A 90% confidence interval for the mean height of a population is This result is based on a sample of size 169. Construct a 99%confidence interval.

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.

at 99%, confidence interval is [ 65.2, 67.8 ]

Step-by-step 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 ]

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