Question

A random sample of n measurements was selected from a population with unknown mean μ and standard deviation σ=35 for each of the situations in parts a through d. Calculate a 95​% confidence interval for muμ

for each of these situations.

a. n=75​, x overbarx=22

b. n=150​,x overbarxequals=110

c. n=90​, x overbarxequals=18

d. n=90​,x overbarxequals=4.69

e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through​ d? Explain.

Answer 1

Solution :

Given that,

Population standard deviation = = 35

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

(a)

Sample size = n = 75

Point estimate = sample mean = = 22

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

= 1.96 * (22 / 75)

= 4.98

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

- E < < + E

22 - 4.98 < < 22 + 4.98

17.02 < < 26.98

(17.02 , 26.98 )

(b)

Sample size = n = 150

Point estimate = sample mean = = 110

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

= 1.96 * (22 / 150)

= 3.52

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

- E < < + E

110 - 3.52 < < 110 + 3.52

106.48 < < 113.52

(106.48 , 113.52)

(c)

Sample size = n = 90

Point estimate = sample mean = = 18

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

= 1.96 * (22 / 90)

= 4.55

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

- E < < + E

18 - 4.55 < < 18 + 4.55

13.45 < < 22.55

(13.45 , 22.55 )

(d)

Sample size = n = 90

Point estimate = sample mean = = 4.69

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

= 1.96 * (22 / 90)

= 4.55

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

- E < < + E

4.69 - 4.55 < < 4.69 + 4.55

0.14< < 9.24

(0.14 , 9.24)

(e)

Yes because sigma is known and sample size > 30 .