Pick one method to be used: Chi-Square Test Mann Whitney U Test Spearman Rank and Friedman Test Mc Nemar’s Test for Correlated Proportions Kendall’s Coefficient of Concordance W Kolmogorov-Smirnov Test
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Pick one method to be used:
- Chi-Square Test
- Mann Whitney U Test
- Spearman Rank and Friedman Test
- Mc Nemar’s Test for
Correlated Proportions - Kendall’s Coefficient of Concordance W
- Kolmogorov-Smirnov Test
After picking one method to be used, answer the following:
What test method to be used:
Computed Value:
Tabular Value:
Conclusion:
Step by step
Solved in 4 steps
- Do various occupational groups differ in their diets? A British study of this question compared 98 drivers and 83 conductors of London double-decker buses. The conductors' jobs require more physical activity. The article reporting the study gives the data as "Mean daily consumption (+ se)." Some of the study results appear below. Drivers Total calories 2821 +44 Alcohol (grams) 0.24 0.06 Conductors 2844 +48 0.39 ± 0.11 Construct a 99% confidence interval for the difference in mean daily alcohol consumption between drivers and conductors.A researcher wants to determine the sensitivity of mammograms to determine how effective they are at diagnosing women who have breast cancer. Assume the researcher obtained the above results from a study, calculate and interpret the sensitivity of mammograms for detecting breast cancer. Frequency of Breast Cancer Cases Frequency of Non-Cancer Cases Frequency of Individuals Who Screened Positive 17 5 Frequency of Individuals Who Screened Negative 8 77 Group of answer choices: a. A total of 66.67% of individuals who have breast cancer test positive for breast cancer when using a mammogram as the primary diagnostic test for breast cancer. b. A total of 68% of individuals who have breast cancer test positive for breast cancer when using a mammogram as the primary diagnostic test for breast cancer. c. A total of 70.59% of individuals who have breast cancer test positive for breast cancer when using a mammogram as the primary diagnostic test for breast cancer. d. A total of…The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Percent of U.S.Households Observed Numberof Households inthe Community Married with children 26% 92 Married, no children 29% 121 Single parent 9% 31 One person 25% 102 Other (e.g., roommates, siblings) 11% 65 A) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.) B)Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)
- Find the standardized test statistic t for a sample with n= 15, x= 10.2, s 0.8, and a= 0.05 if H: H 2 9.9. Round your answer to three decimal places. A. 1.728 B. 1.452 C. 1.631 D. 1.312 Click to select your answer. P Type here to search a +66 hp 2) 立When determining which is significant for part a -make sure to include df, p value (greater than or less than .05), and t statisticView the attatched photo and select the correct answer for A. Then look at the questions below (B and C) and answer those using the data oe information in the photo. A) Answer A which is in the photo. B) Calculate the standardized test statistic. T = ? C) Calculate the P-value. P = ?
- Determine the P-value for a one sided test with t = 2.52 and n=21 O a. 0.0075 O b. 0.0175 OC. 0.01 O d. 0.005A survey was given to 18 employees. One question asked about the one-way distance the employee had to travel for work. The results, in kilometers, are shown below. Kilometers Traveled for Work 12 10 18 3 4 28 5 32 26 10 41 85 1 7 8 5 10 15 Find the first quartile of the data.We collected data for 300 people at a movie theater. For every person, we know what genre of movie they saw and whether or not they bought snacks.(we would expect 100% of people to bring snacks) Here are the results Type of Movie Snacks No Snacks Action 26 34 Comedy 47 23 Family 90 30 Horror 30 20 chi square test of independence
- To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material. Compute the values below (to 2 decimals, if necessary). Sum of Squares, Treatment Sum of Squares, Error a. Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use a = 0.05. Mean Squares, Treatment 0000 Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). 1 17 23 21 19 What conclusion can you draw after carrying out this test? - Select your answer - Manufacturer 2 28 26 31 27 + 3 18 17 21 20 The p-value is - Select your answer - What is your conclusion? - Select your answer - b. At the a = 0.05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3. Calculate Fisher's LSD Value (to 2 decimals).The data on the number of chocolate chips per bag for 42 bags of Chips Ahoy! cookies were obtained by the students in an introductory statistics class at the United States Air Force Academy in response to the Chips Ahoy! 1,000 Chips Challenge sponsored by Nabisco, the makers of Chips Ahoy! Use the data collected by the students to answer the following questions and to conduct the analyses required in each part. They found æ = 1261.6 and s = 117.6 a. Determine a 95% confidence interval for the mean number of chips per bag for all bags of Chips Ahoyl cookies. b. Interpret your result in words. C. Can you conclude that the average bag of Chips Ahoy! Cookies contain at least 1000 chocolate chips? O yes Ono Ads in 4 d. Explain why or why not. + SAVE MacBook Pro FR.IEN DS DII DD F12 F11 F8 F9 F10 F7 F5 ) 7 8 9. %3D dela { } Y U P [A poll classified respondents by gender and political party, as shown in the table. A researcher wonders if gender and party affiliation are not independent. (Assume a significance level of α = 0.05) Complete parts a) through e) below. d) Find the P-value for this test. Calculate the chi-square statistic. x² = 2.339 (Round to three decimal places as needed.) Find the P-value for the test. Choose the correct answer below. 0.0823 0.3104 0.003 DRIO 40 40 25 Male Female 49 35 18 e) State an appropriate conclusion. Choose the correct answer below. O A. Fail to reject the null hypothesis. There is sufficient evidence to support the claim. B. Reject the null hypothesis. There is not sufficient evidence to support the claim. C. Reject the null hypothesis. There is sufficient evidence to support the claim. O D. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim.