252Y0611 10/18/06 (Open in Print Layout Format)
252y0611 10/18/06 (Open in ‘Print Layout’ format)
ECO252 QBA2 Name ___KEY______
FIRST HOUR EXAM Hour of class registered _____
October 17-18 2005Class attended if different ____
Version 1
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not usually acceptable.
I. (8 points) Do all the following.
1.
-3.61 .4998 and .1786 acceptable.
For make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between -3.62 and -0.92. Because this is completely on the left of zero, we must subtract. If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 6. Indicate the mean by a vertical line! Shade the area between -17.5 and 0. These numbers are both to the left of the mean (6), so we subtract.
2.
For make a diagram. Draw a Normal curve with a mean at 0. Shade the area between zero and 0.15. Because this is completely on the left of zero and touches zero, we can simply look up our answer on the standardized Normal table. If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 6. Shade the area between 6 and 7. This area includes the mean (6), but does not include any points to the right of the mean, so that we neither add nor subtract.
3.
For make a diagram. Draw a Normal curve with a mean at 0. Shade theentire area below 0.65. Because this is on both sides of zero, we must add. If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 6. Shade the entire area below 10.22. Since the area is on both sides of the mean, we add.
4. (Do not try to use the t table to get this.) For make a diagram. Draw a Normal curve with a mean at 0. is the value of with 3% of the distribution above it. Since 100 – 3 = 97, it is also the 97th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that the probability between and zero is 97% - 50% = 47% or The closest we can come to this is (1.89 is also acceptable here.) So To get from to , use the formula , which is the opposite of . . If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 6. Show that 50% of the distribution is below the mean (6). If 3% of the distribution is above , it must be above the mean and have 47% of the distribution between it and the mean.
Check:
II. (5 points-2 point penalty for not trying part a.) In order not to violate the truth in labeling law a teabag must contain at least 5.5 oz of tea. A sample of 9 items is taken from a large number of tea bags. The data below is found. (Recomputing what I’ve done for you is a great way to waste time.) b) and d) require statistical tests.
a. Compute the sample standard deviation,, of the waiting times. Show your work! (2)
b. Is the population mean significantly below 5.5 (Use a 95% confidence level)? Show your work! (3) [13]
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
d. Assume that the population standard deviation is 0.10 and create a 94% confidence interval for the mean. (2)
e. (Extra Credit) Given the data, is 0.10 a reasonable value for the population standard deviation?
1
252y0611 10/18/06 (Open in ‘Print Layout’ format)
Original data
Row
1 5.36 28.7296
2 5.17 26.7289
3 5.40 29.1600
4 5.38 28.9444
5 5.46 29.8116
6 5.50 30.2500
7 5.62 31.5844
8 5.35 28.6225
9 5.42 29.3764
48.66263.2078
Solution:
a) ,,
I got .0149 if I did not round the mean.
1
252y0611 10/18/06 (Open in ‘Print Layout’ format)
. I got 0.1222 without rounding.
b) This material is edited from the solution to problem 9.54. Given: and So Note that and .
Note that ‘Below 5.5’ does not contain an equality, so that it must be an alternative hypothesis. Our hypotheses are and . There are three ways to do this.
(i) Test Ratio: . The smaller the sample mean is, the more negativewill be this ratio. We will reject the null hypothesis if the ratio is smaller than .Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone below -1.860. Since the test ratio is below -1.860, we reject .
(ii) Critical value: We need a critical value for below 5.5. Common sense says that if the sample mean is too far below 5.5, we will not believe . The formula for a critical value for the sample mean is , but we want a single value below 5.5, so use . Make a diagram showing an almost Normal curve with a mean at 5.5 and a shaded 'reject' zone below 5.4252. Since is below5.4252, we reject .
(iii) Confidence interval: is the formula for a two sided interval. The rule for a one-sided confidence interval is that it should always go in the same direction as the alternate hypothesis. Since the alternative hypothesis is , the confidence interval is . Make a diagram showing an almost Normal curve with a mean at and, to represent the confidence interval, shade the area below5.4815in one direction. Then, on the same diagram, to represent the null hypothesis,, shade the area above5.5 in the opposite direction. Notice that these do not overlap. What the diagram is telling you is that it is impossible for and to both be true. (If you follow my more recent suggestions, it is actually enough to show that 5.5 is not on the interval.) So we reject .
c) (Extra credit) If you wish to use a p-value approach, since this is a left-side test, the p-value is the probability of getting a value below5.4067 when the null hypothesis is true. If we use the test ratio above we get. . To find this approximately, go to the line of the t table. 2.3209 is between and This means that and . Since -2.309 is between these values, we can conclude that If we want to use this in b), we can say that since the p-value is below we reject the null hypothesis.
Note that this is confirmed by Minitab, with more accurate computations. The column that you are using is t1, its mean is 5.40667, a confidence interval is , the t-ratio is equal to -2.29 and it gives a p-value of .026.
