252y0552 10/31/05 (Open in ‘Print Layout’ format)
ECO252 QBA2 Name _ Key______
FIRST HOUR EXAM Hour of class registered _____
October 17-18 2005Class attended if different ____
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. If you do not use the standard Normal table, explain!
Make diagrams. For draw a Normal curve with a vertical line in the center at 4.5. For draw a Normal curve with a vertical line in the center at zero.
1.
2.
3.
4. There is no excuse for not being able to do this at this point, since you should have done it 3 times in the take-home.
To findmake a Normal diagram for showing a mean at 0 and 50% above 0, divided into 6% above and 44% below . So The closest we can come is or . So use .
II. (5 points-2 point penalty for not trying part a.) A dealer wishes to test the manufacture’s claim that the Toyota Caramba gets 28 miles per gallon. The data below is found. (Recomputing what I’ve done for you is a great way to waste time.)
a. Compute the sample standard deviation,, of themiles per gallon. Show your work! (2)
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg? Show your work! (3)
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
1
252y0552 10/31/05 (Open in ‘Print Layout’ format)
Row
1 23.56 555.074
2 26.09 680.688
3 27.55 759.003
4 26.92 724.686
5 29.20 852.640
6 29.02 842.160
7 22.48 505.350
8 27.45 753.503
9 25.90 670.810
10 25.81 666.156
sum 263.98 7010.070
Solution: a. Compute the sample standard deviation,, of the miles per gallon. Show your work! (2)
.
1
252y0552 10/31/05 (Open in ‘Print Layout’ format)
Note that excessive rounding can throw this answer way off. Using , I got and .
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg? Show your work! (3)
Given: and and .
We are testing . Use only one of the following!
Confidence Interval: or 21.861 to 27.935.
Since 28 is not on the confidence interval, it is significantly different from the sample mean.
Critical Value: or 26.46 to 29.53.
Since 26.398 is not between the critical values, it is significantly different from the population mean.
Test Ratio:
. The ‘accept’ zone is between -2.262 and 2.262. Since -2.197is not between these values, reject the null hypothesis.
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
2.358 is between and . Since , it is between .02 and .05.
III. Do all of the following problems (2 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.
[10]
1.For sample sizes greater than 100, the sampling distribution of the mean will be approximately normally distributed
a)*Regardless of the shape of the population.
b)Only if the shape of the population is symmetrical.
c)Only if the standard deviation of the samples are known.
d) Only if the population is normally distributed.
2.When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimated to be the ______the sample size required.
a)Smaller
b)*Larger
c)Sample size is not affected.
d)The effect cannot be determined from the information given.
- Which of the following would be an appropriate null hypothesis?
a)The population proportion is less than 0.65.
b)The sample proportion is less than 0.65.
c)*The population proportion is at most 0.65.
d)The sample proportion is at most 0.65.
- A Type I error is committed when
a)*We reject a null hypothesis that is true.
b)We don't reject a null hypothesis that is true. Note that b) and c) are not errors.
c)We reject a null hypothesis that is false.
d)We don't reject a null hypothesis that is false
- If we are performing a two-tailed test of whether = 100, the power of the test in detecting a shift of the mean to 105 will be ______its power detecting a shift of the mean to 96.
a)*Less than96 is closer than 105 to 100.
b)Greater than
c)Equal to
d)Not comparable to
- From a sample of 100 students we find that the mean expenditure for books is $316.40 with a sample standard deviation of $43.20. You are asked to test whether the (population) mean expenditure is above $314 using a 10% significance level.
a)What are the null and alternative hypotheses? (2)
b)What is the ‘rejection zone’ (in terms of ) (2)
c) What is your conclusion? (2)[16]
Solution: This is basically the revised version of Exercise 9.54 [9.52 in 9th] (9.50 in 8th edition) that I warned you about.
a) Since the problem asks if the mean is above $314, and this does not contain an equality, it must be an alternate hypothesis. Our hypotheses are , (The average cost of textbooks per semester at a large university is no more than $314.) and (The average cost of textbooks per semester at a large university is more than $314.) This is a right-sided test.
b) Given: and So Note thatWe need a critical value for above $314. Common sense says that if the sample mean is too far above $314, we will not believe . The formulafor a critical value for the sample mean is ,but we want a single value above 314, so use . Make adiagram showing an almost Normal curve with a mean at 314 and a shaded 'reject' zoneabove 319.57.
c) Since is not in the ‘reject’ zone, we do not reject the null hypothesis and cannot conclude that the mean is above $314.
