WILL MARK BRAINLIEST IF YOU GET THE CORRECT ANSWER

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 with a mean equal to 148 lb. Assume that the standard deviation of the weights of all 15-year old boys in the city is known to be 16.1 lb. Use the traditional method of testing hypotheses.

149 141 160 151 134 189 157 144 175 127 164

Choose the correct conclusion associated with this experiment.

I am pretty sure it is either B or C but I am torn on which one to choose so that is why I am only going to include two answer choices...

Edit: I think I figured it out on my own, I am going to go with B, I am pretty sure that is the correct answer

B) Do not reject H0; at the 1% significance level, there is not sufficient evidence to warrant rejection of the claim that these sample weights come from a population with a mean equal to 148 lb.

C) Do not reject H0; At the 1% significance level there is sufficient evidence to warrant rejection of the claim that these sample weights come from a population with a mean greater than 148 lb.

I'm pretty sure this is correct

Step-by-step 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.

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide