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Test the null hyopthesis at the .01 level of significance that 
the disrubution of blood types for college students complies 
with the proportins described in the blood bank bulletin, 
namely, .44 for O, .41 for A, .10 for B and .05 for AB. Now 
however, assume that the results are available for a random 
sample of only 60 students. The results are as follows 27 
for O, 26 for A, $ for B and 3 for AB.

Note: The expcexted frequency for AB (.05)(60) =3, is less 
than 5, the smallest permissible frequency. Create a sufficently 
large expected frequency by combing B ans AB.

I understand Level of singificance would be x2 at a critical value 
of 9.21 with df=2, and I get the Ho Po= .44, Pa=.41 and 
Pb=A+AB=.15

So, x2 would look like (27-?/?)2+ (26-?/?)2+(7-?/?)2=
I also know that each set of ? would have different values, but 
not sure of how to get those values.

The probability of an individual person contracting the H1N1 virus is .12.  If we select four people at random, what is:
a) the probability that 2 of them contracts it?
b) the probability that at least one of them contracts it?

I have to use the values of the linear correlation 
coefficient to calculate the coefficient of determination 
and explain the variation of the data about the 
regression line and the unexplained variation

Can you tell me what formula is used to determine 
the values of a linear correlation coefficient? How 
do I calculate the coefficient of determination?

r=  0.350
r= -0.275
r= -0.891
r=  0.964

Listed below are measures of pain intensity before 
and after using the proprietary drug Duragesic.  The 
data are listed in order by row, and corresponding 
measures are from the same subject before and 
after the treatment.  For example, the first subject 
had a meausre of 1.2 before the treatment and a 
measure of 0.4 after the treatment.  Each pair of 
measurments is from the one subject, and the 
intensity of pain was measured using the standard 
visual analog score.
 
Pain Intensity Before Duragesic Treatment
 
1.2  1.3  1.5  1.6  8.0  3.4  3.5  2.8  2.6  2.2  
3.0  7.1  2.3  2.1  3.4  6.4  5.0  4.2  2.8  3.9
5.2  6.9  6.9  5.0  5.5  6.0  5.5  8.6  9.4  10.0
7.6
 
Pain Intensity After Duragestic Treatment
 
0.4  1.4  1.8  2.9  6.0  1.4  0.7  3.9  0.9  1.8  
0.9  9.3  8.0  6.8  2.3  0.4  0.7  1.2  4.5  2.0
1.6  2.0  2.0  6.8  6.6  4.1  4.6  2.9  5.4  4.8
4.1
 
Analyzing the Results
 
1) Use the given data to construct a scatter plot, then 
use the methods of section 10-2 to test for linear 
correlation between the pain intensity before the 
treatment and after the treatment.  If there is a 
significant linear correlation, does it follow that the 
drug treatment is effective?
 
2)  Use the given data to find the equation of the 
regression line.  Let the response (Y) variable be the 
pain intensity after the treatment.  What would be the 
equation of the regression line for a treatment having 
absolutly no effect?
 
3)  The methods of section 9-3 can be used to test the 
claim that two populations have the same mean.  Identify 
the specific claim that the treatment is effective, then use 
the methods of section 9-3 to test the claim.  The methods 
of section 9-3 are based on the requirments that these 
samples are independent.  Are they independent in this case?
 
4)  The methods of section 9-4 can be used to test a claim 
about matched data.  Identify the specific claim that the 
treatment is effective, then use the methods of section 
9-4 to test the claim.
 
5)  Which of the preceeding results is best for determining 
weather the drug treatment is effective in reducing pain?  
Which of the preceding reults is least effective in determining 
whether the drug treatment is effective in reducing pain?  Based 
on the preceding results, does the drug appear to be effective?

Assume IQ has a Normal Distribution with a mean of 130 and a standard deviation of 20. Find:
a) The probability of being above 135
b) The IQ which represents the 90th percentile
c) The probability that the average of 25 people will be below 128

For "a", this is straight forward:  P(X > 135) = P(Z > [135 - 130]/20) = P(Z > 0.25) = 0.599
For "b", we work backwards.  We are given the probability and want the IQ, so, we want to find the value "?" such that P(X < ?) = .9, or
  P(Z < [? - 130]/20) = .9.  From the normal table we know that a z-score of 1.28 gives that probability (that is P(Z < 1.28) = .9), so we 
   want the quantity [? - 130]/20 to equal 1.28.  Set equal and solve:  ? = 1.28*20 + 130 = 155.6
For "c", this is now a new random variable:  Xbar. It has a Normal distribution with a mean of 135 and a std error of 20/sqrt{25}, or 20/5 = 4.
  So P(Xbar < 128) = P(Z < [128-130]/4) = P(Z < -.5) = 0.31

