Justin is interested in buying a digital phone. He visited 9 stores at random and recorded...

Justin is interested in buying a digital phone. He visited 9 stores at random and recorded the price of the particular phone he wants. The sample of prices had a mean of 337.63 and a standard deviation of 30.04.

(a) What t-score should be used as the multiplier 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 ___

Solutions

Expert Solution

Solution :

Given that,

= 337.63

s =30.04

(A)

n = Degrees of freedom = df = n - 1 = 9- 1 =8

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,8 = 2.306

(B)

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

= 2.306* ( 30.04/ $\sqrt$ 9)

= 23.09

The 95% confidence interval is,

- E < < + E

337.63 - 23.09 < < + 337.63+23.09

314.54 < < 360.73

314.54 to 360.73

milcah answered 6 months ago

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