252y0211 2/27/02 ECO252 QBA2 Name _____Key______
FIRST HOUR EXAM Hour of class registered _____
February 19, 2002 Class attended if different ____
Show your work! Make Diagrams! How many of you looked at "Things You Should Never Do" before this exam?
I. (14 points) Do all the following.
1.
2.
3.
4.
5. (The cumulative probability up to 17)
6. A symmetrical interval about the mean with 58% probability.
We want two points and , so that. From the diagram,
if we replace x by z, . The closest we can come is
. So , and ,
or 4.14 to 13.86. To check this note that
7. We want a point , so that. (This is the 90.5 percentile)
From the diagram, if we replace x by z, . The closest we can come is
. So , and , or 16.86 .
To check this note that
252y0211 2/21/02
II. (6 points-2 point penalty for not trying part a.)
A new product is tried on seven patients. Their breathing capacity after using the product is shown below (Note: You may want to move the decimal point to the left and work in thousands.).
Patient capacity
1 2850
2 2380
3 2800
4 2860
5 2300
6 2650
7 2640
a. Compute the sample standard deviation, , of the breathing capacity. Show your work! (3)
b. Compute a 90% confidence interval for the mean breathing capacity, .(3)
2
Solution: a)
2
Original data
Row
1 2850 8122500
2 2380 5664400
3 2800 7840000
4 2860 8179600
5 2300 5290000
6 2650 7022500
7 2640 6969600
18480 49088600
or .
Data divided by 1000
Row
1 2.85 8.1225
2 2.38 5.6644
3 2.80 7.8400
4 2.86 8.1796
5 2.30 5.2900
6 2.65 7.0225
7 2.64 6.9696
18.48 49.0886
(thousands)
or .
2
b. From the problem statement . From Table 3 of the syllabus supplement, if the population variance is unknown and
2
So or 2475.4 to 2804.6.
So
or 2.4754 to 2.8046 (thousands).
2
252y0211 2/21/02
III. Do at least 3 of the following 4 Problems (at least 10 each) (or do sections adding to at least 30 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.
1. The population mean for similar patients to those mentioned on the previous page who had not used the new medicine was 2628. For your convenience the data are repeated below.
Patient capacity
1 2850
2 2380
3 2800
4 2860
5 2300
6 2650
7 2640
Test to see if the mean breathing capacity is now above 2628 using the sample mean and standard deviation you found in part II.
a. State the null and alternate hypothesis (2)
b. Find a critical value appropriate for this problem, using a confidence level of 90%.(3)
c. Use your critical value to test the hypothesis. State clearly whether you reject the null hypothesis. (2)
d. Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2). Each time state clearly whether you reject the null hypothesis and why.
e. Do a 90% two - sided confidence interval for the variance. (3)
f. (Extra credit) Assume that the data does not come from a normal distribution. (i) State a confidence interval for the median using the highest and lowest values and give the confidence level.(4) (ii) Do the same using the second highest and second lowest values and give the confidence level. (3)
Solution: From Table 3 of the Syllabus Supplement:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (s
Unknown) /
/ / /
a) You were asked to see if mean breathing capacity is now above 2628 (or 2.628 thousand). This gives us Since this does not contain an equality, it must be an alternative hypothesis, so we have and
b) Our facts, from the previous page are: or better, (or 0.0847 thousand). In addition, and Since our alternate hypothesis is and is one-sided, our one -sided critical value must be above 2628. Our value of is or 2.750 thousand .
c) Make a diagram . Show an almost Normal curve with a center at 2628 and a 10% 'reject' zone above 2750.0. Since is not in the 'reject' zone, do not reject .
252y0211 2/21/02
d) (i) The test ratio is Make a diagram . Show an almost Normal curve with a center at 0 and a 10% 'reject' zone above Since is not in the 'reject' zone, do not reject . (ii) The confidence interval will have the same direction as the alternate hypothesis, so it will be or . Make a diagram . Show an almost Normal curve with a center at 2640 and a shaded confidence interval above 2518. If you now represent by shading the area below 2628, you will have double-shaded the area between 2518 and 2628, so the confidence interval and the null hypothesis test do not contradict one another, and we do not reject .
e ) From the outline, the small-sample confidence interval is or from the formula table, . The table says and We know that or 0.050233 million. So
or . In millions, these limits would be 0.023 and 0.184.
f) (i) We can use the binomial tables to answer this. The numbers in order are 2300, 2380, 2640, 2650, 2800, 2850 and 2860. Since and the probability that all 7 numbers are above the median or that all seven numbers are below the median is So (ii) The probability that 6 or more number are above the median or six or more are below the median is So
252y0211 2/21/02
2. You have heard that, though the mean wealth held by US households is at least 270 (thousand dollars), the median wealth held by households is no more than 61 ( thousand dollars). You take a sample of 300 households in your state and find that the mean wealth is 210(thousand) with a sample standard deviation of 707(thousand), Out of the 300 households, 161 have wealth above 61 (thousand dollars). Use a 95% confidence level.