One-Sample T: t1
Test of mu = 5.5 vs < 5.5
95%
Upper
Variable N Mean StDev SE Mean Bound T P
t1 9 5.40667 0.12217 0.04072 5.48239 -2.29 0.026
d) In this section, we assume that the population standard deviation is 0.10 and create a 94% confidence interval for the mean. Here We do this by observing that the formula for a 2-sided confidence interval when the population standard deviation is . Fortunately, we found out on Page 1 that The confidence interval is thus or 5.34 to 5.47.
e) (Extra Credit) Given the data, we want to see if 0.10 a reasonable value for the population standard deviation? and . We will assume and use the chi – squared formula, We test this against and Since it is between them we cannot reject the null hypothesis.
III. Do as many of the following problems as you can. (2 points each unless marked otherwise adding to 18+ points). Show your work except in multiple choice questions. (Actually – it doesn’t hurt there either.) If the answer is ‘None of the above,’ put in the correct answer.
- Which of the following is a Type 2 error?
a)Rejecting the null hypothesis when the null hypothesis is false.
b)Rejecting the null hypothesis when the null hypothesis is true.
c)Not rejecting the null hypothesis when the null hypothesis is true.
d)*Not rejecting the null hypothesis when the null hypothesis is false.
e)All of the above
f)None of the above.
- A librarian provides a confidence interval estimate for the mean number of books checked out daily. The estimate is 229 to 741. The point estimate that this interval is based on is.
a)229.
b)*485
c)741
d)970
e)None of the above
f)There is not enough information to tell.
Explanation: Since the sample mean sits in the middle of a confidence interval, it must be an average of the two end points.
- (BLK8.30) If we want a 95% confidence interval for the average income of in a town, and the population standard deviation is known to be $1000. We have taken a sample of size 50 earlier and found that the sample mean is $14800. What sample size should we take if the width of the interval is to be no more than $100?
a)1537
b)385
c)50
d)40
e)20
f)None of the above.
Explanation: The formula in the outline is This is always rounded up – in this case to 1537.
- An entrepreneur is considering the purchase of a coin-operated laundry. The present owned claims that over the past 6 years the average daily income was $700. A sample is taken of daily revenue over a period of 28 days. Statistics are computed from the sample. If we want to test the statement that the mean is $700, which of the following tests is most appropriate? (1 point changed to 2!)
a)-test of a population mean.
b)-test of a population proportion.
c)*-test of a population mean.
d)-test of a population variance.
e)-test.
f)All of the above could be used.
g)We do not have enough information.[7]
- We are trying to estimate the median income in a region. We wish to test if the median is over $30000. We do not know the population variance. We can compute a statistic or statistics from a sample of 27 incomes. To do this test, the statistic or statistics we need is (are)
a)*The number of incomes in the sample that are above $30000.
b)The sample mean,
c)The sample mean, and the sample variance
d)The sample median .
e)The sample variance
f)The proportion of incomes that are above the sample mean,
g)The proportion of incomes that are above the sample median .[9]
- A study of child support says that the average support paid by noncustodial fathers is $370/month. We hope that fathers in our city are paying more than the national average. A random sample of 324 custodial mothers is taken. The results are a sample mean of $385.46 and a sample standard deviation of $35. Test whether this mean is significantly above $370.
a)State your null and alternative hypotheses. (1)
b)Test your hypothesis in a) by finding a critical value for the sample mean. Can we say that the result of the sample is above $370? Why? (3)
c)Do a 2-sided confidence interval for the mean. (2)
d)Do a 2-sided confidence interval for the variance. (2)[17]
e)(Extra credit) Explain in as much detail as reasonable how you would find a confidence interval for the median. (2)
Solution: This material is Part II all over again.
a) Note that ‘above 370’ does not contain an equality, so that it must be an alternative hypothesis. Or hypotheses are and .
Given: and So . Note that degrees of freedom are not important here. . (1.65 would probably be better.) There is no serious reason not to use instead of
b) We need a critical value for above370. Common sense says that if the sample mean istoo far above370, we will not believe . The formula for a critical value for the sample mean is , but we want a single value above 370, so use . Make a diagram showing an almost Normal curve with a mean at 370and a shaded 'reject' zone above373.20. Since is above 373.20, we reject .
c) Confidence interval for the mean: is the formula for a two sided interval. We need or maybe something a little larger (but not larger than 1.972). . We can say that .
d) Confidence interval for the variance: The outline says that for small samples the formula is , but if the degrees of freedom are too large for the chi-square table use . Using the second formula . Since , we have , which becomes and finally.
e) Confidence interval for the median: The formula given in the outline for large samples is . To be conservative, round this down to 144. Put the numbers in order by size as to . The interval will be to . (Note that the index of the second number is 324 + 1 – 144 = 181.
