A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x overbar equals 120.7 and the sample standard deviation is found to be s=12.1. Construct a 99% confidence interval for the population mean. The lower bound is ________ (Round to two decimal places as needed.) The upper bound is ________ (Round to two decimal places as needed.)
Solution :
Given that,
= 120.7
s =12.1
n =40
Degrees of freedom = df = n - 1 = 40- 1 = 39
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,39 = 2.708
Margin of error = E = t/2,df * (s /n)
= 2.708 * ( 12.1/ 40)
= 5.18
The 95% confidence interval is,
- E < < + E
120.7-5.18- < < 120.7+ 5.18
lower bound 115.52
upper bound 125.88
A simple random sample of size n=40 is drawn from a population. The sample mean is...
A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x overbar equals 120.7 and the sample standard deviation is found to be s=12.1. Construct a 99% confidence interval for the population mean. The lower bound is ________ (Round to two decimal places as needed.) The upper bound is ________ (Round to two decimal places as needed.)
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