- From a sample of 10 students we find that the mean expenditure for books is $316.40 with a sample standard deviation of $43.20. You are asked to test whether the (population) mean expenditure is below some number. You compute a t ratio and find that it is 1.960. The p-value is
a)Between .025 and .05
b)Between .05 and .10
c) Between .95 and .975
d)Between 1.90 and 1.95
e) Exactly .025
f) None of the above – provide a more suitable answer.[18]
Explanation: For this see the writeup that I announced two weeks ago. If our alternate hypothesis is that the mean expenditure is below some number, we want . There are degrees of freedom. The table says that and . Thus we can see that , and we can conclude that .
- An employer states that the median wage in a factory is $45000. You believe that it is lower. From a sample of 300 employees you find that 170 have wages below $45000. The median for the sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test should reflect your beliefs and, if you use in a hypothesis, you must define it. Your hypotheses are (3)
a)
b)
c)
d) *and If is the proportionabove .
e)and
f) and
g) *and If is the proportionbelow .
h)and
i) and [21]
Your belief is that the median is below 45000. This is an alternate hypothesis because it does not contain an equality. The null hypothesis must be that the median is at least 45000. The correct hypothesis involving the proportion can be found in the following table from the outline.
Hypotheses aboutHypotheses about a proportion
a medianIf is the proportionIf is the proportion
above below
9.(Extra credit) An employer states that the median wage in a factory is $45000. You believe that it is lower. From a sample of 300 employees you find that 170 have wages below $45000. The median for the sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test should reflect your beliefs and, if you use in a hypothesis, you must define it. Finish the problem. (3)
Solution:The proportion below $45000 is This is the most natural way to define the proportion here. So our hypotheses are and . We also will assume so that , and we know that and This is a right-side test. From Table 3, we can use one of the methods..
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion /
/ / /
We can say and
(i) Critical Value: = .5475. The rejection region is above .5475 and the observed proportion, .5667 is in that region, so we reject the null hypothesis.
(ii) Test Ratio: = 2.3104. The rejection region is above 1.645 and this value of is in that region, so we reject the null hypothesis.
(iii) Confidence Interval: = .5196. There is no way that the proportion can be both above .5196 and below .5 as stated in the null hypothesis, so we reject the null hypothesis.
10. (Extra Credit – Nasty but not that hard – problem due to Prem S. Mann.) Use . A quality control inspector is checking machines that are supposed to produce packages of cookies with a mean weight of 32 oz and a standard deviation of .015 oz. She takes a sample of 40 packages and finds a sample standard deviation of .029 oz. Do a one-sided test on the variance to see if the machine is out
of control. Do the test by finding a critical value for the sample variance (3) (You can do it in a more familiar way, but you won’t get as much credit!) [27]
Solution:,, and . Table 3 says . If we are doing a right sided test we want a critical value above .015, we use . . This is a right-side test, so that the ‘reject’ region is the area under the curve to the right of 0.0203. Since is above this value, reject the null hypothesis and fix the machine.
ECO252 QBA2
FIRST EXAM
October 17-18 2005
TAKE HOME SECTION
-
Name: ______
Student Number and class: ______
IV. 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 are not usually acceptable. Neatness counts!
1. (Prem S. Mann - modified)
Personalize these results as follows. Change 45% by replacing 5 by the last digit of your student number. We will call your result the ‘proportion of interest.’
(Example: Seymour Butz’s student number is 976512, so he changes 45% to 42%; 42% is his proportion of interest.)
a. State the null and alternative hypotheses in each case and find a critical value for each case. What is the ‘reject’ region? Compute a test ratio and find a p-value for the hypothesis in each case. (12)
(i) The state government wants to test that the fraction of people who favored higher taxes for health insurance is below the proportion of interest
(ii) The state government wants to test that the fraction of people who favored higher taxes for health insurance is above the proportion of interest.
