ECO 252 Second Graded Assignment

252solngr2-08(revised!!!) 2/23/08

ECO 252 Second Graded Assignment

R. E. Bove

Name:

Class days and time:

Student Number:

Please include class and student number on what you hand in! Papers should be stapled. Your writeup should state clearly what you did and concluded. In diagrams of confidence intervals, shade the interval. In diagrams of test ratios and critical values for a sample statistic, shade the ‘reject zone.

Problem 1: Which of the following could be null hypotheses? Which could be an alternate hypothesis? Which could be neither? Why? If some of these are valid null hypotheses, state the alternate. If some of these are valid alternative hypotheses, state the null. Label and clearly. (i), (ii), (iii), (iv), (v), (vi), (vii), (viii) , (ix) (x)

Problem 2: A man walks into a bar. He drinks bottles of beer. These bottles are supposed to contain 12 ounces of beer with a population standard deviation of 0.4 ounces. On the basis of the man's condition when he leaves the bar, we conclude that the total amount he drank was ounces. Personalize the data by replacing by the second to last digit of your student number. If the second to last digit is zero, . For example, Ima Badrisk has student number 012345, so and she claims that he drank 14 bottles of beer and a total of . She then uses the total amount he drank and the total number of bottles to compute a sample mean for the bottles.

Test the hypothesis that the population mean for these bottles was 12 ounces. Assume that the confidence level is 95%.

a) State your null and alternative hypotheses.

b) Find critical values for the sample mean and test the hypothesis. Show your ‘reject’ region on a diagram.

c) Find a confidence interval for the sample mean and test the hypothesis. Show your results on a diagram.

d) Use a test ratio for a test of the sample mean. Show your ‘reject’ region on a diagram.

e) Find a p-value for the null hypothesis using your test ratio and the Normal table. Use the p-value to test the null hypothesis. Will you reject the null hypothesis with a confidence level of 10%, 5%, 1%, .002? No answer will be accepted without a brief explanation.

Problem 3: Continue with your results from problem 2.

a) Use the test ratio that you found on problem 1 to find an approximate p-value using the t-table and assuming that 0.4 ounces was a sample standard deviation? Does this change any of your results in e)?

b) Assume once again that 0.4 is a population standard deviation. Test the hypothesis that the population mean is 12 ounces. Show your ‘reject region on a diagram. Find a p-value.

Note: b) as written was a typo. It should have read ‘0.4 is a sample standard deviation. This error may appear in c) too. This is what I get for writing the solution before writing the problem!!!

c) Assume that 0.4 is a i) population ii) samplestandard deviation. Test the hypothesis that the population mean is above 12 ounces using the test ratio that you found in 1d). State your null and alternative hypotheses. Show your ‘reject region on a diagram.

d) Assume that 0.4 is a population standard deviation. Test the hypothesis that the population mean is above 12 ounces using a critical value for the sample mean. State your null and alternative hypotheses. Show your ‘reject region on a diagram.

e) Assume that 0.4 is a population standard deviation. Test the hypothesis that the population mean is above 12 ounces using a one-sided confidence interval. Shade the confidence interval in yourdiagram.

Problem 4: (Steinberg) Supposedly at least 36% of college students change their major between freshman year and graduation. You take a survey of 630 graduating seniors and find that changed their majors, where is the last digit of your student number. For example, Ima Badrisk has student number 12345, so and she claims that changed their majors. Test the hypothesis.

a) State your null and alternative hypotheses.

b) Find a test ratio for a test of the proportion.

c) Make a diagram showing the rejection region for the test ratio if you use a 99% confidencelevel.

d) Find a p-value for this ratio and use it to test the hypothesis at a 1% significance level

Extra Credit Problem 5:

a) Assume that the sample size is 20 2 changed majors and you are testing the statement that at least 40% changed majors. Do the test at the 90% confidencelevelwithout using the Normaldistribution.

The following data can be used in b) and c).

Make sure that your student number is easily readable. Choose one of the columns below using the second to last digit of your Student Number. Example: Ima Badrisk has student number 123456; so she picks column x5. Forget about the rest of the columns!

