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A simple random sample of 15 year old boys from one city is obtained in their weights are listed below use a 0.01 significance level to test the claim the these sample ways come from a population with a mean equal to 148 pounds. Assume that the standard deviation of the weight of a 15-year-old boys in the city is known to be 16.1 pounds. Use the traditional method of testing hypotheses.

A simple random sample of 15 year old boys from one city is obtained in their weights-example-1

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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.

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