(iii) The state government wants to test that the fraction of people who favored higher taxes for health insurance is equal to the proportion of interest. [12]
b. (Extra credit) Find the power of the test if the true proportion is 50% and:
(i) The alternate hypothesis is that the fraction of people is above the proportion of interest. (2)
(ii) The alternate hypothesis is that the fraction of people does not equal the proportion of interest. (2)
c. Assuming that the proportion of interest is correct, is the sample size given above adequate to find the true proportion within .005 (1/2 of 1%)? (Don’t say yes or no without calculating the size that you actually need!) (2)
d. If the alternate hypothesis is that the fraction of people is above the proportion of interest, create an appropriate confidence interval for the hypothesis test. (2) [20]
Solution: a) From the formula table we have:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion /
/ / /
Only the solution for Version 0 is given here. The solutions for the remainder will be in 252y0551s or 252y0551h. Note that my rule on grading parts like (ii) below is to assume that your alternate hypothesis was correct and to ask if the critical values agree with it.
Version 0 a)
(i) The state government wants to test that the fraction of people who favored higher taxes for health insurance is below the proportion of interest
and . .
The critical value must be below .40. = .3430. The ‘reject’ zone is below .3430.
(ii) The state government wants to test that the fraction of people who favored higher taxes for health insurance is above the proportion of interest.
and . .
The critical value must be above .40. = .4570 . The ‘reject’ zone is above .4570.
(iii) The state government wants to test that the fraction of people who favored higher taxes for health insurance is equal to the proportion of interest. [12]
and . .
The critical value must be on either side of .40. . The ‘reject’ zone is below .3369 and above .4631.
b) (i)
(ii)
c) This is above 400, so the sample size is inadequate.
d) If the null hypothesis is we must reject it.
Minitab Calculation of critical values and test ratios for all Versions of question 1 of the Take-home exam.
————— 10/17/2005 7:01:49 PM ————————————————————
Available in 252y0551s
2. Customers at a bank complain about long lines and a survey shows a median waiting time of 8 minutes.
Personalize the data as follows: Use the second-to-last digit of your student number – subtract it from the last digit of every single number. (Example: Seymour Butz’s student number is 976512, so he changes {6.7 6.3 5.7 ….} to {6.6 6.1 5.6 …}) You may drop any number in the data set that you use for this problem that is exactly equal to 8. Use . Do not assume that the population is Normal!
a. Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
b. (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2)[24]
Solution: The data sets for this problem are below. Let be the count of the sample and be the number of numbers below 8.
Row
1 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 4.9
2 5.4 5.3 5.2 5.1 5.0 4.9 4.8 4.7 4.6 5.5
3 5.6 5.5 5.4 5.3 5.2 5.1 5.0 4.9 4.8 5.7
4 6.2 6.1 6.0 5.9 5.8 5.7 5.6 5.5 5.4 6.3
5 6.6 6.5 6.4 6.3 6.2 6.1 6.0 5.9 5.8 6.7
6 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 6.1 7.0
7 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 7.1
8 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 7.2
9 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
10 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
11 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 7.4
12 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
13 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
14 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
15 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
16 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.1
17 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.2
18 8.1 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 8.4
19 8.3 8.2 7.9 7.8 7.7 7.6 7.5 7.4 7.3 8.4
20 8.3 8.2 8.1 8.1 7.9 7.8 7.7 7.6 7.5 8.4
21 8.3 8.2 8.1 8.5 7.9 7.8 7.7 7.6 7.5 8.5
22 8.4 8.3 8.1 8.6 7.9 7.8 7.7 7.6 7.5 8.9
23 8.8 8.7 8.2 8.7 8.4 7.9 7.8 7.7 7.6 9.0
24 8.9 8.8 8.6 8.9 8.5 8.3 8.2 8.1 8.1 9.1
25 9.0 8.9 8.7 9.1 8.6 8.4 8.3 8.2 8.2 9.3
26 9.2 9.1 8.8 9.2 8.8 8.5 8.4 8.3 8.4 9.5
27 9.4 9.3 9.0 9.3 9.0 8.7 8.6 8.5 8.6 9.6
28 9.5 9.4 9.2 9.4 9.1 8.9 8.8 8.7 8.7 9.7
29 9.6 9.5 9.3 9.9 9.2 9.0 8.9 8.8 8.8 9.8
30 9.7 9.6 9.4 9.3 9.1 9.0 8.9 8.9 10.3
31 10.2 10.1 9.5 9.8 9.2 9.1 9.0 9.4
32 10.0 9.7 9.6 9.5
31 31 32 29 31 32 32 32 31 30
17 18 19 19 22 23 23 23 23 15
a) Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
Hypotheses aboutHypotheses about a proportion
a medianIf is the proportionIf is the proportion
above below
It seems easiest to let be the proportion below 8. If you defined as the proportion above .8, the p-values should be identical. Our Hypotheses are and . According to the outline, in the absence of a binomial table we must use the normal approximation to the binomial distribution. If is our observed proportion, we use . But for the sign test, . So
(For relatively small values of , a continuity correction is advisable, so try , where the + applies if , and the applies if . ) Values of and are repeated below for all ten possible cases. Both and the more correct (except when ) are computed. Since the alternative hypothesis is the probabilities in the two right columns are p-values. If you used , the values of below are correct. The exact probabilities were computed by Minitab in the last column.