Amounts of Coke:  Assume that cans of coke are filled so that the
actual amounts have a mean of 12.00 oz and a standard deviation of 0.11
oz

a. Find the probability that a sample of 36 cans will have a mean amount of at least 12.19 oz, as in Data Set in Appendix B

b. Based on the result from part(a), is it reasonable to believe
that the cans are actually filled with a mean of 12.00 oz?  If the mean
is not 12.00 oz, are the consumers being cheated?

The amount in a SINGLE coke can has a mean of 12 and a std dev of 0.11.  
The AVERAGE amount in a SAMPLE of coke cans has a normal distribution 
(why?) with a mean of 12 and a std dev of:  0.11/6  (let me know if you 
don't know why - this part is key to the problem).  So the question is 
find the probability that the average fill amount is 12.19 or higher, or:

P(x-bar >= 12.19).  But x-bar is the sample mean, so it has the 
distribution we stated above (Normal, mean=12, std dev = 0.11/6), so we 
solve this the way we always do for random variables with a Normal 
distribution:

P(x-bar >= 12.19) = P(Z >=   [12.19 - 12.00] / [0.11/6]  ) = P(Z >= 
.19/.0183) = P(Z >= 10.36).

Normally we would then look that number up in a table.  However, we know 
that 99% of the probability in a standard Normal distribution is between 
-3.5 and 3.5, so 10.36 is VERY unlikely, so the probability is going to 
be VERY small.

For part b, what they are trying to get you to see is that 10.36 is VERY 
unlikely, so the fact that there's a dataset that produces that result 
means that the true mean is probably not 12.00, because if it were, you 
would NOT get a sample mean of 12.19. It's just too unlikely.  However, 
consumers are probably not being cheated because it's likely the mean is 
higher than 12.00 - that's the only way a sample mean of 12.19 could happen.

The competing hypotheses for the distribution of time are:

Ho: The population of values is represented by a rectangle with base 3 to 7 and height 1/4
H1: The population of values is represented by a rectangle with base 0 to 4 and height 1/4

Reject Ho if service time is 3.2 or more extreme. Shade the region the corresponds to the significance level and clearly label the region.

I know how to draw the picture but I do not know what way to shade for Ho.

HOWEVER, and this is really important: how far to the right of 3.2 do you shade? Do you shade all
the way to 7? NO!!! If H1 is really true, then the distribution is really the one stated in H1, and the only possible values are ones in that distribution. So you shade from 3.2 up to...what?

Question:

Type: Hypothesis Test, 1-tailed vs. two-tailed

Highway safety engineers test new road signs, hoping that increased reflectivity will make them more visible to drivers. Volunteers drive through a test course with several of the new and old style signs and rate which kind shows up best.

a) Is this a one-tailed or two-tailed test? Why?
b) In this context, what would a Type I error be?
c) In this context, what would a Type II error be?
d) In this context, describe what is mean by the power of the test
e) If the hypothesis is tested at the 1% level of significance instead of 5%, how will this affect the power of the test?

Answer:

a) we use two-tailed test when we don't have any pre-existing ideas of what the outcome will be. That's not the case here, is it? We're assuming that the new signs won't be any worse than the old ones - they'll either be the same or they'll be better, right?

b and c) A Type II error is failing to reject a false null hypothesis, right? So that would be: the new signs are truly better, but the test doesn't demonstrate it. Can you answer the one about a Type I error now?

d) Power is 1 minus the probability of a Type II error. So it's the probability of rejecting a false null hypothesis, so it's the probability of what in terms of the old signs and new signs?

e) This is a good question, but difficult. If we move the level of significance from 5% to 1%, we are lowering the probability of a type I error. That makes it more difficult to reject the Null Hypothesis, even if it really is false, therefore INCREASING our probability of making a type II error. If the probability of a type II error goes UP, then the power goes DOWN.

Question:

Type: Normal Distribution, Percentile

The lifetime of ZZZ batteries are normally distributed with a mean of 265 hours and a standard deviation
of 10 hours. Find the number of hours that represent the 40th percentile

Answer:

The 40th percentile is the point at which there is .40 probability below it (or to the left, on a graph). So, in a regular question they'll ask "how much probability is to the left of X" while here they are saying "What is X so that the probability is .40?"

You want to find X so that when you standardize it into a z-score it corresponds to the point where P(z < (x - 265)/10 ) = .40

So find the z-score that corresponds to that probability, say it was -1.5 (it's not, but say that it was), then you'd want your z-score to be -1.5, so you'd solve:

(x - 265)/10 = -1.5 => x = -15 + 265 = 250.

Question:

Type: Confidence Interval for a Sample Mean

A sample of the math test scores of 35 fourth-graders has a mean of 82 with a standard deviation of 15.

a. Find the 95% confidence interval of the mean math test scores of all fourth-graders.
b. Find the 99% confidence interval of the mean math test scores of all fourth-graders.
c. Which interval is larger? Explain why.

Answer:

The whole idea of confidence intervals is to give some precision to the estimate you're making. So, someone says, "what's the average score of all fourth-graders" and you say "well, when I looked at only 35 of them I got a mean of 82", that doesn't mean much without some knowledge of how spread out they are. Maybe someone got a 40 or a 50, and others got 100. Or maybe everyone got between 75 and 85. So the idea is to give some level of precision. Keep that in mind when we answer part c.

I'm not sure where you are in your course, so I don't know if you've only talked about the "Standard Normal" distribution, or whether you've covered the "t-distribution" yet. Since the problem doesn't say anything about the population's distribution, typically we'd use the t-distribution (let me know if that doesn't make sense). So in general, the formula should look something like:

x-bar ± t-score * (s/sqrt{n}) where the "t-score" is something from a table of the t-distribution, right?

x-bar is always your sample mean: 82. "s" is the sample standard deviation: 15. And "n" is the size of your sample: 35. The t-score is the point where there is 95% of the probability inbetween, and only 5% outside of it. In a Standard Normal distribution, that number is 1.96. That is, 95% of the probability is between -1.96 and +1.96. In this problem, however, you have to use a t-distribution with (n-1) degrees of freedom. Let me know if you have trouble doing that.

Now you have all of the things you need (by the way, "sqrt{n}" means "the square root of n"). For the left-hand side of the interval, you'll have:

82 - t-score*(15/sqrt{35}), and for the right-hand side you'll have 85 + t-score*(15/sqrt{35})

The only difference between parts a and b will be the t-score you use. In part b you want the one where 99% of the probability is between (so will that be larger or smaller than the one for 95%? And what will that do to the endpoints of your interval? Make them further out or closer in?).

If you can answer those last questions, you have your answer to part c. If not, ask yourself this: say for example I told you a 95% confidence interval was (75, 85) (that is, "75 to 85"). If you wanted to be MORE sure the interval included the real mean, would it have to be wider or smaller? That is, would you need to include MORE possibilities or fewer?

Question:

Type: Hypothesis Test, Two Means

The department at a university is concerned that dual degree students may be receiving
lower grades than the regular MBA students. Two independent random samples have been
selected. 100 observations from population 1 (dual degree students) and 100 from
population 2 (MBA students). The sample means obtained are X1(bar)=85 and X2(bar)=88.
It is known from previous studies that the population variances are 4.0 and 5.0 respectively.
Using a level of significance of .10, is there evidence that the dual degree students are
receiving lower grades? Fully explain your answer.

Answer:

The first thing is to set up the hypothesis you are testing. For the Null Hypothesis (Ho) we always assume that nothing is going on, or no association, or no difference, or whatever. So in this case, you assume that the mean grades for the two groups of students are equal:

Ho: MEAN of dual degree students = MEAN of mba students

For the alternative hypothesis, in this case it's only one-sided, and that's that the dual degree students' grades are lower:

Ha: MEAN dual degree < MEAN mba

Notice that for Ho, if the means were equal, their difference would be zero, and if the dual degree students mean was lower, MEAN dual degree minus MEAN mba would be negative. So you are testing the difference between two sample means. You should have a formula for that in your notes or book, it will look something like:

(xbar1 - xbar2) / sqrt{ some big mess }

and that big mess is a combination of population variances and sample sizes, right? sigma1 / n1 + sigma2 / n2, does that look familiar? You have all of the information: you have xbar1 and xbar2 (the sample means), sigma1 and sigma2 (the population variances) and n1 and n2 (the sizes of each sample).

When you plug those in you'll get a (negative) z-score, and then it's a matter of comparing that to the critical value (which will also be negative), and will be the point of the t-distribution where .10 is to the left of it. (Why?) Also, can you calculate the degrees of freedom? Isn't it n1 + n2 - 1?

