Answer:
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Explanation:
What you are testing is a distribution of means, because you have a sample of eleven individuals (N=11), a population mean, and the population variance. Basically, the problem is asking: what is the likelihood that the sample mean (153.45 lbs) could have been obtained from a population where (M=149) if the null hypothesis is true?
p1: 15 year old boys with a mean weight of 153.45 lbs from City "Unknown".
p2: 15 year old boys with a mean weight of 149.00 lbs from the population.
H1: The sample of boys from City "Unknown" was not drawn from a population where the average weight of 15 year old boys = 149lbs.
H0: The sample of boys from City "Unknown" was drawn from a population where the average weight of 15 year old boys = 149lbs.
Mean = 149 [μM = μ = 149]
Variance = 16.2squared/11 = 262.44/11 = 23.86 [σM2 = σ2/N]
Standard Deviation = 4.88 [σM = √σM2 = √(σ2/N)]
Shape = normal
Using .01 level of significance for a two-tailed test, the cutoff sample score is +/- 2.575
Z = (M-µ) / σM = (153.45 - 149) / 4.88 = .91
.91 < 2.575
σM or the Standard Error of the Mean is 4.88 so...
for 99% confidence interval,
lower limit = 153.45 + (-2.575)(4.88) = 140.88
upper limit = 153.45 + (2.575)(4.88) = 166.02
the 99% confidence interval = 140.88 — 166.02 (lbs)
the population mean (µ=149.00) is included in the interval, which confirms results of Z test.
Conclusion: retain null, there is <.01 probability that the sample from City "Unknown" was drawn from a population where the mean weight is NOT = 149.00lbs.