Iposted questions but nobody is answering them gabriella owned so many cds that she gave 64 of them away. after donating the cd she still has 147 left. write and solve an equations to find the number of cd c gabriella had originally.
c+64=147; 83 cds
c + 147 = 64; 83 cds
c+ 147 = 64; 211 cds
c-64=147; 211 cds

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.