Week 3 Assignment PDF

Preview

Summary

BST 322...

Description

Week 3 Assignment

1.

(3 pts) Using StatCrunch, construct a labeled contingency table (with row, column, and total %s), as well as the contribution to the chi-square statistic, for the following set of data for 300 people:

Had flu shot Didn’t have flu shot Total

Group A 20 80 100

Group B 30 70 100

Group C 38 62 100

Total 88 212 300

Is the value of the chi-square statistically significant? Contingency table results: Rows: Had flu shot Columns: None

Cell format Count (Row percent) (Column percent) (Percent of total) Had Flu Shot?

Group A Group B Group C

Yes % Row % Column %Total No % Row % Column %Total Total

20 30 38 (22.73%)(34.09%) (43.18%) (20%) (30%) (38%) (6.67%) (10%) (12.67%) 80 70 62 (37.74%)(33.02%) (29.25%) (80%) (70%) (62%) (26.67%)(23.33%) (20.67%) 100 100 100 (33.33%)(33.33%) (33.33%) (100%) (100%) (100%) (33.33%)(33.33%) (33.33%)

Chi-Square test:

Statistic DF Value P-value Chi-square 27.8473413 0.0198

The value of the chi-square is statistically significant because the p-value is less than .05. Meaning we can reject the null hypothesis of the relationship between getting the flu shot and not.

2. (1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 1. Research shows that most people do not get the influenza vaccine. This study was conducted to identify the difference among the three groups of 300 people who did or did not get the flu shot. The null hypothesis I tested is that there is a difference between getting a flu not and not getting one. The alternative hypothesis is that there is no difference. The results were significant because the p-value is less than .05. The null hypothesis was also rejected which means there is no difference between the groups getting their flu shot and not.

3.

(3 pts) Using StatCrunch, construct a labeled contingency table (with row, column, and total %s), as well as the contribution to the chi-square statistic, for the following set of data for 180 people undergoing a knee replacement treatment with a drug supplement: Treatment with drug X Treatment without Drug X Had > 8 wk rehab 18 30 Had < 8 wk rehab 70 62 Total 88 92 Is the value of the chi-square statistically significant at the 0.05 level?

Contingency table results: Rows: Rehab Columns: None

Cell format Count (Row percent) (Column percent) (Percent of total) Treatment Rehab

Treatment w/ Drug X Treatment w/o Drug X Total 48 18 30 (37.5%) (62.5%) (100%) (20.45%) (32.61%) (26.67%) (16.67%) (26.67%) (10%) 132 62 70 Had 8wk rehab %within amount of rehab %within treatment

Chi-Square test:

Statistic DF Value P-value Chi-square 13.3976374 0.0653

Total 48 132 180

The chi-square is not significant because the p-value is greater than 0.05. This means we have to accept the null hypothesis. 4.

(1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 3.

This study tested a sample of people getting a knee replacement to see if drug X had an impact on how much time they spent in rehab, whether it was more than 8 weeks or less than 8 weeks. In conclusion, we have to accept the null hypothesis because the data is not statically significant due to the p-value being greater than 0.5. Based off of this study, there is no relationship between time in rehab and drug X.

5.

a. b. 6.

(1 pt) Given each of the following circumstances, determine whether the calculated values of chi-square are statistically significant: χ2 = 3.02, df = 1, α = 0.05= NS χ2 = 8.09, df = 2, α = 0.05= S

c. d.

χ2 = 12.67, df = 3, α = 0.01= S χ2 = 9.01, df = 2, α = 0.01= NS

(1 pt) Match each of the nonparametric tests in Column A with its parametric counterpart in Column B A. Nonparametric Test B. Parametric Test 1. Mann-Whitney U-test: C a. Paired t-test 2. Friedman test: D b. One-way ANOVA 3. Kruskall-Wallis test: B c. Independent groups t-test 4. Wilcoxon signed-ranks test: A d. Repeated measures ANOVA

7.

(1 pt) Using the information provided, indicate which statistical test you think should be used for each of the following situations (a-d):

a.

Independent variable: normal birth weight vs. low birth weight infants; dependent variable: breathing rate (in breaths per minute): Mann-Whitney Independent variable: time of measurement of same patient (before and after surgery); dependent variable: heart rate: T-test Independent variable: time of measurement (before, during, and after intervention); dependent variable: did vs did not exercise regularly: Cochran’s Q Independent variable: infertility treatment A vs infertility treatment B vs control condition; dependent variable: did vs did not become pregnant: Chi-square

b. c. d.

8.

(3 pts) The relationship between cigarette smoking and major depressive disorder has been studied for decades. In the assignment area, you will find an excel spreadsheet of data from a study on smoking and depression by the St Louis Epidemiologic Catchment Area Survey of the National Institute of Mental Health. The file is labeled “Catchment Area Survey”. Using this data set in StatCrunch, calculate the chisquare statistic and degrees of freedom for the items “smoker” and a measure of depression, “FeltDown”. Construct a labeled contingency table for this data as you did for Exercise 1 and 3 above. Be sure to include and label the row, column and total %’s as well as the expected counts. Is the value of the chisquare statistically significant at the 0.05 level? Any reason to use a correction to the chi-square test here given your expected counts?

