In: Statistics and Probability

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?

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} =
t_{0.005,176} = 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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3) The brand manager for a brand of toothpaste must plan a
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Here are summary statistics for randomly selected weights of
newborn girls:
nequals=211211,
x overbarxequals=27.127.1
hg,
sequals=6.76.7
hg. Construct a confidence interval estimate of the mean. Use
a
9898%
confidence level. Are these results very different from the
confidence interval
25.725.7
hgless than<muμless than<29.329.3
hg with only
1414
sample values,
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hg?What is the confidence interval for the population mean
muμ?
nothing
hgless than<muμless than<nothing
hg (Round to one decimal place as needed.)
Enter your answ

Here are summary statistics for randomly selected weights of
newborn girls:
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x overbarxequals=30.130.1
hg,
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hg. Construct a confidence interval estimate of the mean. Use
a
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confidence level. Are these results very different from the
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27.427.4
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1414
sample values,
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hg, and
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nothing
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hg (Round to one decimal place as needed.)

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