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I have some practice homework questions. I did the first one I will go over the steps please tell me If I am doing it right.

a) As mentioned earlier, it is claimed that 70% of households in Ontario now own large-screen TVs. You would like to verify this statement for your class in mass communications. If you want your estimate to be within 5 percentage points, with 99 per cent level of confidence, how large of a sample must be acquired?

So for this i knew the targe parameter was p so the standard error is $\sqrt{\frac{pq}{n}}$ and that is equal to 0.05 and since we are finding the 99% C.I. our $Z_\alpha /2$ = 2.575

so my formula became

$0.05 = 2.575 \cdot \sqrt{\frac{(0.7)(0.3)}{n}} \Rightarrow n = 556.07 = 557$

b) You are to conduct a sample survey to determine the mean annual family income in a rural area. The question is, “How many families should be sampled?” In a pilot study of just 10 families, the standard deviation of the sample was $\$500$. The sponsor of the survey wants to use the 95 per cent confidence level. The estimate is to be within $\$100$. How many families’ should be interviewed?

For this I kinda got confused because it said how many family should be interviewed but we are given 10 families? but I was going to follow my procedure as in a but I didnt know where the n would come from because we are give the S.D. ?

or would it be this

$100 = 1.96 \cdot \frac{500}{\sqrt{n}}$ and solve for n afterwards?

c) A student conducted a study and reported that an 80% confidence interval ranged from 48 to 52. He was sure the sample standard deviation was 16 and that the sample was at least 30, but could not remember the exact number (i.e., the size of the sample). Can you help him out?

for this since we are finding the 80% CI our area is 0.9 and our critical value now would be 1.285.

ans since we know it ranged from 48 to 52 our mean is $\frac{48+52}{2} = 50$ and then that is equal to

$50 = 1.285 \cdot \frac{16}{\sqrt{n}}$ and then solve for n only confusion was is that number 50 right?

Thanks alot for your help appreciate it

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a) is correct.

b) The problem is that the formula for the minimum required sample size is $$ n=\frac{z^2\cdot p\cdot (1-p)}{d^2} $$ where $z=z_{\alpha/2}$ and $d$ is half the length of the corresponding confidence interval. So in order to determine $n$ you need to know the true value of $p$, which is what you want to estimate. Therefore, one often makes a pilot study (a much smaller test study) only to obtain knowledge of the parameter $p$ so that one can decide a proper sample size. In your case you're told that the standard deviation of the sample $\sqrt{p(1-p)}$ is $500$ and so your approach is correct.

c) I think your approach is not correct here. In your equation you have the average/mean on the left hand side, where in a) and b) you had half the length of the confidence interval $d$. In this case $d=(52-48)/2=2$ and so your minimum required sample size is $$ n=\frac{1.2815\cdot 16^2}{2^2}=105.1. $$

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