Row x0 x1 x2 x3 x4 x5 x6 x7 x8 x9

1 2.3 8.8 12.1 5.7 16.0 11.4 3.3 -2.5 6.7 3.0

2 7.9 2.9 3.6 5.7 3.5 -4.6 3.5 1.6 1.7 5.3

3 11.5 2.5 1.8 2.8 -1.7 4.7 12.2 2.7 2.0 1.1

4 2.1 6.5 9.9 2.9 6.8 -2.7 -1.6 -0.8 10.5 6.0

5 9.1 3.4 9.8 1.2 8.9 -3.6 7.1 -0.3 4.6 8.9

6 5.8 -3.6 6.8 6.8 4.1 -0.5 4.1 3.0 11.3 7.4

7 2.7 3.4 3.9 0.3 -2.9 -1.2 0.1 0.3 6.7 1.2

8 12.0 -1.5 9.9 9.6 15.5 -2.6 7.1 10.6 4.1 0.6

9 0.6 11.5 8.2 3.8 1.9 13.4 -1.5 4.9 5.2 4.2

10 -8.7 16.1 14.1 0.7 -0.9 0.5 1.0 3.3 3.7 7.1

11 4.1 4.6 6.8 9.3 3.0 3.2 6.2 14.8 9.9 5.0

12 -0.6 1.4 13.9 9.4 2.8 9.1 1.7 17.4 4.9 -2.5

13 -0.4 4.4 6.7 6.1 4.1 0.8 -2.3 9.5 9.7 8.8

b) Compute a sample standard deviation for your column and test the hypothesis

c) Using the same data, test the hypothesis that the median is below 4.

e) Use Minitab to check your answer to problem 4. Do this three ways.

First: Enter Minitab. Use the Editor pull-down menu to enable commands. Then enter the commands below.

Pone 630 220+b;(Replace 220+b with the number you used.)

Test 0.36;

Conf 99;(Sets a 99% confidence level)

Alter -1;(Makes H1 ‘less than.’)

useZ.(Uses normal approx. to binomial)

Second: Use the Stat pull-down menu. Choose ‘Basic Stat’ and then ‘1 proportion.’

Check ‘summarized data’ and enter your for ‘number of trials’ and 220+b for ‘number ofevents.’ Check ‘perform hypothesis test.’ Set ‘hypothesized proportion’ as 0.36. Press Options. Set‘confidence level’ as 99, alternative hypothesis as ‘less than’ and check ‘Normal distribution.’ Go.

Third: Use the pull-down menu again. But before you start put 220+b yeses and 410+b noes incolumn 1. (You should have 630 rows of data. You can do this entry very rapidly if you use only y or 1 for yes and n or 0 for no. Just enter y 220+b times and then n until you get to row 630. Even better, put a zero in the first row and use the fill handle on the bottom right of the cell to enter 1 an appropriate number of times then put a zero in the rest of the column the same way.) Uncheck ‘summarized data’ and let Minitab know that the data are in column 1 (C1). Other options are unchanged.

(Fourth: repeat your third try, but uncheck ‘Normal distribution.’)

Problem 1 Solution: Remember the following:

α) Only numbers like (the population mean, proportion, variance, standard deviation and median) that are parameters of the population can be in a hypothesis; (the sample mean, proportion, variance, standard deviation and median) are statistics computed from sample data and cannot be in a hypothesis because a hypothesis is a statement about a population;

β) The null hypothesis must contain an equality;

γ) must be between zero and one;

δ) A variance or standard deviation cannot be negative.

(i)can be since it contains a parameter and an equality. would be .

(ii)can be since it contains a parameter and an inequality. would be .

(iii) can be since it contains a parameter and an equality. would be .

(iv) could not be or because is a sample statistic, not a parameter.

(v)could not be or because is a sample statistic, not a parameter.

(vi)could not be or since it contains an unreasonable value for a parameter. A probability or proportion must be between zero and one.

(vii) could not be or because is a sample statistic, not a parameter.

(viii)can be since it contains a parameter and an inequality. would be .

(ix)could not be or since it contains an unreasonable value for a parameter. A variance or standard deviation must be positive.

(x)can be since it contains a parameter and an equality. would be .

Learn to make and call it ‘mu.’ It’s not a ‘u’ and you are too young to be unable to adjust to using a Greek letter!

Problem 2: A man walks into a bar. He drinks bottles of beer. These bottles are supposed to contain 12 ounces of beer with a population standard deviation of 0.4 ounces. On the basis of the man's condition when he leaves the bar, we conclude that the total amount he drank was ounces. Personalize the data by replacing by the second to last digit of your student number. If the second to last digit is zero, . For example, Ima Badrisk has student number 12345, so and she claims that he drank 14 bottles of beer and a total of . She then uses the total amount he drank and the total number of bottles to compute a sample mean for the bottles.

Test the hypothesis that the population mean for these bottles was 12 ounces. Assume that the confidence level is 95%.

a) State your null and alternative hypotheses.

