Calculating The Chi-Square Statistic Assignment Paper
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Unit 4 Assignment- Word Document(18.1 KB)
Vaccination Data Q5 Unit 4 Assignment- Excel File(12 KB) Calculating The Chi-Square Statistic Assignment Paper
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Question 1: Calculation of Chi-square and degrees of freedom for a set of data of 711 breast cancer survivors.
Based on the results from the cross tabulation table, the p-value is higher than 0.05, therefore, the result of the chi-square is not statistically significant at the 0.05 level (specifically, the p-value is 0.2894). This indicates that there isn’t enough data to rule out the null hypothesis, which is typically that there isn’t a relationship between the Piper Fatigue Scale (PFS-12) and race which are the variables under investigation.
Question 2: Explanation of Results in Question 1 above
The chi-square test was used in Exercise 1’s analysis to see if there was a significant relationship between breast cancer survivors’ PFS-12 scores and race. The counts of breast cancer survivors in each racial group and PFS-12 score category were used to produce the contingency table. With 4 degrees of freedom, the chi-square statistic was determined to be 4.9798051, and the corresponding p-value was discovered to be 0.2894. We were unable to reject the null hypothesis since the p-value was higher than the significance level of 0.05, indicating that there was no significant correlation between race and PFS-12 score among breast cancer survivors. This implies that there may be other variables that have a greater impact on predicting PFS-12 scores among breast cancer survivors.
Question 3: Calculation of Chi-square and degrees of freedom for a set of data of 180 people undergoing a knee replacement treatment with a drug supplement: Calculating The Chi-Square Statistic Assignment Paper
The p-value (0.0526) is higher than the significance level of 0.05, hence the value of the chi-square is not statistically significant at the 0.05 level. The result is often regarded as statistically significant if the p-value is less than or equal to the significance level.
Question 4: Summary of Results in Exercise 3
A sample of 180 participants who underwent knee replacement surgery with a medication supplement were the subject of the chi-square study. Using two degrees of freedom and a chi-square statistic of 5.89, the results had a p-value of 0.0526. The results are not statistically significant because the p-value is higher than 0.05, indicating that there is no proof of a connection between the type of medication supplement used and the success rate of knee replacement surgery. Yet, because the p-value is so near to the level of significance, it is possible that a weak association exists that could be found with a bigger sample size or a different statistical test. Calculating The Chi-Square Statistic Assignment Paper
Question 5: Vaccination Coverage
Because the p-value (0.2449) is higher than 0.05, the result of the chi-square is not statistically significant at the 0.05 level. In other words, at the significance level of 0.05, we are unable to reject the null hypothesis.
Question 6
The MMR vaccine data set underwent chi-square analysis, which revealed a chi-square statistic of 1.35 and 1 degree of freedom with a p-value of 0.2449. The null hypothesis, which states that there is no significant correlation between poverty level and MMR vaccination status among adolescents aged 13 to 17 in the tiny high school in Northern California, cannot be disproved because the p-value is greater than 0.05. The findings thus imply that the prevalence of poverty may not be a key determinant of MMR immunization in this community.
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Question 7
The provided p-value of 0.2426 exceeds the level of significance. As a result, we might draw the conclusion that there is no statistically significant association between the variables (high rate of UTI and a certain staff member’s shift) under study and that there is insufficient evidence to reject the null hypothesis.
Question 8: Explanation on Exercise 7 Results
We cannot rule out the null hypothesis that there is no correlation between the variables under study based on the findings of the chi-square test with a chi-square statistic of 35 and 30 degrees of freedom. The observed data are not statistically different from the expected values, as the p-value of 0.2426 is larger than the conventional significance level of 0.05. Consequently, we can draw the conclusion that there is no proof of a causal link between the employee’s shift and the frequency of UTI infections. Calculating The Chi-Square Statistic Assignment Paper
Question 9
- For χ2 = 3.02 and df = 1, the critical value of chi-square at α = 0.05 is 3.84. Since 3.02 < 3.84, the result is not statistically significant at the 0.05 level.
- For χ2 = 8.09 and df = 4, the critical value of chi-square at α = 0.05 is 9.49. Since 8.09 < 9.49, the result is not statistically significant at the 0.05 level.
- For χ2 = 10.67 and df = 3, the critical value of chi-square at α = 0.01 is 11.34. Since 10.67 < 11.34, the result is not statistically significant at the 0.01 level.
- For χ2 = 9.88 and df = 2, the critical value of chi-square at α = 0.01 is 13.82. Since 9.88 < 13.82, the result is not statistically significant at the 0.01 level.
Question 10
- Nonparametric Test B. Parametric Test
- Mann-Whitney U-test Independent groups t-test
- Friedman test Repeated measures ANOVA
- Kruskall-Wallis test One-way ANOVA
- Wilcoxon signed-ranks test Paired t-test
Question 11
- A chi-square test of independence can be used to examine the association between the independent variable (normal birth weight vs. low birth weight infants) and the dependent variable (sleep apnea issues – yes/no).
- A repeated-measures ANOVA can be used to compare the means of the dependent variable (heart rate) at several periods in time for the independent variable (before, during, and after operation).
- We can apply a McNemar’s test for matched pairs to compare the proportions of the dependent variable (did vs. did not exercise regularly) at various time points of the independent variable (before, during, and after intervention).
