Question

In: Statistics and Probability

Here are summary statistics for randomly selected weights of newborn​ girls: nequals177​, x overbarequals33.4 ​hg, sequals6.4...

Here are summary statistics for randomly selected weights of newborn​ girls: nequals177​, x overbarequals33.4 ​hg, sequals6.4 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 31.3 hgless thanmuless than35.1 hg with only 13 sample​ values, x overbarequals33.2 ​hg, and sequals2.2 ​hg?

Solutions

Expert Solution

Solution :

Given that,

= 33.4 ​hg

s = 6.4 hg

n = 177​

Degrees of freedom = df = n - 1 = 177 - 1 = 176

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,176 = 2.604

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

= 2.604* ( 6.4/ 177)

= 1.25

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

- E < < + E

33.4 - 1.25 < < 33.4 + 1.25

32.15 < < 34.65

32.2 < < 34.7

(32.2,34.7)  


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