Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals153​, x overbarequals31.5 ​hg, sequals7.1 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 30.4 hgless thanmuless than32.8 hg with only 15 sample​ values, x overbarequals31.6 ​hg, and sequals2.7 ​hg?

Answer 1

yes it is little different from the confidence interval (30.4 ≤μ≤ 32.8) changes statistics

90% confidence interval estimate of the mean is

(30.1048 , 33.0952)

Explanation:

Step(I):-

Given sample size 'n' = 153

Given mean of the sample x⁻ = 31.5

Sample standard deviation 'S' = 7.1 h g

90% confidence interval estimate of the mean is determined by

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ) )`

Degrees of freedom

ν = n-1 = 153-1 =152

t₀.₀₅ =1.9757

Step(ii)

90% confidence interval estimate of the mean is

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ) )`

`(31.5 - 1.9757 } (7.1)/(√(153) ) , 31.5 + 1.9757 (7.1)/(√(153) ) )`

( 31.5 - 1.1340 , 31.5 + 1.1340)

(30.366 , 32.634)

90% confidence interval estimate of the mean is

(30.4 , 32.6)

b)

Given sample size 'n' = 15

Given mean of the sample x⁻ = 31.6

Sample standard deviation 'S' = 2.7 h g

90% confidence interval estimate of the mean is determined by

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ) )`

Degrees of freedom

ν = n-1 = 15-1 =14

t₀.₀₅ =2.1448

90% confidence interval estimate of the mean is

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ) )`

`(31.6 - 2.1448 } (2.7)/(√(15) ) , 31.6 + 2.1448 (2.7)/(√(15) ) )`

( 31.6 - 1.4952 , 31.6 + 1.4952)

(30.1048 , 33.0952)

Conclusion:-

yes it is little different from the confidence interval (30.4 ≤μ≤32.8)