In a random sample of six mobile devices, the mean repair cost was $75.00 and the standard deviation was $13.0013.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Interpret the results.
The 95% confidence interval for the population mean μ is
Solution :
Given that,
Point estimate = sample mean =
= $75.00
sample standard deviation = s = $13.00
sample size = n = 6
Degrees of freedom = df = n - 1 = 6 - 1 = 5
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t
/2,df = t0.025,5 = 2.571
Margin of error = E = t/2,df
* (s /
n)
= 2.571 * (13.00 /
6)
Margin of error = E = 13.64
The 95% confidence interval estimate of the population mean is,
- E <
<
+ E
75.00 - 13.64 <
< 75.00 + 13.64
61.36 <
< 88.64
The 95% confidence interval for the population mean μ is (61.36 , 88.64)
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