<p>So my teacher gave me the 2007 ap statistics answers but i think some of the answers might be wrong. Specifically for #5, does anyone know the right answer? It goes as follows:</p>
<p>A safety group claims that the mean speed of drivers on a highway exceeds the posted speed limit of 65 miles per hour. To investigate the safety group’s claim, which of the following statements is appropriate?</p>
<p>A. The null hypothesis is that the mean speed of drivers on this highway is less than 65 mph
B. The null hypothesis is that the mean speed of drivers on this highway is greater than 65 mph
C. the alternative hypothesis is that the mean speed of drivers on this highway is greater than 65 miles per hour
D. the alternative hypothesis is that the mean speed of drivers on this highway is less than 65 miles per hour
E. the alternative hypothesis is that the mean speed of drivers on this highway is greater than equal to 65 mpg</p>
<p>my answer was C, but my teachers key says E. any idea why?</p>
<p>I just looked at the problem you have posted and I got C. The alternative hypothesis would be in support of their claim that the mean speed EXCEEDS 65 mph. It could not be equal to 65 as well if it is to exceed 65. So yes, the answer has to be C.</p>
<p>Can anybody help me with this problem? (2002 MC)</p>
<p>A quality control inspector must verify whether a machine that
packages snack foods is working correctly. The inspector will randomly
select a sample of packages and weigh the amount of snack food in each.
Assume that the weights of food in packages filled by this machine have
a standard deviation of 0.30 ounces. An estimate of the mean amount of
snack food in each package must be reported with 99.6 percent confidence
and a margin of error of no more than .12 ounce. What would be the
minimum sample size for the number of packages the inspector must
select?</p>
<p>The z critical value is found at the bottom of the t-distribution critical values chart, at the infinity value right above 99.5. 2.88 is an approximate value between the value at 99.5% and 99.8%, which is the range where 99.6 falls. Standard deviation is given as .3, so plug that value into the formula. Margin of error is given as .12, which you plug into E in the formula.</p>
<p>n= ((2.88^2)(.30^2))/.12^2</p>
<p>From this calculation, you get 51.84, and you have to round up to 52, since you cannot have a partial package, so the answer is D.</p>
<p>okay one more problem from the 2007 someone please help!</p>
<p>"Julie generates a sample of 20 random integers between 0 and 9 inclusive. She records the number of 6’s in the sample. She repeats this process 99 more times, recording the number of 6’s in each sample. What kind of distribution has she simulated?</p>
<p>C. the binomial dist. with n=20 and p =.1
D. he binomial dist. with n=100 and p =.1
E. he binomial dist. with n=20 and p =.6
"</p>
<p>Again i got D, but the answer thing said C. any idea why?</p>
<p>^^ look at my post, I edited it to explain where the z-critical value comes from. It is an approximation of 99.6%, since it is between 99.5% and 99.8%.</p>
<p>Recognizing is a binomial distribution allows you to narrow it down to C, D and E. There are 10 numbers with equal chance of being chosen; p = .1. The sample size is 20. Although she does the process 100 times, they are each of size n = 20.</p>
<p>You could also use InvNorm in the way it was just shown. I have more practice in just looking things up on the chart, so use whatever your preference is.</p>
<ol>
<li>When performing a test of significance about a population mean, a t-distribution, instead of a normal
distribution, is often utilized. Which of the following is the most appropriate explanation for this?
(A)The sample size is not large enough to assume that the population distribution is normal.
(B)The sample does not follow a normal distribution.
(C)There is an increase in the variability of the test statistic due to estimation of the population
standard deviation.
(D)The sample standard deviation is unknown.
(E)The population standard deviation is too large.</li>
</ol>
<p>Ahhh that makes sense. Could you help me with this one also?</p>
<p>A simple random sample produces a sample mean, x(bar), of 15. A 95 percent confidence interval for the corresponding population mean is 15 +/- 3. Which of the following statements must be true?</p>
<p>(A) Ninety-five percent of the population measurements fall between 12 and 18.
(B) Ninety-Five of the sample measurements fall between 12 and 18.
(C) If 100 samples were taken, 95 of the sample means would fall between 12 and 18.
(D) P(12<= x(bar) <= 18) = 0.95
(E) If mu = 19, this x(bar) of 15 would be unlikely to occur.</p>
<p>Answer’s E, but I absolutely don’t get why…</p>
<p>It is E because the mean of the actual population is 19 and there is a 95% chance that the mean will fall between 16 and 22. 15 is outside this range, so is very unlikely. Although, I do not know why it says mu.</p>
<p>@chocolatecricket - The answer is D. You use the t-distribution if the population standard deviation is unknown.</p>
<p>@zzxjoanw3 The margin of error is 3, so 19-3 would be 16, and 19+3 is 22. You are 95% confident that the true mean is between 16 and 22, so it is unlikely that any value out of that range will be the true mean. 15 is out of that range, so the answer is E.</p>
<p>^You beat me to it, but this just confirms the answer. And it says mu because that is the symbol for population mean. It is the greek-letter pronounce "mu’.</p>
<p>15 is D because it is the only answer choice which refers to the confidence interval. The other choices allude to the endpoints of the interval, which cannot be determined for sure.</p>
<p>23 uses the formula moe = t* multiplied by sqrt (p (1-p) /n). We arbitrarily assign the value of p = 0.5 when it isn’t given.</p>
<p>t* = 1.645, p = 0.5, 1-p = 0.5, n is our unknown, moe = 0.05.</p>