252y0753 10/19/07 (Open in ‘Print Layout’ format)
ECO252 QBA2 Name:_KEY______FIRST EXAM Class hour: ______
October 4 and 8, 2007
Version 3 Student number: ______
Show your work! Make Diagrams! Include a vertical line in the middle! Exam is normed on 50 points. Answers without reasons are not usually acceptable.
I. (8 points) Do all the following.
1. = .1406 + .5 = .6406
2.
=.4996 - .0714 = .4282
3. = .2673
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 7.5% of the distribution above it. Since 100 – 7.5 = 92.5, it is also the .925 fractile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that the probability between and zero is 92.5% - 50% = 42.5% or If we check this against the Normal table, the closest we can come to .4250 is So This is the value of z that you need for a 85% confidence interval. 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 4. Show that 50% of the distribution is below the mean (4). If 7.5% of the distribution is above , it must be above the mean and have 42.5% of the distribution between it and the mean.
Check: This is identical to the way you normally get a p-value for a right-sided test.
II. (9 points-2 point penalty for not trying part a.)
Monthly incomes (in thousands) of 6 randomly picked individuals in the little town of Rough Corners are shown below.
2.5 7.3 3.1 2.6 2.4 3.0
a. Compute the sample standard deviation,, of expenditures. Show your work! (2)
b. Assuming that the underlying distribution is Normal, compute a 99% confidence interval for the mean. (2)
c. Redo b) when you find out that there were only 50 people living in Rough Corners. (2)
d. Assume that the population standard deviation is 2 and create an 85% two-sided confidence interval for the mean. (2)
e. Use your results in a) to test the hypothesis that the mean income is above 2.3(thousand) at the 99% level. (3) State your hypotheses clearly!
f. (Extra Credit) Given the data, test the hypothesis that the population standard deviation is below 2. (3)
Solution: a) Compute the sample standard deviation,, of expenditures.
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252y0753 10/19/07 (Open in ‘Print Layout’ format)
Row
1 2.5 6.25 -0.98333 0.9669
2 7.3 53.29 3.81667 14.5669
3 3.1 9.61 -0.38333 0.1469
4 2.6 6.76 -0.88333 0.7803
5 2.4 5.76 -1.08333 1.1736
6 3.0 9.00 -0.48333 0.2336
20.9 90.67 0.0 17.8682
The first two columns are needed for the computational (shortcut) method. The first, third and fourth are needed for the definitional method. Using (both methods or) the definitional method wastes time.
, , (a check), and .
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252y0753 10/19/07 (Open in ‘Print Layout’ format)
. If you used the definitional method, you would have gotten 1.8904. There seems to be a lot of potential for rounding error here. Note that the column, even though it carries an extra place, does not quite add to the expected zero but to .00002.
b) Assuming that the underlying distribution is Normal, compute a 99% confidence interval for the mean. (2) or 0.371 to 6.595.
c) Redo b) when you find out that there were only 50 people living in Rough Corners. (2)
or 116.02 to 155.98
d) Assume that the population standard deviation is 2 and create an 85% two-sided confidence interval for the mean. (2) (2) We found on the last page. We have , , and or 2.3075 to 4.6591.
e) Use your results in a) to test the hypothesis that the mean income is above 2.3(thousand) at the 99% level. (3) State your hypotheses clearly! The statement that the mean is above 2.3 does not contain an equality, so it must be an alternate hypothesis. We have the following information. , , and . Since this is a one-sided hypothesis we will use . Needless to say, because of the small sample size, we are assuming that the parent distribution is Normal.
Our hypotheses are so . Since we are worrying about the mean being too large, this is a right-sided test.
There are three ways to do this. Do only one of them.
(i) Test Ratio: . This is a right-sided test - the larger the sample mean is, the more positive will be this ratio. We will reject the null hypothesis if the ratio is larger than . Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone above 3.365. Since the test ratio is below 3.365, we cannot reject .
