A random sample of size 18 taken from a normally distributed population revealed a sample mean of 100 and a sample variance of 49. The 95% confidence interval for the population mean would equal: Group of answer choices
93.56 - 103.98
97.65 – 107.35
95.52 -104.65
96.52 – 103.48
Solution :
t /2,df = 2.110
Margin of error = E = t/2,df * (s /n)
= 2.110 * (7 / 18)
Margin of error = E = 3.48
The 95% confidence interval estimate of the population mean is,
- E < < + E
100 - 3.48 < < 100 + 3.48
96.52 < < 103.48
(96.52 , 103.48)
96.52 – 103.48
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