153.45 is the mean for those eleven weights given.
Hi,
This is my first post to MHF. I'm taking an introductory stats course at college, and I have a problem from a sample exam that's giving me some difficulty. I was hoping that someone might be able to offer some assistance. Thank you.
Problem: A simple random sample of 15-year-old boys from one city is obtained and their weights (in pounds) are listed below. Use a 0.01 significance level to test the claim that these sample weights come from a population w/ a mean equal to 149 lbs. Assume that the standard deviation of the weights of all 15-year-old boys in the city is known to be 16.2 lbs. Identify the null hypothesis, alternative hypothesis, test statistic, critical value(s), and conclusion.
147, 138, 162, 151, 134, 189, 157, 144, 175, 127, 164
So, as I understand this question, this is what I have so far:
n = 11
x̄ = 153.45
s = 16.2
α = 0.01
Ho: p = 149 (original claim)
H1: p ≠ 149 (counter claim)
Za/2 = -2.575 and 2.575
Test Statistic: Z = p̂p / √pq / n
I understand that p̂ = x/n, but I don't understand what x is in this problem. Any help would be greatly appreciated. Thank you!
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.
Thank you so much! I see where I was looking at this in the wrong light. Clearly, I need to work on understanding what the problem is asking for. That seems to be my weakest area. I find that I'm often unclear on what is being of me in the initial problem, and therefore I'm looking at the wrong equation(s) to solve the problem. Again, thank you!