Question
In: Math
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 = t0.025,8 = 2.306
(B)
Margin of error = E = t/2,df
* (s /
= 2.306* ( 30.04/
= 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 1 year agoRelated Solutions
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