Contingency table results:

Rows: FeltDown Columns: smoker

Cell format Count (Row percent) (Column percent) (Percent of total) (Expected count) Smoker? Felt Down

No

Yes Total 8 22 30 Most of the time (26.67%)(73.33%) (100%) %within Felt Down %within Smoker Status (20%)(37.29%) (30.3%) %Total (8.08%) (22.22%) (30.3%) Expected Count (12.12) (17.88) None of the time 12 13 25 %within Felt Down (48%) (52%) (100%) %within Smoker Status (30%)(22.03%) (25.25%) %Total (12.12%)(13.13%) (25.25%) Expected Count (10.1) (14.9) 44 20 24 Some of the time (45.45%)(54.55%) (100%) %within Felt Down %within Smoker Status (50%)(40.68%) (44.44%) %Total (20.2%) (24.24%) (44.44%) Expected Count (17.78) (26.22) 40 59 99 Total %within Felt Down (40.4%) (59.6%) (100%) %within Smoker Status (100%) (100%) (100%) %Total (40.4%) (59.6%) (100%) Chi-Square test:

Statistic DF Value P-value Chi-square 23.416339 0.1812

The value of the chi-square is not significant because the p-value is greater than 0.05. We don’t need to use a correction test because all of the expected counts are above 10.

9. (1 pt) Write a brief paragraph summarizing the results of the analysis in Exercise 8. What can you conclude about smoking and depression (“Feeling Down”) in this sample? This study was conducted to testify whether there is a relationship between smoking and depression. With the data provided we have to accept the null hypothesis because the p-value is greater than 0.05. Based on these results there is no statistical significance and there is no relationship between smoking and depression.

10. (4 pts) Using the data set from this week’s discussion thread, question 2 from the Pew Research Study on Aging (Nashville), load the data set into StatCrunch and calculate the chi-square statistic and degrees of

freedom for the items “Marital” and a measure of dementia, “Forgot date”. Construct a labeled contingency table for this data as you did for Exercise 8 above. Be sure to include and label the row, column and total %’s as well as the expected counts. Is the value of the chi-square statistically significant at the 0.05 level? Any reason to use a correction to the chi-square test here given your expected counts? What hypotheses did you test and what can you conclude about marriage status and dementia in this sample? Contingency table results: Rows: Forgot Date Columns: Marital

Cell format Count (Row percent) (Column percent) (Percent of total) (Expected count) Marital Status Forgot Date

Married Never married Total No 108 97 205 %within Forgot Date (52.68%) (47.32%) (100%) %within Marital Status (56.54%) (46.41%) (51.25%) %Total (27%) (24.25%) (51.25%) Expected Count (97.89) (107.11) 83 112 195 Yes (57.44%) (100%) %within Forgot Date (42.56%) (53.59%) (48.75%) %within Marital Status (43.46%) (20.75%) (28%) (48.75%) %Total Expected Count (93.11) (101.89) 191 209 400 Total (52.25%) (100%) %within Forgot Date (47.75%) %within Marital Status (100%) (100%) (100%) %Total (47.75%) (52.25%) (100%) Chi-Square test:

Statistic DF Value P-value Chi-square 14.1013697 0.0428

Null Hypothesis: There is no relationship between marital status and dementia Alternative Hypothesis: There is a relationship between marital status and dementia. The chi-square is significant because the p-value is less than 0.05. There is no reason to use a correction to the chi-square because the expected counts are all above 10. We can reject the null hypothesis, meaning there is a difference between marital status and dementia.

11. (1 pt) Below are two (a and b) sets of cells of 2 X 2 contingency tables including the expected frequencies (Totals not included). Identify which statistical procedure would be appropriate for each, using the most conservative approach.

a.

Chi-Square

b. Fisher’s Exact Test As p i r i nT a k e r ?

Ca n c e r

No

Y e s

Y e s %wi t h i nCa n c e r %wi t h i nAs p i r i n T a k e r Ex p e c t e dCo u n t s

2 8 ( 5 8 . 3 3 %) ( 2 9 . 1 7 %) 2 4 . 9 1

2 0 ( 4 1 . 6 7 %) ( 2 2 . 4 7 %) 2 3 . 0 9

No %wi t h i nCa n c e r %wi t h i nAs p i r i n T a k e r Ex p e c t e dCo u n t s

6 8 ( 4 9 . 6 4 %) ( 7 0 . 8 3 %) 7 1 . 0 9

6 9 ( 5 0 . 3 6 %) ( 7 7 . 5 3 %) 6 5 . 9 1

Smo k e r Di a b e t e s

No

Y e s

Y e s %wi t h i n Di a b e t e s %wi t h i nSmo k e r Ex p e c t e dCo u n t s

2 ( 2 5 %) ( 4 . 8 7 8 % ) 4

6 ( 7 5 %) ( 1 4 . 6 3 % ) 4

No %wi t h i n Di a b e t e s %wi t h i nSmo k e r Ex p e c t e dCo u n t s

3 9 ( 5 2 . 7 %) ( 9 5 . 1 2 % ) 3 7

3 5 ( 4 7 . 3 %) ( 8 5 . 3 7 % ) 3 7...