A simple random sample of 16 students from our class was taken. The mean height was 66.9 inches with a standard deviation of 2.8 inches. You will be asked some questions about confidence intervals for the actual mean height of all the students in this class.
Interpret the 99% confidence interval.
)solution
Given that,
= 66.9
s =2.8
n = 16
Degrees of freedom = df = n - 1 = 16- 1 = 15
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005,15= 2.947 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.947 * ( 2.8/ 16) = 2.06
The 99% confidence interval is,
- E < < + E
66.9 - 2.06 < < 66.9+ 2.06
64.84 < < 68.96
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