If you wish to find a p-value for your hypothesis, note that the t-ratio is 1.5332. The p-value will be the probability that is above 3.365. The line of the t table for 5 degrees of freedom is below. {ttable}
.45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
5 0.132 0.267 0.408 0.559 0.727 0.920 1.156 1.476 2.015 2.571 3.365 4.032 5.893
What this tells us, among other things, is that and . Since 1.5332 lies between 1.476 and 2.015, the probability that lies above 1.5332 must be between .05 and .10. . This is above our significance level of .01, so we will not reject the null hypothesis.
(ii) Critical value: We need a critical value for above 2.3. Common sense says that if the sample mean is too far above 2.3, we will not believe . The formula for a critical value for the sample mean is , but we want a single value above 2.3, so use . Make a diagram showing an almost Normal curve with a mean at 2.3 and a shaded 'reject' zone above 4.8970. Since is not above 4.8970, we do not 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 or . Make a diagram showing an almost Normal curve with a mean at and, to represent the confidence interval, shade the area above 0.8863 in one direction. Then,
on the same diagram, to represent the null hypothesis, , shade the area below 2.3 in the opposite direction. Notice that these overlap. What the diagram is telling you is that it is possible for and to both be true. (If you follow my more recent suggestions, it is actually enough to show that 2.3 is on the confidence interval.) So we do not reject .
f.) (Extra Credit) Given the data, test the hypothesis that the population standard deviation is below 2. (3)
This is an alternate hypothesis, . The null hypothesis is Remember , . Table 3 says that the test ratio is .
Recall The first paragraph of the chi-squared table appears below. If we look at the 5 column, we see that the lower 1% of values of chi-squared are cut off by , so that the reject region is below 0.5543.
Degrees of Freedom
1 2 3 4 5 6 7 8 9 0.005 7.87946 10.5966 12.8382 14.8603 16.7496 18.5476 20.2778 21.9550 23.5893
0.010 6.63491 9.2103 11.3449 13.2767 15.0863 16.8119 18.4753 20.0902 21.6660
0.025 5.02389 7.3778 9.3484 11.1433 12.8325 14.4494 16.0128 17.5346 19.0228
0.050 3.84146 5.9915 7.8147 9.4877 11.0705 12.5916 14.0671 15.5073 16.9190
0.100 2.70554 4.6052 6.2514 7.7794 9.2364 10.6446 12.0170 13.3616 14.6837
0.900 0.01579 0.2107 0.5844 1.0636 1.6103 2.2041 2.8331 3.4895 4.1682
0.950 0.00393 0.1026 0.3518 0.7107 1.1455 1.6354 2.1674 2.7326 3.3251
0.975 0.00098 0.0506 0.2158 0.4844 0.8312 1.2373 1.6899 2.1797 2.7004
0.990 0.00016 0.0201 0.1148 0.2971 0.5543 0.8721 1.2390 1.6465 2.0879
0.995 0.00004 0.0100 0.0717 0.2070 0.4117 0.6757 0.9893 1.344 1.7349
Computer output for parts b) d) e) and f) follows
————— 10/5/2007 1:39:55 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0751-23.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x0751-23.MTW'
Worksheet was saved on Wed Oct 03 2007
Results for: 252x0751-23.MTW
MTB > describe x
Descriptive Statistics: x
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x 6 0 3.483 0.772 1.890 2.400 2.475 2.800 4.150 7.300
MTB > onet c1; Part b)
SUBC> conf 99.
One-Sample T: x
Variable N Mean StDev SE Mean 99% CI
x 6 3.48333 1.89041 0.77176 (0.37149, 6.59517)
MTB > onez c1; Part d)
SUBC> sigma 2;
SUBC> conf 85.
One-Sample Z: x
The assumed standard deviation = 2
Variable N Mean StDev SE Mean 85% CI
x 6 3.48333 1.89041 0.81650 (2.30796, 4.65871)
MTB > Onet c1; Part e)
SUBC> Test 2.3;
SUBC> Confidence 99;
SUBC> Alternative 1.
One-Sample T: x
Test of mu = 2.3 vs > 2.3
99%
Lower
Variable N Mean StDev SE Mean Bound T P
x 6 3.48333 1.89041 0.77176 0.88642 1.53 0.093
MTB > Save "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0751-23.MTW";
SUBC> Replace.
