Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals193​, x overbarequals32.2 ​hg, sequals6.4 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 31.1 hgless thanmuless than33.1 hg with only 15 sample​ values, x overbarequals32.1 ​hg, and sequals2.1 ​hg?

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

Are the results between the two confidence intervals very​ different?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 32.2

sample standard deviation = s = 6.4

sample size = n = 193

Degrees of freedom = df = n - 1 = 193 - 1 = 192

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,192 = 1.653

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

= 1.653 * (6.4 / 193)

Margin of error = E = 0.8

The 90% confidence interval estimate of the population mean is,

- E < < + E

32.2 - 0.8 < < 32.2 + 0.8

( 31.4 hg < < 33.0 hg )

Yes, because the confidence interval limits are not similar