- We can apply a chi-square test of independence to examine the association between the independent variable (infertility treatment A vs. infertility treatment B vs. control condition) and the dependent variable (did vs. did not become pregnant)Calculating The Chi-Square Statistic Assignment Paper.
Question 12
a). To ascertain whether there is a significant correlation between illness and room type, a chi-squared test would be acceptable.
b). To evaluate whether there is a substantial correlation between diabetes and smoking, a Fisher’s exact test would be appropriate.
References
Black, K. (2019). Business statistics: for contemporary decision making. John Wiley & Sons.
Keller, G. (2022). Statistics for management and economics. Cengage Learning.
Unit 4 Assignment NSG 623
- (3 pts) Using StatCrunch, calculate the chi-square statistic and degrees of freedom for the following set of data for 711 breast cancer survivors that were categorized by their score on the Piper Fatigue Scale (PFS-12) and their race: Calculating The Chi-Square Statistic Assignment Paper
Hispanic Caucasian African-American Total
Low PFS-12 99 159 112 370
High PFS-12 79 175 87 341
Total 178 334 199 711
Copy and paste your crosstabulation table into your WORD document submission for full credit. Is the value of the chi-square statistically significant at the 0.05 level?
- (1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 1.
- (3 pts) Using StatCrunch, calculate the chi-square statistic and degrees of freedom 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 Total
Had > 8 wk rehab 20 35 55
Had < 8 wk rehab 70 55 125
Total 90 90 180
Copy and paste your crosstabulation table into your WORD document submission for full credit. Is the value of the chi-square statistically significant at the 0.05 level?
- (1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 3.
- (3 pts) According to the US Health Department, the Estimated Vaccination Coverage for MMR = (measles, mumps, and rubella) among adolescents Aged 13-17 Years and living above the poverty level is 90.8%. A small high school in Northern California reported a much lower vaccination rate and a researcher collected data on the poverty and MMR vaccination status. Using StatCrunch, calculate the chi-square statistic and degrees of freedom for the vaccination data set supplied in the assignment area regarding vaccination and poverty. Copy and paste your crosstabulation table into your WORD document submission for full credit. Is the value of the chi-square statistically significant at the 0.05 level? Calculating The Chi-Square Statistic Assignment Paper
- (1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 5.
- (3 pts) An Administrator at a large urban hospital is concerned that the UTI (urinary track infection) rate is too high when a certain staff member is working. UTI infections doubled when the staff member started their 3-day shift on a Saturday:
UTIs Reported
Sunday 29
Monday 19
Tuesday 18
Wednesday 30
Thursday 28
Friday 15
Saturday 30
Using StatCrunch, calculate the chi-square statistic and degrees of freedom for the above set of data. Copy and paste your result table into your WORD document submission for full credit. Is the value of the chi-square statistically significant at the 0.05 level?
- (1 pt) Write a short paragraph summarizing the results of the analysis in Exercise 7.
- (1 pt) Given each of the following circumstances, determine whether the calculated values of chi-square are statistically significant: Calculating The Chi-Square Statistic Assignment Paper
- χ2 = 3.02, df = 1, α = 0.05
- χ2 = 8.09, df = 4, α = 0.05
- χ2 = 10.67, df = 3, α = 0.01
- χ2 = 9.88, df = 2, α = 0.01
- (1 pt) Match each of the nonparametric tests in Column A with its parametric counterpart in Column B
- Nonparametric Test B. Parametric Test
- Mann-Whitney U-test a. Paired t-test
- Friedman test b. One-way ANOVA
- Kruskall-Wallis test c. Independent groups t-test
- Wilcoxon signed-ranks test d. Repeated measures ANOVA
- (1 pt) Using the information provided, indicate which statistical test you think should be used for each of the following situations:
- Independent variable: normal birth weight vs. low birth weight infants; dependent variable: Issues with sleep apnea (yes/no)
- Independent variable: time of measurement (before, during, and after surgery); dependent variable: heart rate.
- Independent variable: time of measurement (before, during, and after intervention); dependent variable: did vs did not exercise regularly.
- Independent variable: infertility treatment A vs infertility treatment B vs control condition; dependent variable: did vs did not become pregnant.
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- (1 pt) Below are two (a and b) sets of cells of 2 X 2 contingency tables including the expected frequencies. Identify which statistical procedure would be appropriate for each, using the most conservative approach. Calculating The Chi-Square Statistic Assignment Paper
Room Type | ||
Infection | Concrete | Linoleum |
Yes
% within Infection % within Room Type % of Total Expected Counts |
2 (9.091%) (5.882%) (1.626%) 6.08 |
20 (90.91%) (22.47%) (16.26%) 15.9 |
No
% within Infection % within Room Type % of Total Expected Counts |
32 (31.68%) (94.12%) (26.02%) 27.9 |
69 (68.32%) (77.53%) (56.1%) 73.1 |
b.
Smoker | ||
Diabetes | No | Yes |
Yes
% within Diabetes % within Smoker % of Total Expected Counts |
2 (25%) (4.878%) (2.439%) 4 |
6 (75%) (14.63%) (7.317%) 4 |
No
% within Diabetes % within Smoker % of Total Expected Counts |
39 (52.7%) (95.12%) (47.56%) 37 |
35 (47.3%) (85.37%) (42.68%) 37 |
Calculating The Chi-Square Statistic Assignment Paper
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