Solution to Graded Assignment 2
1. Which of the following could be null hypotheses? Which could be an alternate hypothesis? Which could be neither? Why? If it is , what is ? If it is , what is ? (i) (ii) (iii) (iv) (v)(vi) (vii)(viii) (ix), (x) .
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’t be either or since the sample proportion is not a parameter and a proportion can’t be above 1.
(ii) can’t be either since a proportion can’t be negative
(iii) could be since it contains a parameter and an inequality. would be .
(iv) can’t be either or since the sample mean is not a parameter.
(v) could be since it contains a parameter and an equality. would be .
(vi) can’t be either since the sample standard deviation is not a parameter.
(vii) can’t be or because the population standard deviation can’t be negative.
(viii) can be because it contains a parameter and an inequality. would be .
(ix) could be since it contains a parameter and an equality. would be .
(x) . can be because it contains a parameter and an inequality. would be
Before we start hypothesis testing here are three important points:
α) You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion.
β) It is not enough to know what a critical value is. You should be able to explain on what side of the critical value the ‘reject’ zone lies.
γ)The rule on p-value says that 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.
δ) Make sure that I know what formulas you are using.
2. Make sure that I know what formulas you are using.
Seymour Butz has invented a new golf ball. He claims that it will increase your distance off the tee by more than 20 feet. 40 golfers test the ball and the additional distances that the balls go are given on the next page. Use a 10% significance level. Is the ball a success?
a) State your null and alternative hypotheses.
b) Find critical values for the sample mean and test the hypothesis.
c) Find a confidence interval for the population mean and test the hypothesis.
d) Use a test ratio for a test of the mean
e) Find an approximate p-value for the test ratio using the table and use the p-value to test the hypothesis.
Solution: We were given the following data: Data for Problem 2.
7
Row
1 13.37 178.757
2 28.72 824.838
3 12.60 158.760
4 30.77 946.793
5 24.00 576.000
6 20.67 427.249
7 24.12 581.774
8 16.95 287.303
9 23.73 563.113
10 20.92 437.646
11 17.51 306.600
12 30.44 926.594
13 20.15 406.023
14 24.60 605.160
15 24.11 581.292
16 29.91 894.608
17 12.27 150.553
18 27.75 770.063
19 23.25 540.563
20 26.40 696.960
21 14.01 196.280
22 28.86 832.900
23 23.34 544.756
24 25.00 625.000
25 17.01 289.340
26 30.50 930.250
27 25.16 633.026
28 13.40 179.560
29 22.03 485.321
30 18.99 360.620
31 21.61 466.992
32 27.01 729.540
33 26.59 707.028
34 30.22 913.248
35 17.06 291.044
36 26.16 684.346
37 24.13 582.257
38 16.45 270.603
39 18.25 333.063
40 21.99 483.560
900.01 21399.380
7
Given the nature of the data, it seems that we need to test thee mean, so we look at our formula table under means.
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (s
known) / / / /
Mean (s
unknown) /
/ / /
a) State your null and alternative hypotheses: What the problem says: ‘Seymour Butz has invented a new golf ball. He claims that it will increase your distance off the tee by more than 20 feet. 40 golfers test the ball and the additional distances that the balls go are given on the next page. Use a 10% significance level. Is the ball a success?’ -----This definitely implies that ” goes somewhere, but it does not contain an equality, so it must be an alternate hypothesis. The other clue is that the alternate hypothesis is often an ‘action’ hypothesis, while the null hypothesis generally says ‘keep things as they are.’ So the null hypothesis is and the alternate hypothesis is .
What else the problem says: A sample of golfers is taken. . The following information was computed for you. and .
So and This implies that (It makes no sense to put 22.50 in the hypotheses when you know is true.)
b) Find critical values for the sample mean and test the hypothesis: Degrees of freedom are We must use because we do not know . Because the alternate hypothesis is , we need a critical value for above 20. Since this means a one-sided test, we use The two-sided formula , , becomes . Make a diagram showing an almost Normal curve with a mean at 20 and a shaded 'reject' zone above 21.119. Since is above 21.119, we reject .
c) Find a confidence interval for the population mean and test the hypothesis: Because the alternate hypothesis is we need a confidence interval. becomes or . Make a diagram showing an almost Normal curve with a mean at and the confidence interval above 21.381 shaded. Since is below 21.054 and thus not in the confidence interval, we reject .
d) Use a test ratio for a test of the mean: The test ratio is . Since this is a one-sided, right-tail test, pick from the table. The ‘reject’ zone is the area above 1.304. Since 2.913 is in the ‘reject’ zone, reject the null hypothesis.
e) Find an approximate p-value for the test ratio and use the p-value to test the hypothesis: . On the 39 line of the t-table, note that 2.913 is between 2.708 and 3.313. Because 2.708 is in the .005 column and 3.313 is in the .001 column, the table is telling us that and or , so which means Since these p-values are below the 10% significance level, reject .
