Question

A survey of 49 retail stores revealed that the average price of a laptop was $650 with a standard deviation of $50

a) What is the 99% confidence interval to estimate the true cost of the tablet?

b) What sample size would be needed to estimate the true average price of table with an error of + or - 5$ and 95% confidence?

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = $\bar x$ =  $650

sample standard deviation = s = $50

sample size = n = 49

Degrees of freedom = df = n - 1 = 49 - 1 = 48

At 99% confidence level

$\alpha$ = 1 - 99%  

$\alpha$ = 1 - 0.99 =0.01

$\alpha$/2 = 0.005

Z_$\alpha$/2 = t_0.005,48 = 2.682

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.682 * (50 / $\sqrt$ 49)

Margin of error = E = 19.16

The 99% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 650 ± 19.16

= ( 630.84, 669.16 )

b) margin of error = E = 5

sample size = n = [t$\alpha$/2,df* s / E]²

n = [2.682 * 50 / 5 ]²

n = 719.31

Sample size = n = 720