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A simple random sample of birth weights of 30 girls has a standard deviation of 829.5 hg. Use a 0.01 significance level to te
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Answer #1

Answer:

a)

Option A

b)

Option C

c)

alpha = 0.01

degree of freedom = n-1 = 30 - 1 = 29

\chi ^{2}(df, 1-alpha/2) = 13.121

\chi ^{2}(df , alpha/2) = 52.336

d)

test statistic = (n-1)s^2/\sigma^2

substitute values

= (30-1)*829.5^2 / 660.2^2

= 45.78

e)

P value = 0.024674

= 0.025

Here we observe that, p value > alpha, so we fail to reject Ho.

So there is no sufficient evidence to support the claim.

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