MA 180 (Statistics) - Chap.11 Homework

Winning team data were collected for teams in different​ sports, with the results given in the table below. Use the​ TI-83/84 Plus results at a 0.05 level of significance to test the claim that​ home/visitor wins are independent of the sport. ___________________________________________ ​TI-83/84 PLUS χ2−Test χ2=2.178552556 P=0.5361851944 df=3 ___________________________________________ ............................Basketball............ Baseball............ Hockey............Football Home team wins.......125...................... 70 .....................68 ...................66.... Visiting team wins.......80 .......................53 .....................51 ...................35...

(Fail to reject) the null hypothesis that​ home/visitor wins are independent of the sport. It appears that the​ home-field advantage (does not) depend on the sport.

A random sample of 784 subjects was asked to identify the day of the week that is best for quality family time. Consider the claim that the days of the week are selected with a uniform distribution so that all days have the same chance of being selected. The table below shows​ goodness-of-fit test results from the claim and data from the study. Test that claim. Num Categories -----> 7 Test​ statistic, χ^2 -----> 12.411 Degrees of freedom -----> 6 Critical χ^2 -----> 12.592 Expected Freq -----> 112.0000 ​P-Value -----> 0.0534 A. Determine the null and alternative hypotheses. B. Identify the test statistic. C. Identify the critical value. D. State Conclusion

A. H0​: All days of the week have an equal chance of being selected. H1​: At least one day of the week has a different chance of being selected. B. χ2 = 12.411 C. χ2= 12.592 D. (Fail to reject)H0. There (is not) sufficient evidence to warrant rejection of the claim that the days of the week are selected with a uniform distribution. It (does appear) that all days have the same chance of being selected.

The accompanying table summarizes successes and failures when subjects used different methods when trying to stop smoking. The determination of smoking or not smoking was made five months after the treatment was begun. If we test the claim that success is independent of the method​ used, the technology provides a​ P-value of 0.003(rounded). What does the​ P-value tell us about that​ claim? ..............Nicotine Gum.....Nicotine Patch.....Nicotine Inhaler Smoking........185...................275.................... 97............ Not Smoking....56.....................58.................... 44...........

ANSWERS ARE IN BRACKETS: Because the​ P-value of 0.003 [is] small​ (such as 0.05 or​ lower), [reject] the null hypothesis of independence between the treatment and whether the subject stops smoking. This suggests that the choice of treatment [appears to make a difference].

The table below lists days of the week selected by a random sample of 1007 subjects who were asked to identify the day of the week that is best for quality family time. Consider the claim that the days of the week are selected with a uniform distribution so that all days have the same chance of being selected. If we test the claim using the​ goodness-of-fit test, what is actually​ tested? Sun 529 Mon 19 Tues 9 Wed 21 Thurs 12 Fri 44 Sat 373 Choose the correct answer below. A. The test is to determine whether the observed frequency counts agree with the claimed​ chi-square distribution so that the frequencies for at most three days are equally likely. B. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for the different days are equally likely. C. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for at least two days are equally likely. D. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for only two days are equally likely.

B. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that the frequencies for the different days are equally likely.

For a recent​ year, the following are the numbers of homicides that occurred each month in a city. Use a 0.05 significance level to test the claim that homicides in a city are equally likely for each of the 12 months. Is there sufficient evidence to support the police​ commissioner's claim that homicides occur more often in the summer when the weather is​ better? Month Number Jan. 37 Feb. 30 March 46 April 39 May 47 June 48 July 47 Aug. 49 Sep. 49 Oct. 41 Nov. 38 Dec 38 Determine the null and alternative hypotheses. Calculate the test statistic, χ². Calculate the​ P-value. What is the conclusion for this hypothesis​ test? Is there sufficient evidence to support the police​ commissioner's claim that homicides occur more often in the summer when the weather is​ better?

Determine the null (H0) and alternative (H1) hypotheses H0​: Homicides occur with equal frequency in the different months. H1​: At least one month has a different frequency of homicides than the others. Calculate the test statistic, χ² n= 37+30+46+39+47+48+47+49+49+41+38+38 = 509 E = (1/12)(509) = 42.417 χ²= ∑((O-E)^2/E) = 9.64 Calculate the​ P-value. if you don't have StatCrunch, use http://www.socscistatistics.com/pvalues/chidistribution.aspx and input The Test Statistic and the Degrees of freedom to find... P-value = 0.5630 What is the conclusion for this hypothesis​ test? Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that homicides in a city are equally likely for each of the 12 months. Is there sufficient evidence to support the police​ commissioner's claim that homicides occur more often in the summer when the weather is​ better? There is not sufficient evidence to support the police​ commissioner's claim that homicides occur more often in the summer when the weather is better.