Solution: The question asks us to test the hypothesis that the mean is 12 ounces. This is and must be a null hypothesis because it contains an equality. Our hypotheses are , so .

b) Find critical values for the sample mean and test the hypothesis. Show your ‘reject’ region on a diagram.

Solution: The general formula for a critical value for the sample mean when the population standard deviation is known is {Table 3}. Because this is a two-sided hypothesis, we need two critical values.

Computation of the mean

We have been told that . . Thus . We have been told that

Computation of Standard Errors

. For this would be .

Finding z

The significance level is given as 95%., and . From the bottom of the t-table, we find . {ttable}

The following table gives us the values of the sample size, the total amount of beer consumed, the sample mean and the standard error.

Row

1 11 128.5 11.68180.12060

2 12 142.0 11.8333 0.11547

3 13 155.5 11.9615 0.11094

4 14 169.0 12.0714 0.10690

5 15 182.5 12.1667 0.10328

6 16 196.0 12.2500 0.10000

7 17 209.5 12.3235 0.09701

8 18 223.0 12.3889 0.09428

9 19 236.5 12.4474 0.09177

10 20 250.0 12.5000 0.08944

For we have or we can say that the lower and upper critical values are and . We compare the sample mean of 11.6818. Since it lies below our two critical values, we can reject the null hypothesis.

For we have or we can say that the lower and upper critical values are and . We compare the sample mean of 12.5000. Since it lies above our two critical values, we can reject the null hypothesis.

Make a diagram. Center a Normal curve at 12. Mark the two critical values and . To represent the two ‘reject’ regions shade the area above and below . Indicate where your sample mean falls on the diagram. If your sample mean falls between and , do not reject the null hypothesis. If your sample mean does not fall between and , reject the null hypothesis. In both cases we can say that there is a significant difference between our sample mean and 12. This will not be true for all values of .

c) Find a confidence interval for the sample mean and test the hypothesis. Show your results on a diagram.

Solution: The general formula for a confidence interval for the population mean when the population standard deviation is known is {Table 3}. We have essentially reversed the sample and population means in the critical value formula, so we should get exactly the same results.

For we have or we can say . We compare the hypothesized population mean of 12. Since 12 lies above our confidence interval, we can reject the null hypothesis.

For we have or we can say . We compare the hypothesized population mean of 12. Since 12 lies below our confidence interval, we can reject the null hypothesis.

Make a diagram. Center a Normal curve at the sample mean. Mark the two limits of the confidence interval and shade the area between them. Indicate where your hypothesized population mean falls on the diagram. If your population mean falls within the confidence interval, do not reject the null hypothesis. If your sample mean does not fall within the confidence interval, reject the null hypothesis.

d) Use a test ratio for a test of the sample mean. Show your ‘reject’ region on a diagram.

Solution: The general formula for a test ratio for the population mean when the population standard deviation is known is {Table 3}. This is a rearrangement of the terms in critical value formula, so we should get exactly the same results.

For we have . We compare this with a 95% interval around zero, this interval is between and . Since -2.6385 lies below both values of z, we can reject the null hypothesis.

For we have . We compare this with a 95% interval around zero, this interval is between and . Since 3.9667 lies above both values of z, we can reject the null hypothesis.

e) Find a p-value for the null hypothesis using your test ratio and the Normal table. Use the p-value to test the null hypothesis. Will you reject the null hypothesis with a confidence level of 10%? 5%? 1%? .002 (0.2%)? No answer will be accepted without a brief explanation.

Solution: We already know that, if the population standard deviation is known, {Table 3}. For a two sided test, the p-value is defined as the probability of getting results as extreme or more extreme than those actually observed. If we call the value of that we actually obtained and is negative, . If is positive, . In either case, can be found on the Normaltable {norm}.

For we have. If we round this to 2.64 because of the limitations of Table 17, we find . We compare this with , , and . Since .0082 is below .10, .05 and .01, we reject the null hypothesis at the 90%, 95% and 99% confidence levels. But, since .0082 is above .002, we cannot reject the null hypothesis at the 99.8% confidence level.

For we have . If we round this to 5.59 because of the limitations of Table 17, we find . We compare this with , , and . Since 0 is below .10, .05, 01 and .002, we reject the null hypothesis at the 90%, 95%, 99% and 99.8% confidence levels. If we had a much better table, we could find a confidence level between 99.8% and 100% at which we could not reject the null hypothesis.

Though Minitab does allow you to state that the population variance is known, the routine I used assumed that 0.4 was a sample variance. Thus the values of z that it computed were correct, but the p-values were wrong. In the table below, Minitab computed z, but I did the p-value.

