This question is incomplete, the complete question is;
A 90% confidence interval for the mean height of a population is 65.7 < μ < 67.3
This result is based on a sample of size 169.
Construct a 99%confidence interval.
Answer: at 99%, confidence interval is [ 65.2, 67.8 ]
Explanation:
Given that;
confidence level = 90%
∝ = 1 - 90% = 1 - 0.9 = 0.1
the mean height of a population is 65.7 < μ < 67.3
so
x" - t_∝/2, n-1{standard Error] = 65.7 ----let this be equation 1
x" + t_∝/2, n-1{standard Error] = 67.3 -----let this be equation 2
where t∝/2, n-1 = t_0.1/2, 169 - 1 = t_0.05, 168 = 1.654 (t table)
now we add equation one and equation two
2x" = 65.7 + 67.3 = 133
x" = 66.5
next we subtract equation one from equation two
(x" + t_∝/2, n-1{standard Error]) - (x" - t_∝/2, n-1{standard Error])
= 67.3 - 65.7 = 1.6
so
2 × t_∝/2,n-1 × standard Error = 1.6
standard Error = 1.6 / (2 × t_∝/2,n-1)
we substitute
Standard Error = 1.6 / ( 2 × 1.654 ) = 1.6 / 3.308 = 0.48367
Now 99% confidence interval for μ is given as;
∝ = 1 - 0.99 = 0.01
x" ± t_∝/2, n-1{standard Error]
(x" ± t_0.01/2, 169-1 [standard Error])
where t_0.005, 168 = 2.6054 (t table)
now for Lower Bound: (x" - t_0.005, 168 [standard Error])
Lower Bound : ( 66.5 - (2.6054 × 0.48367))
= 66.5 - 1.26015 = 65.2398
Upper Bound : ( 66.5 + (2.6054 × 0.48367))
= 66.5 + 1.26015 = 67.7602
Therefore at 99%, confidence interval is [ 65.2, 67.8 ]