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain. The confidence interval constructed in part (a) ▼(Choose one)( would OR would not)  be valid since the sample size ▼(Choose one)( is OR is not ) large

Answer 1

Solution :



Point estimate = sample mean = = 50.35

Population standard deviation = = 8.9

Sample size = n = 66

a) At 98% confidence level

= 1-0.98% =1-0.98 =0.02

/2 =0.02/ 2= 0.01

Z/2 = Z0.01 = 2.326

Z/2 = 2.326
Margin of error = E = Z/2 * ( /n)

= 2.326 * ( 8.9 / 66 )

= 2.548

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

- E < < + E

50.35 - 2.548 <   < 50.35+ 2.548

47.80 <   < 52.90

(47.80 ,52.90 )

b) The confidence interval constructed in part (a) would be valid the sample size is large.