Question

Here are summary statistics for randomly selected weights of newborn​ girls: n=239, x=33.2hg, s=6.6 hg. Construct a confidence interval estimate of the mean. Use a 99% confidence level. Are these results very different from the confidence interval 31.3hg <μ<34.5 hg with only 15 sample​ values, x=32.9 hg, and s=2.1 hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 33.2

sample standard deviation = s = 6.6

sample size = n = 239

Degrees of freedom = df = n - 1 = 239 - 1 = 238

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,238 = 2.5966

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

= 2.5966 * (6.6/ 239)

Margin of error = E = 1.11

The 99% confidence interval estimate of the population mean is,

- E < < + E

33.2 - 1.11 < < 33.2 + 1.11

32.09 < < 34.31

(32.09 , 34.31) , These results not very different from first case results are (31.3 , 34.5 ) for samples 15 and for this case results are (32.09 , 34.31)