Question

Justin is interested in buying a digital phone. He visited 12 stores at random and recorded the price of the particular phone he wants. The sample of prices had a mean of 342.25 and a standard deviation of 9.67.

(a) What t-score should be used for a 95% confidence interval for the mean, ?, of the distribution?
t* =

(b) Calculate a 95% confidence interval for the mean price of this model of digital phone:
(Enter the smaller value in the left answer box.)
to

Answer 1

Solution :

Given that,

$\bar x$ = 342.25

s =9.67

n =12

Degrees of freedom = df = n - 1 =12 - 1 = 11

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

  $\alpha$/ 2= 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,11 = 2.201 ( using student t table)

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

= 2.201 * (9.67 / $\sqrt$ 12)

= 6.1441

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

342.25 - 6.1441 < $\mu$ < 342.25+ 6.1441

336.1059 < $\mu$ < 348.3941

(336.1059 to  348.3941 )