If you "reject" or "fail to reject" Ho, be sure to explain it in terms of the problem. That is "given my results, there is evidence (or there is not evidence) that dual degree students have a lower mean grade than MBA students".

Question:

Type: Normal Distribution, Emperical Rule

In a set of 90 ACT scores, where the mean is 25 and the standard deviation is 4.69,

(a) how many scores are expected to be lower than 20.31 (one standard deviation below the mean)?
(b) How many of the 90 scores are expected to be below 34.38 (two standard deviations above the mean)?

Answer:

One thing that it doesn't say is what type of distribution the scores have. Can I assume they have a Normal distribution? I'll start as if I can, but at the bottom I'll say something about if I can't assume that.

For all of these, you want to know "how many standard deviations above or below the mean is the number they are giving?" That's because in a Standard Normal (mean=0 and std dev=1) you know what proportion of the data lies above or below any point. And notice that since the mean = 0 and the std dev = 1, the point 1.34, for example, is EXACTLY 1.34 standard deviations above the mean.

It turns out that in this problem they actually do this for you. So to find the answer to "a", you first have to find the PROPORTION of scores more than 1 standard deviation below the mean. For a Normal distribution that's the area to the left of -1.0. Find that proportion, and then multiply it by the total number of scores you have (90) to get the NUMBER of scores (and be sure to round that to a whole number). part "b" is the same, except that it's the area to the left of -2.0.

IF you can't assume the distribution is Normal, then you should have some general rules from your notes or the book like: "in any distribution, about X% is within 1 standard deviation of the mean, and about Y% is within 2 standard deviations of the mean", right? Let's say that it's actually 67% of the distribution is within 2 standard deviations of the mean. That means how much is OUTSIDE of 2 std devs? 100% - 67%, or 33%. And if the distribution is symmetric, that means half of that is more than 2 std devs ABOVE the mean, and the other half is more than 2 std devs BELOW the mean. So what proportion is more than 2 std devs below the mean? Half of 33%, or 16.5%. So you'd expect 16.5% of the 90 scores to be below 34.38.

Question:

Type: Normal Distribution, Mean and Sample Mean, Hypothesis Test

scores for men on the verbal portion of the SAT-I test are normally distibuted with a mean of 509 and the standard deviation of 112. randomlly selected men are given the columbia review course before taking the SAT test. assume that the course has no effect.

A) if 1 of the men is randomly selected find the probability that his score is at least 590
B) if 16 men are randomly selected find the probability that there mean score is 590
C) in finding the probability for part b why can the central limit theorem be used even thought the sample is less then 30
D) is the ramdom sample of 16 men does result in a mean score of 590 is there strong evidence to suppost the claim that the course is actually effective?why/why not?

Answer:

So, the population mean (mu) is 509 and sigma is 112. Men are given a review course, but we're to assume initially that the course has no effect. Again, that's our typical null hypothesis: assume there is no association, no effect, or in general the status quo.

A) P(X > 590) (notice this isn't x-bar because it's not a sample mean. It's just one person from the overall population.) Just standardize this and calculate a p-value: P(X > 590) = P(Z > (590-509)/112) = ...

B) I assume the question is actually "...that their mean score is AT LEAST 590" right? That's similar to part A, except that now we are dealing with a sample mean. So instead of X it's X-bar, so when we standardize to a z-score, we don't divide by 112, we divide by...what? There's a square root of "n" in there, right? Where does that go?

C) The CLT says that the distribution for a sample mean is Normal if one of two things is true: if either the sample is large enough, or...what? Something about the population's distribution...

D) This will come from your p-value in part B. Your previous question showed that you understand p-values, so it's just a matter of making sure you have the right hypothesis test down. Remember that our Null Hypothesis is that there is no effect, or Ho: mean = 509. The alternative is that there is an effect, or that Ha: mean > 509.

I tried to not completely do the problem for you, but instead give you some help to get you there.

Question:

Type: Hypothesis Test, Sample Mean

106 body temperatures with the mean of 98.20F assume standard devation is known as 0.62F,consider the body temperature of the population is less then 98.06 the signifigance level is .05
A) what is the test statistic
B) whats the critical value
C) what is the p value
D) what is the conclusion

Answer:

First of all, keep in mind that for Ho we always assume that status quo, or that there is no difference or no effect. Therefore they would look like:

Ho: mean = 98.6 (that is, that the sample of 106 people came from the population who has a mean body temp of 98.6)

Ha: mean < 98.6

And you're given that the population standard deviation is 0.62, and that in a sample of 106 people you got a sample mean of 98.2, right?