Since we rejected the null hypothesis the ball is a success.
Note: Many people incorrectly said that the null hypothesis was , so that the alternative was . If this had been correct, the critical value would be below 20. Nevertheless, some concluded that they should reject their null hypothesis. I don’t think that you need to know any statistics at all to see that cannot possible be evidence against
3. Assume, in problem 2, that the population standard deviation was 10ft. State the test ratio and find its p-value. Using this p-value, would we reject the null hypothesis a) If the significance level is 1%, b) If the significance level is 5% and c) If the significance level is 10%?
Solution: Note!!! The only thing that has changed from problem 2 is that the sample standard deviation has been replaced by the population standard deviation of 10 – not the sample mean, not the hypotheses. The only reason for this section was because some find e) above difficult, and this is easier since you can use the Normal table.
We still have the null hypothesis is and the alternate hypothesis .
What else the problem says: Assume, in problem 2, that 10 was a population standard deviation. Everything else is unchanged, except that there is no significance level at first. So and But and We still have a right-tailed test. The table says that the test ratio is So If we use the Normal table, we find . To do the last part of this problem, make a diagram of the Normal distribution with a mean of zero and shade the area above 1.58.
a) If since the p-value of .0571 is not below the significance level, we do not reject .
b) If since the p-value of .0571 is above the significance level, we do not reject .
c) If since the p-value of .0571 is below the significance level, we reject .
You did not answer this question if you did the problem 3 different ways and never found the p-value!
4. Seymour is elected the mayor of the Borough of Pineapplnham and boasts to a reporter that at least 35% of the town has internet access. The reporter takes a fast survey and finds that 13 out of a sample of 53 have internet access.
a) State your null and alternative hypotheses.
b) Find a test ratio for a test of the proportion
c) Find a p-value for the test ratio and use the p-value to test the hypothesis at the 5% significance level.
d) Restate the problem as a two sided hypothesis, find a p-value for the null hypothesis and use it to test your hypothesis at a 5% significance level.
Solution: Since 35% is a proportion, we look up the test for a proportion in the formula table.
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion /
/ / /
a) What the problem says: ‘Seymour …. boasts to a reporter that at least 35% of the town has internet access’ 35% is a proportion. There is nothing here about the mean. The problem implies that the reporter is skeptical. Since ‘at least 35% is , there is an equality in this question, it must be an alternate hypothesis. So the null hypothesis is and the alternate hypothesis is What else the problem says: A sample of 53 households found that 13 had internet access. So and . This implies that and
b) Find a test ratio for a test of the proportion: The table says and This means that . (It makes no sense to use the test ratio, , if you have already said the null hypothesis is about the mean.)
c) Find a p-value for the test ratio and use the p-value to test the hypothesis at the 5% significance level: Note that the alternate hypothesis says We are worried about the proportion being too low. We use the Normal table to find that the . So, if since the p-value is above the significance level, we do not reject and cannot call Seymour a liar.
d) Restate the problem as a two sided hypothesis, find a p-value for the null hypothesis and use it to test your hypothesis at a 5% significance level: The hypotheses are and Since our value of remains the same, . So, if since the p-value is above the significance level, we do not reject .
5. (Extra credit) a) Finish 4 by testing the original hypothesis using a critical value for the proportion and an appropriate confidence interval.
b) Were we reasonable assuming that in question 3? Use the sample standard deviation that you found in question 3 to test this using a test ratio and a confidence interval.
Solution: Note: All of you should be able to do this problem using a critical value for or a confidence interval.
Critical Value: a) The table says , but this is a one-sided test and we want one critical value below so we use . Make a diagram showing a Normal curve centered at and a shaded 'reject' zone below .2422. Since is not in the ‘reject’ zone, we cannot reject .
Confidence Interval: The table says , where, because and , . Since implies a 1-sided test we want a 1-sided confidence interval in the same direction as the alternate hypothesis. This would be . Make a diagram showing a Normal curve centered at .2453 with the confidence interval below .3424 shaded. Since is above .3424 and thus not in the confidence interval, we reject . This is another one of those rare cases where the confidence interval contradicts the test ratio and critical value results.
b) Were we reasonable assuming that in question 3? Use the sample standard deviation that you found in question 3 to test this using a test ratio and a confidence interval: Go back to the formula table.
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueVariance-
Small Sample / / / /
Variance-
Large Sample / / / /
Test Ratio: Remember that and Degrees of freedom are This is a 2-sided test of and You cannot find for 39 degrees of freedom on your table, so use
. . Your diagram would show a Normal curve with a center at zero and ‘reject’ zones above and below . Since -8.2956 is in the lower ‘reject’ zone, reject .