a. Test the statement about median wealth using a critical value. (4)
b. Test the same statement using a test ratio and a p-value (3)
c. Test the statement about the mean wealth using a critical value (3)
d. Test the same statement using a test ratio and a p-value. (3)
e. Do a 95% 2-sided confidence interval for the proportion of houses with wealth above 61 (thousand dollars.) (3)
f. How large a sample would I need, if I wanted the proportion in the confidence interval to be correct to (3)
Solution: a) The outline explains that hypotheses about a median are hypotheses about a proportion. The correspondence is below:
Hypotheses about Hypotheses about a proportion
a median If is the proportion If is the proportion
above below
Since we are told that the median wealth is no more than 61 and the proportion above 61 is 161 out of 300, Our hypotheses are
and Note that so that
From the formula table we have:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion /
/ / /
And we can also use the sign test formula, so we use
The standard deviation of the sample proportion is , so the critical value is Make a diagram of a normal curve with a mean at .5 and a reject zone above .5475. Since is not in the 'reject' zone, do not reject .
b) If we use a test ratio we get or To get a p-value use Since do not reject .
252y0211 2/21/02
c) Our hypotheses are now and . The problem says and so we can use in place of .
From the formula table we have:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (s
unknown) /
/ / /
Because of the alternate hypothesis, we want a critical value below 270, so we use Make a diagram of a normal curve with a mean at 270 and a reject zone below 202.85. Since is not in the 'reject' zone, do not reject .
d) If we use a test ratio we get To get a p-value use Since do not reject . It would also be acceptable to note that 1.470 is between and so that the p-value is between .05 and .10.
e) Using the confidence interval formula from a) above with and . or .484 to .590.
f) From the outline . So use 9553 or more. It would also be acceptable to use in this formula to get a slightly larger sample size.
Most of you did not state hypotheses in this problem, making grading it a nightmare!
252y0211 2/21/02
3. In the previous problem you tested the mean wealth held by US households is at least 270 (thousand dollars), and, based on the fact that the sample standard deviation is 707 (thousand dollars), created a critical value for the sample mean. Since the sample was so large, you can assume that 707 is the population standard deviation. ( The sample mean is still 210.)
a. Assume that the mean wealth is actually 260 (thousand dollars), and using the critical value you found on the last page, what is the power of the test? (3)
b. Find the power curve for your test showing the necessary calculations.(6)
c. Do a 2-sided 95% confidence interval for the mean in the problem on the last page and figure out how large a sample you would need to get the error part of the confidence interval down to (thousand dollars). (4)
d. Do a 2-sided 58% confidence interval for the mean, assuming that your sample of 300 came from a population of 1000. (3)
e. (Extra credit) If you place the 300 values in the sample in order, which numbers would you use to find a 95% confidence interval for the median? (4).
Solution: a) Our hypotheses were and . The problem said and so we can use in place of .
Because of the alternate hypothesis, we wanted a critical value below 270, so we used We will not reject if . This problem says that From the outline Then
Pathetic!
b) Half of the distance between 270 and 202.85 is, roughly, 34, so I found the power at and
Note that, in general, for this one-sided hypothesis .
From previous examples we know that
From previous examples, we know that at the critical value
You now have the power for 5 points above or equal to 270. Make a diagram.
252y0211 2/21/02
c) From the last question and . and Again, we are using in place of because of the large number of degrees of freedom. or 130.0 to 290.0. From the outline . So use a sample of size 19203 or larger.
d) The confidence level is 58% and . From the first page From the outline if use . Since and we compute or 182.3 to 237.7.
e) The formula from the outline was We will put the numbers in order and use the 134th from the bottom and the 134th from the top (the 167th number).
252y0211 2/21/02
4. According to Ronald Weirs, the Veterans Administration closed the cardiac units of VA hospitals that had mortality rates above 5%. In one hospital, their were 7 deaths in 102 operations (Consider this a sample proportion). Using a 99% confidence level, can you say that the hospital's mortality rate (proportion that died) was significantly above 5%? (Be specific about your significance level, especially in parts e and f.)
a. State your null and alternate hypotheses (2)
b. Do the problem using a critical value for the proportion that died. State clearly whether you reject the null hypothesis and why. Does this mean that the mortality rate was significantly above 5%? (3)
c. Do the problem using a test ratio. (2)
d. Do the problem using an appropriate confidence interval. (3)
e. According to Weirs, a high school running back can expect an average of 8 injuries per 100 games. A Poisson distribution applies. You believe that your league is particularly rough. In the last 100 games there were 15 injuries. Formulate a test of your belief and tell me if you are right. Explain why. (4)
f. If, instead, you had data for 1000 games, what would be a critical value for the number of injuries? (3)
Solution: a) In parts a) through d) we are doing tests on a proportion. The material from the formula table is quoted in question 2. The important words in the question for hypothesis formulation are " Can you say that the hospital's mortality rate (proportion that died) was significantly above 5%?" This translates as It is an alternate hypothesis because it does not contain an equality. The null hypothesis is thus The problem says that so that . This is a one-sided test and