A) The test statistic will look like:

{x-bar - mu} / {std dev/sq root(n)} where "x-bar" is the sample mean (which is what?), "mu" is the hypothesized population mean (which is what?), and "std dev" is the known standard deviation (and "n" is the sample size). When I put in the numbers you provide, I get a test statistic of -6.64, which is a huge number...but that's what I get with the info you've provided.

B) 1.96 would actually be the critical value at alpha = .05 if this were a two-sided alternative. It's not. It's one sided, so you need to find a different critical value. Secondly, because the alternative is "less than", the critical value is actually going to be negative. Does that make sense?

C) Assuming the test statistic I got in part A is correct (again, I'm not convinced) the p-value would be so small that it won't be on your Normal Table. the p-value would be < .0001.

D) Instead of just saying "reject" or "fail to reject", you should also state what the conclusion is in terms of the problem. Do you have evidence that this sample did (or did not?) come from a population who has a mean body temp of 98.6?

Question:

Type: Hypothesis Testing

ABC Casino claims to have the loosest slots around. According to ABC their $1 slots have an average payout of 96 cents with a standard deviation of $15. On my last visit to ABC, I played the $1 slot 10,000 times and lost $3500. Do I have sufficienct evidence (at ALPHA= .05) to file a complaint with the Gaming Commission?

Answer:

This is a neat question, and requires some thought.

First of all, what would be sufficient evidence to file a complaint? If there is evidence that their "claim" is false, right? So, what you have to determine is whether your success rate is reasonable given their claim, or unreasonable (or unlikely). This is a classic hypothesis test.

So, what is the hypothesis test? Ho: mean = 0.96 vs. H1: mean < 0.96 (it's one-sided because you would only have a complaint if the average payout is lower than they claim - if it's higher than they claim you'd be ok with that!)

The key here is to realize what info you have. You have the hypothesized mean of THE POPULATION, and you also have a SAMPLE MEAN (that is, the mean from one sample). Those are two different things, right? "mu" and "x-bar". You need to think about what's true between a population (and its mean and standard deviation) and the distribution of possible SAMPLE MEANS.

If Ho were true, what is the likelihood you'd get the sample mean you did? (Which is what? you put in 10,000 and lost 3500, which means you got paid 6500 out of 10,000, or 6500/10000 = .65, right?).

The POPULATION standard deviation is 15, but what's the standard deviation of the sample mean? It's 15/sqrt(10000) = 0.15. You now have the hypothesized mean and standard deviation of the distribution of the sample mean. How do you determine if something is likely to occur in that distribution? You standardize it to a Standard Normal, sometimes it's called creating a z-score.

Remember that in general, a z-score = (x - mu)/std dev . What are "x" (or x-bar in this case), "mu", and "std dev" in this case? When you get that z-score, how do you determine if it's likely or unlikely? You compare it to the cut-off for an unlikely z-score. In this case, since alpha = 0.05, you compare it to a z-score of -2.545 (why is it negative?). You are going to determine that your result is unlikely if your z-score is (greater or less than?) that value. And based on that you'll say whether or not you have sufficient evidence to file a complaint.

Question:

Type: Confidence Interval

Suppose you take a single random sample of size 100 from a bell-curved population with a mean of 77 and a std dev of 5, and you get a mean test score in that sample of 76. Is this something you would have expected? Use the standardized score in your answer.

Which would be wider, a 90% confidence interval or a 95% confidence interval? (Assume both of them were calculated using the same sample data.) Explain your answer.

Answer:

The first one deals with "sample means" which have a Normal Distribution if the population they are drawn from has a Normal Distribution (which is the case here). Basically: if you took all of the possible samples of 100 and plotted the means of all of those samples, they would have a Normal Distribution centered at 77, with a standard deviation of...what? not 5, but 5/sqrt(100), or 5/10 = 1/2. So, is 76 a sample mean that is possible from such a distribution? well, the "standardized score" is (x-bar - mu)/ (sigma/sqrt(n)) or (76 - 77)/1/2, or -2.0. Is this possible? I'll leave that for you to answer.

For the second one, let me ask you this: if you wanted to be MORE confident that an interval included the right value, would that interval have to be bigger or smaller? That is, what if someone asked you when the next bus was, and you said "um, I think between 3:00 and 3:10" and they replied, "I don't want to miss it - give me an interval you're REALLY sure that if I'm here during that time I won't miss the bus" will that interval be wider or narrower?

OTHER/MISCELLANEOUS

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