Question

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<

P(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

Given that,

= 50.35

= 8.9

n = 66

A ) At 98% confidence level the z is ,

  = 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

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

= 2.326* (8.9 / 66)

= 2.55

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

- E < < + E

50.35 - 2.55< < 50.35+ 2.55

47.80 < < 52.90

(47.80,52.90)