A study was undertaken to investigate the effectiveness of an aquarobic exercise program for patients with osteoarthritis. A convenience sample of 70 individuals with arthritis was selected, and each person was randomly assigned to one of two groups. The first group participated in a weekly aquarobic exercise program for 8 weeks; the second group served as a control. Several pieces of data were collected from the individuals, including their total cholesterol (mg). Determine if there is a significant difference in the mean cholesterol for the two groups (aquarobic & control) at the start of the study using a significance level of 0.05. Difference Sample Diff. Std. Error d Aquarobic - Control -16.1139 7.5751 66.3985 . What hypotheses should be tested? Make sure to select the hypotheses which are written with notation consistent with the type of samples selected. Hold = 0 Ho:1 = /2 Hold =0 Ho: /d = 0 Ho: /1 = /2 Ho: #1 = /2 Ha:Ma > 0 Ha: /2 a = . TS: t= (round to 3 decimal places) . probability = Select an answer v . decision: Select an answer v . What conclusion is reached based upon the decision made in your test? At the 0.05 level, there is not sufficient evidence to conclude there is a difference in the mean cholesterol of individuals participating in the aquarobic program and those in the control group. At the 0.05 level, there is sufficient evidence to conclude there is a difference in the mean cholesterol of individuals participating in the aquarobic program and those in the control group. After the 8-week program, those who participated in the aquarobic program had their ending cholesterol measured, and the change in cholesterol was recorded for each participant. Estimate the mean cholesterol change using 95% confidence. . The formula which should be used for this interval is: Oyatt-ad Vn O(31 - 12 ) + ty/ Si . With % confidence, we estimate that the mean cholesterol before participating in 8 weeks of aquarobics is between mg and mg than the mean cholesterol after participation. Note: Round the limits of your interval to three decimal places. In the last box type the appropriate word - more or less. Think carefully about what positive and negative differences mean about the change in cholesterol based on how the differences were taken. Difference Sample Diff. Std. Error Critical Pt Pre - Post 18.2724 0.9657 2.0322 A 95% confidence interval was also calculated for the change in total cholesterol (pre - post) for the control group. That interval was found to be (-1.011, 3.588). Based on this interval and the one which you calculated for the aquarobic group, what conclusion would you draw? The aquarobic group did not have a significant change in mean cholesterol, while the control group had a significant increase in mean cholesterol. The control group did not have a significant change in mean cholesterol, while the aquarobic group had a significant decrease in mean cholesterol. ONeither group had a significant change in mean cholesterol. Both groups had a significant decrease in mean cholesterol. However, the decrease for the aquarobic group was larger. The mean cholesterol for the control group increased, while the aquarobic group had a significant decrease in mean cholesterol.A factory hiring people to work on an assembly line gives job applicants a test of manual agility. This test counts how many strangely shaped pegs the applicant can fit into matching holes in a one-minute period. The table below summarizes data collected for 90 applicants - 45 men and 45 women: Male Female n 45 45 Mean 20.43 18.48 Std Dev 2.398 4.047 Find separate 90% confidence intervals for the average number of pegs males and females can correctly place (note: these intervals are one-sample intervals from previous material!). The appropriate t-critical point for both intervals is t = 1.68. . The 90% confidence interval for males is: lower limit = , upper limit = . The 90% confidence interval for females is: lower limit =[ , upper limit = . What do these intervals suggest about gender differences in manual agility? The individual intervals suggest no difference in the mean number of pegs placed by males & females because the two intervals overlap. The individual intervals suggest males place a greater mean number of pegs because the entire interval for this gender is above the interval for females. The individual intervals suggest females place a greater mean number of pegs because most of the interval for this gender is above the interval for males. The individual intervals suggest males place a greater mean number of pegs because most of the interval for this gender is above the interval for females. The individual intervals suggest females place a greater mean number of pegs because the entire interval for this gender is above the interval for males. The intervals above didn't take into account a comparison between the two genders like a two-sample interval would have. To estimate the mean difference in the number of pegs correctly placed by the two genders, we can use the following formula: (91 - 32) + t n2 . What type of samples were selected if the formula above is used? O independent samples O dependent samples . Calculate the 90% confidence interval using the t-critical point t = 1.667: lower limit = , upper limit = (Round to 4 decimal places.) . Which of the following conclusions is correct for the interval calculated? With 90% confidence, we estimate the mean number of pegs placed by males is more than the mean placed by females by some amount between the limits calculated above. .With 90% confidence, we estimate the mean number of pegs placed by females is more than the mean placed by males by some amount between the limits calculated above. .With 90% confidence, there is no evidence of a difference in the mean number of correctly placed pegs by the two genders.For each of the following scenarios, determine whether independent or dependent samples were selected. Note: You may only attempt this question 2 times. On your second attempt, a penalty will be assessed. Allison would like to estimate the difference, on average, in the time to swim 100 meters between freestyle and butterfly techniques. She randomly selects 35 swimmers. For each person, she records the time to swim 100 meters using the freestyle technique and the time to swim 100 meters using the butterfly technique. O dependent samples independent samples An economist is studying the average household income between 2005 and 2013. He obtains a random sample of 100 tax returns from 2005. He records both the age for the head of household and the household income. From 2013 tax returns, he locates a head of household who is the same age as one from 2005. For the tax return, he then records the household income for 2013. From this, he will estimate the difference in annual household incomes between 2005 and 2013, on average. independent samples O dependent samples Brandon records the body mass index (BMI) for a random sample of 75 adult men. He then records the BMI for a random sample of 75 adult women. He plans to determine if the mean BMI is different between adult men and adult women. dependent samples independent samples A restaurant manager wants to compare the typical bill for those paying with credit card compared to those paying with cash. A sample of 81 credit card payments is selected and a sample of 51 cash payments is selected. The bill amount for all payments is recorded. O independent samples O dependent samples Jill wants to compare prices of food at Wendy's and at Mcdonald's. She makes a list of 40 randomly selected fast food items for each restaurant. She obtains the prices of the items at Wendy's. She obtains the prices of the same items at McDonald's. She wants to know if the mean price is lower at Wendy's. independent samples O dependent samplesTwo of the hottest smartphones on the market are the newly released iPhone6 and the Samsung Galaxy S6. CNet.com offers online reviews of all major cell phones, including battery life tests. In a review of the iPhone6, the talk-time battery life of 35 iPhones was measured. Similarly, the talk-time battery life of 30 Galaxy S6s was measured. Two outputs are given below. Which is appropriate for analyzing the data collected? Output 1 Output 2 will Mean of iphone6 Up = mean of the paired difference between iphone6 Hz : Mean of Galaxy S6 and Galaxy S6 Difference Sam ple Diff. Std. Err. Difference Sam ple Diff. Std. Err. Hi - HZ 0.71759861 0.189403 iPhone6 - Galaxy S6 -0.754246 0.192151 Possible p-values: 0.0001, 0.0002, 0.9999 Possible p-values: 0.0002, 0.0004, 0.9998 Output 1 O Output 2 Using the StatCrunch output chosen above, determine if there is a difference in the mean battery life for the two phones. Use a significance level of 0.10 when conducting the test. . Select the appropriate hypotheses. Make sure the notation used in the hypotheses agrees with the type of samples selected in the output. Hold = 0 Ho:1 = #2 Ho:ud = 0 Ho:ud = 0 Ho: /1 = #2 Ho: #1 = #2 Ha:Ha # 0 Ha: # #2 HaHa 0 Ha:#1 > #2 Ha: #1