Row p-value

1 11 128.5 11.68180.12060 -2.64 .4959 .0082

2 12 142.0 11.8333 0.11547 -1.44 .4251 .1498

3 13 155.5 11.9615 0.11094 -0.35 .1368 .7264

4 14 169.0 12.0714 0.10690 0.67 .2794 .4412

5 15 182.5 12.1667 0.10328 1.61 .4463 .1074

6 16 196.0 12.2500 0.10000 2.50 .4938 .0124

7 17 209.5 12.3235 0.09701 3.33 .4996 .0008

8 18 223.0 12.3889 0.09428 4.12 .50000.0000

9 19 236.5 12.4474 0.09177 3.68 .4999 .0002

10 20 250.0 12.5000 0.08944 5.59 .50000.0000

To decide whether to reject the null hypothesis use the rule on p-value:

If the p-value is less than the significance level (alpha) reject the null hypothesis; if the p-value is greater than or equal to the significance level, do not reject the null hypothesis.

Problem 3: Continue with your results from problem 2. We are still testing if the mean is 12.

a) Use the test ratio that you found on problem 2 to find an approximate p-value using the t-table and assuming that 0.4 ounces was a sample standard deviation? Does this change any of your results in 2e)?

Solution: If 0.4 is a sample standard deviation, and and . {Table 3}. Our hypotheses are , so . Degrees of freedom are . The relevant part of the t-table follows.{ttable}.

.45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001

10 0.129 0.260 0.397 0.542 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 4.144

11 0.129 0.260 0.396 0.540 0.697 0.876 1.088 1.363 1.796 2.201 2.718 3.106 4.025

12 0.128 0.259 0.395 0.539 0.695 0.873 1.083 1.356 1.782 2.179 2.681 3.055 3.930

13 0.128 0.259 0.394 0.538 0.694 0.870 1.079 1.350 1.771 2.160 2.650 3.012 3.852

14 0.128 0.258 0.393 0.537 0.692 0.868 1.076 1.345 1.761 2.145 2.624 2.977 3.787

15 0.128 0.258 0.393 0.536 0.691 0.866 1.074 1.341 1.753 2.131 2.602 2.947 3.733

16 0.128 0.258 0.392 0.535 0.690 0.865 1.071 1.337 1.746 2.120 2.583 2.921 3.686

17 0.128 0.257 0.392 0.534 0.689 0.863 1.069 1.333 1.740 2.110 2.567 2.898 3.646

18 0.127 0.257 0.392 0.534 0.688 0.862 1.067 1.330 1.734 2.101 2.552 2.878 3.611

19 0.127 0.257 0.391 0.533 0.688 0.861 1.066 1.328 1.729 2.093 2.539 2.861 3.579

20 0.127 0.257 0.391 0.533 0.687 0.860 1.064 1.325 1.725 2.086 2.528 2.845 3.552

If , we have 11 – 1 = 10 degrees of freedom . . For a two sided test, the p-value is defined as the probability of getting results as extreme or more extreme than those actually observed. If we call the value of that we actually obtained and is negative . To find values of , look at the row of the t-table. Note that , and that 2.6385 is between them. Since the table is telling us that, for 10 degrees of freedom, and , we can conclude that . If we double this, we have . We compare this with , , and . Since our p-value is below is below .10 and .05 but above 01 and .002, we reject the null hypothesis at the 90% and 95% confidence levels, but do not reject the null hypothesis at the 99% and 99.8% confidence levels.

Row This is between Approximate p-value

1 11 11.68180.12060 -2.64 and

2 12 11.8333 0.11547 -1.44 and

3 13 11.9615 0.11094 -0.35 and

4 14 12.0714 0.10690 0.67 and

5 15 12.1667 0.10328 1.61 and

6 16 12.2500 0.10000 2.50 and

7 17 12.3235 0.09701 3.33 and

8 18 12.3889 0.09428 4.12 and nothing

9 19 12.4474 0.09177 3.68 and nothing

10 20 12.5000 0.08944 5.59 and nothing

I got lucky. The excessively rounded values of t produced by Minitab were still distinct from the values from the t-table. To decide whether to reject the null hypothesis use the rule on p-value:

If the p-value is less than the significance level (alpha) reject the null hypothesis; if the p-value is greater than or equal to the significance level, do not reject the null hypothesis.

To get greater accuracy, I ran this on the computer. Notice that the p-values are in the expected range.

MTB > Onet 11 11.6818 .4;

SUBC> Test 12.

One-Sample T

Test of mu = 12 vs not = 12