1 17 31 0.53882 0.35921 .5-.2054=.2946 .5-.1404=.3596 0.360050
2 18 31 0.89803 0.71842 .5-.3133=.1867 .5-.2642=.2358 0.236565
3 19 32 1.06066 0.88388 .5-.3554=.1445 .5-.3106=.1894 0.188543
4 19 29 1.67126 1.48556 .5-.4525=.0475* .5-.4319=.0681 0.068023
5 22 31 2.33487 2.15526 .5-.4901=.0099* .5-.4846=.0154* 0.014725
6 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
7 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
8 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
9 23 31 2.69408 2.51447 .5-.4964=.0036* .5-.4940=.0060* 0.005337
10 15 30 0.00000 0.00000 .5 .5 0.572232
Since , and the starred items are below .05, these are the cases in which we reject the hypothesis that the median is, at least 8. It makes about as much sense to reject a hypothesis about a median because the sample median is above or below with no other information as it does to reject a hypothesis about a mean because is above or below without more information.
b) (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2) I am going to assume that you continued to use the data sets above. The outline says that an approximate value for the index of the lower limit of the interval is . In this formula
We use this formula with the following results.
Row sqrtn interval
1 31 5.56776 10.5436 10 22
2 31 5.56776 10.5436 10 22
3 32 5.65685 10.9563 10 23
4 29 5.38516 9.7225 9 23
5 31 5.56776 10.5436 10 22
6 32 5.65685 10.9563 10 23
7 32 5.65685 10.9563 10 23
8 32 5.65685 10.9563 10 23
9 31 5.56776 10.5436 10 22
10 30 5.47723 10.1323 10 21
3. Use the personalized data from Problem 2 (but do not drop any numbers.) test that the mean is below 8. Minitab found the following the original numbers, which were in C4:
Descriptive Statistics: C4
Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
C4 32 7.944 0.228 1.291 4.900 7.225 8.000 8.975 10.300
Sum of squares (uncorrected) of C4 = 2070.94
I do not know the values for your numbers, but the following (copied from last year’s exam) should be useful: . Your value of is negative or zero. Can you say that the population mean is below 8? Use .
If you remembered what we learned last semester, you could have done this calculation with little or no work. The relevant formulas are:
For the mean
For the variance
So that good old Seymour, whose student number is 976512, subtracts 0.1 from every number and gets
Mean
Variance
a. State your null and alternative hypotheses (1)
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results. (1)
e. Test to see if the population standard deviation is 2. (2)
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a power curve for your test. (5)
h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided confidence interval for the mean. (1)
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean is below 8. (1)
j. Using a 96% confidence interval an assuming that the population standard deviation is 2, how large a sample do you need to have an error in the mean of ? (1) [39]
Solution:
a. State your null and alternative hypotheses (1)
Since the problem asks if the mean is below, and this does not contain an equality, it must be an alternate hypothesis. Our hypotheses are
, (The average wait is at least 8 minutes.) and
(The average wait is less than 8 minutes.) This is a left-sided test.
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)