Question

The amount of time that a certain cell phone will keep a charge is known to be normally distributed with a standard deviation σ=15 hours. A sample of 45 cell phones has a mean time of 145 hours. Let μ represent the population mean time that a cell phone will keep a charge.

a. What is the point estimate of μ ?

b. What is the standard error of the point estimate? Round to three decimal places.

c. Suppose that a 90% confidence interval is to be constructed for the mean time. Find the margin of error for this confidence interval (use three decimal places).

e. Construct the 90% confidence interval and interpret your result:

f. What sample size would be necessary so that a 98% confidence interval will have a margin of error of 2 hours?

Answer 1

Solution

Given that,

a ) = 145

= 15

n = 45

The standard error = (/n)

= (15 / 45 )

= 2.24

c ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * (15 / 45 )

= 3.68

e )At 90% confidence interval estimate of the population mean is,

- E < < + E

145 - 3.68 < < 145 + 3.68

141.32< < 148.68

(141.32 , 148.68)

f ) Given that,

standard deviation = = 15

margin of error = E = 2

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

Sample size = n = ((Z/2 * ) / E)2

= ((2.326 * 15 ) / 2 )2

= 304

Sample size = 304