Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals182​, x overbarequals31.3 ​hg, sequals6.5 hg. Construct a confidence interval estimate of the mean. Use a 95​% confidence level. Are these results very different from the confidence interval 30.1 hgless thanmuless than32.3 hg with only 18 sample​ values, x overbarequals31.2 ​hg, and sequals2.3 ​hg?

What is the confidence interval for the population mean μ​?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 31.3

sample standard deviation = s = 6.5

sample size = n = 182

Degrees of freedom = df = n - 1 =182-1=181

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,181 = 1.973

Margin of error = E = t/2,df * (s /n)

= 1.973* (6.5 / 182)

Margin of error = E = 1.0

The 95% confidence interval estimate of the population mean is,

- E < < + E

31.3 - 1.0 < < 31.3 + 1.0

30.3 < < 32.3 ,

(30.3,32.3) ,These results having not very much difference for 1st case results are (30.3, 32.3) and for 2nd case results are (30.1, 32.3) for the sample 18. Lower value is same for two cases and upper value having little bit difference.