The table below includes results from polygraph​ (lie detector) experiments conducted by researchers. In each​ case, it was known if the subjected lied or did not​ lie, so the table indicates when the polygraph test was correct. Use a 0.05 significance level to test the claim that whether a subject lies is independent of the polygraph test indication. Do the results suggest that polygraphs are effective in distinguishing between truth and​ lies? .................................................Did the Subject Actually Lie?......... ........................................No (Did Not Lie)........................ Yes (Lied) Polygraph indicated lie............15...........................................48...... Polygraph indicated no lie.......23...........................................12...... Determine the null and alternative hypotheses. Determine the test statistic. Determine the P-value of the test statistic. Do the results suggest that polygraphs are effective in distinguishing between truth and​ lies?

Determine the null and alternative hypotheses. H0: Whether a subject lies is independent of the polygraph test indication. H1​: Whether a subject lies is not independent of the polygraph test indication. Determine the test statistic. ______________________________________________________ StatCrunch instructions: 1. Stat 2. Tables 3. Contingency 4. With Summary 5. Select column(s) --> "Did Not Lie" and "Lied" 6. Row labels --> "_" 7. Hypothesis tests: --> "Chi-square test for independence 8. COMPUTE ______________________________________________________ Test statistic is χ2 = 16.643 Determine the P-value of the test statistic. (found in StatCrunch results) P-value = 0.0001 Do the results suggest that polygraphs are effective in distinguishing between truth and​ lies? There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.

Results from a civil servant exam are shown in the table to the right. Is there sufficient evidence to support the claim that the results from the test are​ discriminatory? Use a 0.05 significance level. .............................Passed.................. Failed White candidates......17.........................19... Minority candidates...11.........................24... Determine the null and alternative hypotheses. Determine the test statistic. Is there sufficient evidence to support the claim that the results from the test are​ discriminatory?

Determine the null and alternative hypotheses. H0: White and minority candidates have the same chance of passing the test H1: White and minority candidates do not have the same chance of passing the test. Determine the test statistic. ____________________________________ StatCrunch instructions: 1. Stat 2. Tables 3. Contingency 4. With Summary 5. Select column(s) --> "Passes" and "Failed" 6. Row labels --> "_" 7. Hypothesis tests: --> "Chi-square test for independence 8. COMPUTE ____________________________________ Test statistic is χ2 = 1.853 Determine the P-value of the test statistic. (found in StatCrunch results) P-value = 0.1734 Is there sufficient evidence to support the claim that the results from the test are​ discriminatory? There is not sufficient evidence to support the claim that the results are discriminatory.

Many people believe that criminals who plead guilty tend to get lighter sentences than those who are convicted in trials. The accompanying table summarizes randomly selected sample data for defendants in burglary cases. All of the subjects had prior prison sentences. Use a 0.05 significance level to test the claim that the sentence​ (sent to prison or not sent to​ prison) is independent of the plea. If you were an attorney defending a guilty​ defendant, would these results suggest that you should encourage a guilty​ plea? ___________________________________________ ...............................Guilty Plea.........Not Guilty Plea Sent to Prison................373.....................59............ Not Sent to Prison.........566......................20........... ___________________________________________ Determine the test statistic. Determine the P-value of the test statistic. Use a 0.05 significance level to test the claim that the sentence​ (sent to prison or not sent to​ prison) is independent of the plea. If you were an attorney defending a guilty​ defendant, would these results suggest that you should encourage a guilty​ plea?

H0: The sentence​ (sent to prison or not sent to​ prison) is independent of the plea. H1​: The sentence​ (sent to prison or not sent to​ prison) is not independent of the plea Determine the test statistic. ____________________________________ StatCrunch instructions: 1. Stat 2. Tables 3. Contingency 4. With Summary 5. Select column(s) --> "Sent" and "Not Sent" 6. Row labels --> "_" 7. Hypothesis tests: --> "Chi-square test for independence 8. COMPUTE ____________________________________ Test statistic is χ2 = 36.460 Determine the P-value of the test statistic. (found in StatCrunch results) P-value = 0.0001 Use a 0.05 significance level to test the claim that the sentence​ (sent to prison or not sent to​ prison) is independent of the plea. If you were an attorney defending a guilty​ defendant, would these results suggest that you should encourage a guilty​ plea? There is sufficient evidence to warrant rejection of the claim that the sentence is independent of the plea. The results encourage pleas for guilty defendants.

Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively: 25​, 29, 42​, 38​, 27​, 39. Use a 0.025 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die? The test statistic is..? The critical value is..? The P-value is...?

H0: p0+p1+p2+p3+p4+p5 H1: At least one p does not match the claimed value The test statistic is..? O*= Observed E* = Expected n = ∑O = 200 (make this chart↓↓) Expected fraction(p) - O* - E*=np - (O-E)²/E 1/6 25 33.333 2.083 1/6 29 33.333 0.563 1/6 42 33.333 2.253 1/6 38 33.333 0.653 1/6 27 33.333 1.203 1/6 39 33.333 0.963 χ²= ∑((O-E)^2/E) = 7.718 The critical value is..? (use the Chi-squared distribution table) match 6-1 = 5 degrees of freedom with α=0.025 critical value = 12.833 The P-value is... if you don't have StatCrunch, use http://www.socscistatistics.com/pvalues/chidistribution.aspx and input The Test Statistic and the Degrees of freedom and α=0.05 to find... P-value = 0.1725 State the conclusion. (Do not reject) H0. There (is not) sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes (appear) to be equally​ likely, so the loaded die (does not appear) to behave differently from a fair die.

Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person purchased a slot machine and tested it by playing it 1,207 times. There are 10 different categories of​ outcomes, including no​ win, win​ jackpot, win with three​ bells, and so on. When testing the claim that the observed outcomes agree with the expected​ frequencies, the author obtained a test statistic of χ2=8.599. Use a 0.025 significance level to test the claim that the actual outcomes agree with the expected frequencies. Does the slot machine appear to be functioning as​ expected?

The test statistic is : 8.599 The critical value is : (K - one degree of freedom and the given significance level) (K = the number of different categories or outcomes.) K = 10 K -1 = 9 (Use the​ chi-square ​(χ2​) distribution table to find the critical value with 9 degrees of freedom and a significance level of α = 0.05​, rounding to three decimal places.) Critical Value χ2​ = 19.023 The P-value is : if you don't have StatCrunch, use (http://www.socscistatistics.com/pvalues/chidistribution.aspx) P-value = 0.4751 (Do not reject) H0. There (is not) sufficient evidence to warrant rejection of the claim that the observed outcomes agree with the expected frequencies. The slot machine (appears) to be functioning as expected.

Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the conclusion. A person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Use a 0.10 significance level to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first​ category, but do the results support that​ expectation? Cents portion of check......Number ............0-24...........................32.... ...........25-49..........................20.... ...........50-74..........................24.... ...........75-99..........................24.... The test statistic is ___? The critical value is ___ ? The P-value is ___ ? State the conclusion.

The test statistic is... *O= Observed, *E= Expected Total (100)÷(number of columns) = Expected = 25 (make this chart ↓) _________________________________ Cents... O* E* (O-E) (O-E)² (O-E)²/E --------- ------ ------ ------- ---------- ------------- 0 - 24 | 32 | 25 | 7 | 49 | 1.96 | 24 - 49| 20 | 25 | -5 | 25 | 1 | 50 - 74| 24 | 25 | -1 | 1 | 0.04 | 75 - 99| 24 | 25 | -1 | 1 | 0.04 | Test statistic = ∑ ((O-E)²/E) = χ²= 3.04 The critical value is... "k" is the number of columns k-1= degrees of freedom = 3 Use the chi-square table to find that α = 0.10 and degrees of freedom = 3 gives... Critical value is χ²= 6.251 The P-value is... if you don't have StatCrunch, use http://www.socscistatistics.com/pvalues/chidistribution.aspx and input The Test Statistic and the Degrees of freedom to find... P-value = 0.3855 State the conclusion. (Do not reject) H0. There (is not) sufficient evidence to warrant rejection of the claim that the four categories are equally likely. The results (do not appear) to support the expectation that the frequency for the first category is disproportionately high.

The table below summarizes results for randomly selected drivers stopped by police in a recent year. Using​ technology, the data in the table results in the statistics that follow. .............................Black and​ Non-Hispanic............................White and​ Non-Hispanic Stopped by police...................37............................................................135.............. Not stopped by police...........198..........................................................1179............... ​chi-square statistic = 6.044, degrees of freedom =​ 1, ​P-value = 0.014 Use a 0.05 significance level to test the claim that being stopped is independent of race. Based on available​ evidence, can we conclude that racial profiling is being​ used? Can we conclude that racial profiling is being​ used?

Yes, because the P-value is less than the significance level.