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x0751-23.MTW'
Existing file replaced.
Part f)
MTB > %sigtest c1 4 Tests data in column 1 for variance of 4. Packaged Minitab Macro. This is a 2-sided test.
Executing from file: sigtest.MAC
The value of the test statistic is 4.4671.
If the test statistic is less than 0.8312 or greater
than 12.8325 then there is statistical evidence indicating
that your variance does not equal to 4.0000, at alpha =
0.0500.
MTB > %sigtest c1 4; Since I didn’t have time to input a 1- sided test. I ran two-sided tests with a confidence level of 98% and 90% because their lower critical values would be the same as for 1-sided tests with confidence levels of 99% and 95%.
SUBC> alpha 98.
Executing from file: sigtest.MAC
The value of the test statistic is 4.4671.
If the test statistic is less than 4.2789 or greater
than 4.4249 then there is statistical evidence indicating
that your variance does not equal to 4.0000, at alpha =
0.9800.
MTB > %sigtest c1 4;
SUBC> alpha 90.
Executing from file: sigtest.MAC
The value of the test statistic is 4.4671.
If the test statistic is less than 3.9959 or greater
than 4.7278 then there is statistical evidence indicating
that your variance does not equal to 4.0000, at alpha =
0.9000.
III. Do as many of the following problems as you can.(2 points each unless marked otherwise adding to 13+ 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 if possible.
1) If I want to test to see if the mean of is smaller than the given population mean my null hypothesis is:
i)
ii)
iii) *
iv)
v) Could be any of the above. We need more information.
vi) None of the above
Explanation: is our alternate hypothesis since it doesn’t contain an equality. is the opposite, so it must be the null hypothesis.
2) Assuming that you have a sample mean of 100 based on a sample of 36 taken from a population of 300 with a sample standard deviation of 80, the 99% confidence interval for the population mean is
a)
b)
c)
d)
e) *
f)
g)
h)
i)
j)
g) None of the above. Fill in a correct answer.
Explanation: The formula for a confidence interval when the variance is known when the sample is more than 20% of the population was given in the solution to problem A2 as , where 100, 36, 300, , 99% and .
Here and . So
3) Which of the following is a Type 2 error?
a) Rejecting the null hypothesis when the null hypothesis is true.
b) Not rejecting the null hypothesis when the null hypothesis is true.
c) *Not rejecting the null hypothesis when the null hypothesis is false.
d) Rejecting the null hypothesis when the null hypothesis is false.
e) All of the above
f) None of the above.
4) If a random sample is gathered to get information about a population proportion, what do we mean by a p-value?
a) P-value is the probability that, if the null hypothesis was false, that, if we were to repeat the experiment many times, we would get a sample proportion as extreme as or more extreme than the sample proportion actually observed.
b) *P-value is the probability that, if the null hypothesis was true, that, if we were to repeat the experiment many times, we would get a sample proportion as extreme as or more extreme than the sample proportion actually observed.
c) P-value is the population proportion in the null hypothesis.
d) P-value is the population proportion in the alternate hypothesis.
e) P-value is the probability of a type 2 error.
f) P-value is the probability that the alternate hypothesis is true, given the sample proportion actually observed.
g) None of the above is true.
5) If a difference in proportions (in a business-related problem) is called statistically significant at the 1% significance level, this means that
a) *If the null hypothesis is true, the difference in proportions is surprisingly large.
b) There is a 99% chance that the null hypothesis is true.
c) The difference in proportions is large enough so that we must take account of it in our business decisions.
d) All of the above
6) (Wonnacott & Wonnacott) When an industrial process is in control, it produces bolts with a hardness that has a mean of at least 80 and a (population) standard deviation of 8. If the hardness is too far below 80, you must shut down the process. Every hour you take a sample of 16 bolts. How low must the average hardness of these bolts be before we shut the process down? (Use a 10% significance level, and don’t forget to state your hypotheses) (3)