- Return to the problem in Question 6. Let’s say that the sample mean payment of $385.46 and the sample standard deviation of $35 come from a sample of 49. Assume that your null hypothesis is correct and that you get a t-ratio of 3.091. What should you say that the p-value is? (3 points). Note that showing your calculations here could get you partial credit.
a)Exactly .001
b)Exactly .002
c)Between .01 and .005
d)Between .02 and .01
e)*Between .005 and .001
f)***Between .01 and .002
g)Exactly .999
h)Exactly .998
i)Between .99 and .995
j)Between .98 and .99
k)**Between .995 and .999
l)None of the above – Show your answer![20]
Solution: If you have 48 degrees of freedom, the t-table says and This means that and .
*If you said that your hypotheses were and , you have a right sided test and , which will be between .005 and .001.
**However, if you said that your hypotheses were and , you have a left-sided test and , which will be between .995 and .999.
***If you said that your hypotheses were and , you have a two-sided test and , which will be between .01 and .002.
I can only promise to look at your reasoning if you have any other answer.
- (Mann) According to a 1992 survey, 45% of workers say that they would change careers if they could. You wish to show that the proportion of workers in your union that want to change careers is above the national figure. Find the correct set of hypotheses below.
a)
b)
c)
d)
e)*
f) **
g)
h)None of the above. Put in your answer![22]
- (Mann) According to a 1992 survey, 45% of workers say that they would change careers if they could. You wish to show that the proportion of workers in your union that want to change careers is above average. Assume that your null hypothesis in 8 is correct, that you take a sample
of 350 workers and that you compute the ratio . If your confidence level is 90%,
you should do thefollowing.
a)Reject the null hypothesis if the ratio is not between -1.96 and 1.96.
b)Reject the null hypothesis if the ratio is not between -1.645 and 1.645
c)Reject the null hypothesis if the ratio is above 1.645
d)Reject the null hypothesis if the ratio below -1.645
e)Reject the null hypothesis if the ratio is below 1.282
f)Reject the null hypothesis if the ratio is above 1.282
g)Reject the null hypothesis if the ratio is below -1.282
h)None of the above. Put in your answer![24]
Solution: This is clearly a one-sided test and . The table says If you said *, you have a right-sided test and you reject the null hypothesis if
**However, if you said (wrongly), you have a left-sided test and you would reject the null hypothesis if Note that for a t or z ratio, zero is never in the reject region, so that e) could never be true.
- We wish to use a 2-sided confidence interval to test a proportion. The following things may influence the size of a confidence interval: (.5 points for good ones, .5 off for bad ones, at worst zero)
1. Decreasing to make it closer to
2. Increasing from to
3. Changing the null hypothesis to make closer to .5
4. Increasing the sample size. (Assume a large population)
5. Decreasing the sample size. . (Assume a large population)
6. Increasing the confidence level
7. Increasing the significance level
8. Using a continuity correction with a relatively small sample.
9. Decreasing the population size from to
10. Increasing the population size from to
Put down the numbers of the things that make the confidence interval larger.______
Solution:The confidence interval has the following form: , where and, if we include a finite sample correction, the standard error is .
1. Decreasing to make it closer to . This makes closer to .5. It was explained in class that becomes larger as approaches .5. This makes the interval larger.
2. Increasing from to . This makes farther from .5 and makes the interval smaller.
3. Changing the null hypothesis to make closer to .5. does not appear in the formula for a confidence interval, so it has no effect.
4. Increasing the sample size. (Assume a large population.) Since , the sample size, is in the denominator of , it makes the interval smaller.
5. Decreasing the sample size. (Assume a large population.) This makes the interval larger.
6. Increasing the confidence level. If we raise the confidence level , the significance level gets smaller. This gives us a larger value of , so it makes the interval larger.
7. Increasing significance level . This makes the interval smaller.
8 Using a continuity correction. This has the effect of making all intervals slightly larger. It has little effect with a large sample.
9. Decreasing the population size from to . If we use a finite population correction, it has the general effect of decreasing the standard error. Decreasing the population size makes its effects stronger, so it makes the interval smaller.
10. Increasing the population size from to . Let the size of the population be , where is a value from 1 to 20. Then the finite population correction is . This obviously grows as grows. This makes the interval larger.
In Summary:
Larger: 1, 5, 6, 8, 10
Smaller: 2, 4, 7, 9
Neither: 3.
ECO252 QBA2
FIRST EXAM
October 11-12 2006
TAKE HOME SECTION
-
Name: ______
Student Number and class: ______
IV. Do at least 3 problems (at least 7 each) (or do sections adding to at least 20 points - Anything extra you do helps, and grades wrap around) . Show your work! State and where appropriate. You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion.. (Use a 95% confidence level unless another level is specified.) Answers without reasons usually are not acceptable.Neatness and clarity of explanation are expected. This must be turned in when you take the in-class exam.Note that from now on neatness means paper neatly trimmed on the left side if it has been torn and paper written on only one side.
- (Moore, Notz) You are thinking that it may be desirable to start a wellness program for your (large) company. You are told that the company will only start such a program if you can show that the blood pressure of a group of mid-level executives is above normal. The individuals are all between 35 and 44 years old and US statistics show that mean systolic blood pressure for men in that age range is 128. You take a sample of 72 executives and get the following results.