Question

construct a 98% confidence interval to estimate the popuulation mean with x overbar =60 sigma=11

a) n = 39

b) n=41

c) n =69

a) with 98% confidence, when n =39, the population mean is between the lower limit of _ and the upper limit of _

b) with 98% confidence, when n =41, the population mean is between the lower limit of _ and the upper limit of _

c) with 98% confidence, when n =69, the population mean is between the lower limit of _ and the upper limit of _

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 60

Population standard deviation = = 11

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

(a)

Sample size = n = 39

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

= 2.326 * (11 / 39)

= 4.10

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

- E < < + E

60 - 4.10 < < 60 + 4.10

55.90 < < 64.10

With 98% confidence, when n =39, the population mean is between the lower limit of 55.90 and the upper limit of 64.10

(b)

Sample size = n = 41

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

= 2.326 * (11 / 41)

= 4.0

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

- E < < + E

60 - 4.0 < < 60 + 4.0

56.0 < < 64.0

With 98% confidence, when n =41, the population mean is between the lower limit of 56.0 and the upper limit of 64.0

(c)

Sample size = n = 69

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

= 2.326 * (11 / 69)

= 3.08

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

- E < < + E

60 - 3.08 < < 60 + 3.08

56.92 < < 63.08

With 98% confidence, when n =69, the population mean is between the lower limit of 56.92 and the upper limit of 63.08