Activity #3 Worksheet PDF

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Online STA 282QR Activity #3: Hypothesis Testing and Regression Analysis ( 35 points) You can type your answers on this form, cut and paste any output into this document, and submit it via Blackboard when complete. Both the Distance Activity and Hand Size Activity referenced here were done in the previous two activities, so you may review those assignments for any background information. For the Activity #3 data, there are two Activities: The Raisen Activity and the Pulse Rate Activity. For the Raisen Activity, n=150 students were given a ½ oz box of raisens and asked to guess the number of raisens in the box (Column A: Guess). The students then counted the raisens in the box (Column B: Actual). For the Pulse Rate Activity, n=130 students were asked to provide their Gender, Height, and Weight. Body Mass Index (BMI= Weight/Height 2 ) was calculated for them using kg and m. Before Exercise pulse was determined (BEPL) for 60 sec, then recalculated after 1 minute of jumping-jacks (AEPL). The difference, or how much their pulse increased, was also calculated as AEPL – BEPL. 1. (8 points) Assume we wish to conduct the following studies. Simply identify each of the following problems as a matched pair or independent sampling design: (a) For the Hand Size Activity: Is the average student’s hand width is different than their hand length? Match Pair For Distance activity: Are the average Miles for female students different than the average Miles for male students? Independent Sample (c) For the Pulse Rate Activity: Is the average Difference in pulse (AEPL – BEPL) more than 25 beats per minute? Matched Pair (d) For the Pulse Rate Activity: Is the average Height for females different than the average Height for males? Independent Sample (b)

2. (10 points) For the Raisin activity, students were asked to make a ‘guess’ and then count the # of raisins in a ½ Oz box of raisins. (a) If we wished to determine if the average Guess was different than the average ‘Actual’, would this be a matched pair or independent sampling design? Explain why. This would require a matched pair design because each set is dependent on the other one since they are both from the same box of raisins. Also, since it was 2 sets of data over a period of time. (b) Determine the statistics (Statistics>Descriptive Statistics) for both the Guess and Actual variables and paste them here.

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i. Do students tend to under or over guess the number of raisins in the box? Explain. People tend to under guess the number of raisins in the box. This is because the mean value for the guess variable is smaller (24.51) than the mean value for the actual variable (29.89). ii. Do the students’ guesses tend to have more or less variation than the Actual number? Explain.

The students’ guesses tend to have more variation than the actual number. This is because the guess variable has a larger deviation (8.873) than the actual variable (3.040). iii. Why do you believe there might be such a difference in variation? The reason there is a much higher variation in the guess variable than the actual variable is because of perception vs. fact. When looking at a jar of raisins, some people may over predict or under predict solely because they only have their eyes to determine the true number. When it comes to the actual number of raisins, there will be a lot less variation because there is only one number it can be and the people now have the chance to count the raisins, which would lead to less error. Simply stated, when guessing the number, there is a lot less certainty but when counting the raisins, there is more certainty. (c) We wish to determine, at = 5% (95% Confidence), if the average Guess is less than the average Actual number of raisins. State the Null and Alternate Hypothesis for this research. Use 2(a) to help you write the correct hypotheses. Ho: d = 0 Ha: d < 0 (The letter d stands for difference which in this case is guess – actual)

(d) Use CrunchIt to conduct the hypothesis test to test if the average guess is less than the actual average # of raisins per box using ‘Copy and paste the output table here.

(e) Use the output to make the decision (NOTE: use p-value to Reject or Not Reject Ho): Given that the p-value is less than 0.0001 and the alpha level is 0.05, it can be determined that p-value is statistically significant, and we can reject the null hypothesis and assume that the alternative hypothesis is true.

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(f) Based on your decision, what conclusion can you make regarding the relationship between average Guess and average Actual count in the raisin box: Based on our ability to reject the null hypothesis due to a statistically significant p-value, it can be concluded that the guess average is significantly smaller than the actual average. 3. (9 points) Using the Pulse Rate Activity data, we wish to determine if the average BMI for females is different than the average BMI for males. a. If we wished to determine if the average BMI for females is different than the average BMI for males, would this be a matched pair or independent sampling design? Explain why. This would require an independent sampling design because all of the measurements in this study are independent from each other. The males and the females would be the 2 independent groups. b. Determine the summary statistics (Statistics>Descriptive Statistics, use grouped tab) to compare the BMI for both Males and Females. Cut and paste the summary stats here.

i. Based on the sample, who has the higher mean BMI? Based on the sample, males have a higher mean BMI. c. We wish to determine, at = 5% (95% Confidence), if the average BMI for females is different than the average BMI for males. State the Null and Alternate Hypothesis for this research. Use part 3(a) to help you write the correct hypotheses. Ho: d = 0 Ha: d  0 (The letter d stands for difference which in this case is female - male) d. Use CrunchIt to conduct the hypothesis test. ‘Copy and paste the output table here.

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e. Use the output to make the decision (NOTE: use p-value to Reject or Not Reject Ho): Given that the p-value is 0.008134 which is less than the alpha level is 0.05, it can be determined that p-value is statistically significant, and we can reject the null hypothesis and assume that the alternative hypothesis is true. f.

Based on your decision, what conclusion can you make regarding the mean BMI for males versus the mean BMI for females?

Based on our ability to reject the null hypothesis due to a statistically significant p-value, it can be concluded that the average female BMI is different than the average male BMI.

Part II: Regression Inference for Chapter 10 (Section 10.3) We will still be using the Pulse Rate Activity data. Here, we will ask two questions: A. Can we use height to predict a person’s weight? B. Can we use BMI to predict a person’s pulse difference (Difference)? 4. (4 points) Using regression, create a fitted plot and numeric results for Weight as the dependent variable and Height as the independent variable. Paste both the plot and regression results here.

(a) What is the equation of the regression line? weight = -223.8 + 5.660 * height or in y=mx+b form… weight = 5.660 * height – 223.8 (b) Conduct a hypothesis test to determine if Height is a significant predictor of weight. Do this by testing Ho: B1 = 0. If we find the slope=0, we can conclude Height is not significant predictor of weight. Based on your results, can you conclude height is a significant predictor of weight? Explain why or why not. 4

Based on the results, we can conclude that the height is a significant predictor of weight (reject the null hypothesis). This is because the slope was found to be 5.660. Having a slope that is equal to zero would indicate that height is not a significant predictor of weight. Simply stated, since the slope was not equal to zero, height is a significant predictor of weight. 5. (4 points) Using regression, create a fitted plot and numeric results for Difference as the dependent variable and BMI as the independent variable. Paste both the plot and regression results here.

(a) What is the equation of the regression line? difference = 26.58 + 0.07447 * BMI or in y=mx+b form… difference = 0.07447 * BMI + 26.58 (b) Conduct a hypothesis test to determine if BMI is a significant predictor of Difference. Do this by testing Ho: B1 = 0. If we find the slope=0, we can conclude BMI is not a significant predictor of Difference. Based on your results, can you conclude BMI is a significant predictor of Difference? Explain why or why not. Based on the results, we can conclude that BMI is not a significant predictor of difference (fail to reject the null hypothesis). This is because the slope was found to be 0.07447. Having a slope that is equal to zero would indicate that BMI is not a significant predictor of difference. Simply stated, since the slope was equal to zero, BMI is not a significant predictor of difference.

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