Please answer the questions below based of the information I have provided below. Thank you so much
Question:
Please answer the questions below based of the information I have provided below. Thank you so much for your help!
PART A:
1) Pick your variable x or y from the regression project.
2) Give a brief description as to why you chose this variable instead of the other.
3) Calculate the mean and standard deviation of your sample using statcrunch.
4) Compute the 95% Confidence Interval for the population mean using statcrunch.
5) Interpret what step 4 means.
PART B:
1) Use the same variable that was used for part A to come up with a claim. Make sure to explain your reasoning as to why to choose it.
2) Write both hypotheses
3) Tell about the tail of the test, find the test statistic and p value
4) Write the initial conclusion based on the p value approach using alpha = .05
5) Write the final conclusion using a complete sentence in reference to the context of your problem
6) Discuss the error (type 1 or 2) that would be associated with your decision
PART C:
1. Determine how many populations you will be comparing and set up the null and alternative hypothesis
2. Find the test statistic (F stat) and the P value
3. Use alpha = .05 to conclude whether the means are the same or different
Topic: Relationship between Body Mass Index (BMI) and Systolic Blood Pressure (SBP) in College Students.
The additional variable is the academic year.
Independent variable: Body Mass Index (BMI)
Dependent variable: Systolic Blood Pressure (SBP)
Expected finding: We expect to find a positive correlation between BMI and SBP.
This is because previous research has shown that a higher BMI is associated with higher blood pressure due to increased resistance in blood vessels.
Population: College students
Student # | BMI (kg/m^2) | SBP (mmHg) | Academic year | Major | Marriage Status |
1 | 22.1 | 115 | Freshman | Biology | Single |
2 | 23.3 | 120 | Freshman | Psychology | Single |
3 | 25.4 | 126 | Freshman | Computer Science | Single |
4 | 27.8 | 135 | Freshman | Economics | Single |
5 | 24.9 | 129 | Sophomore | Chemistry | Single |
6 | 29.6 | 140 | Sophomore | Education | Single |
7 | 21.5 | 118 | Sophomore | History | Married |
8 | 25.8 | 128 | Sophomore | Sociology | Single |
9 | 26.2 | 132 | Junior | Nursing | Single |
10 | 28.4 | 137 | Junior | English | Single |
11 | 23.6 | 124 | Junior | Physics | Single |
12 | 22.9 | 117 | Junior | Marketing | Married |
13 | 27.3 | 131 | Senior | Anthropology | Single |
14 | 24.2 | 126 | Senior | Music | Single |
15 | 21.6 | 115 | Senior | Political Science | Single |
16 | 26.9 | 130 | Senior | Art | Widowed |
17 | 29.1 | 139 | Senior | Mathematics | Single |
18 | 25.3 | 123 | Freshman | Psychology | Single |
19 | 23.1 | 121 | Freshman | Biology | Single |
20 | 24.8 | 127 | Freshman | Economics | Single |
21 | 28.2 | 134 | Sophomore | Chemistry | Single |
22 | 22.8 | 119 | Sophomore | Sociology | Single |
23 | 26.7 | 133 | Sophomore | Education | Single |
24 | 30.1 | 142 | Junior | Physics | Single |
25 | 27.4 | 136 | Junior | Marketing | Married |
26 | 24.5 | 127 | Junior | Nursing | Single |
27 | 25.6 | 125 | Senior | Political Science | Single |
28 | 21.9 | 116 | Senior | Anthropology | Single |
29 | 23.8 | 122 | Senior | Mathematics | Single |
30 | 28.9 | 138 | Senior | Music | Single |
Solution 6
The correlation coefficient (r) is 0.9713. This indicates a strong positive linear relationship between BMI and SBP. As BMI increases, SBP tends to increase as well.
Solution 7
The coefficient of determination (r2) is 0.9432. This means that approximately 94.32% of the variation in SBP can be explained by variation in BMI. In other words, BMI is a very strong predictor of SBP.
Solution 8 -
Mean Mx: 25.4567,
Mean My: 127.5
Sum ofX= 763.7 Sum ofY= 3825 MeanX= 25.4567 MeanY= 127.5
X values(BMI) | Y Values (SBP) | X- Mx | Y- My | (X- Mx)2 | (X- Mx)(Y- My) |
22.1 | 115 | -3.3567 | -12.5 | 11.2672 | 41.9583 |
23.3 | 120 | -2.1567 | -7.5 | 4.6512 | 16.175 |
25.4 | 126 | -0.0567 | -1.5 | 0.0032 | 0.085 |
27.8 | 135 | 2.3433 | 7.5 | 5.4912 | 17.575 |
24.9 | 129 | -0.5567 | 1.5 | 0.3099 | -0.835 |
29.6 | 140 | 4.1433 | 12.5 | 17.1672 | 51.7917 |
21.5 | 118 | -3.9567 | -9.5 | 15.6552 | 37.5883 |
25.8 | 128 | 0.3433 | 0.5 | 0.1179 | 0.1717 |
26.2 | 132 | 0.7433 | 4.5 | 0.5525 | 3.345 |
28.4 | 137 | 2.9433 | 9.5 | 8.6632 | 27.9617 |
23.6 | 124 | -1.8567 | -3.5 | 3.4472 | 6.4983 |
22.9 | 117 | -2.5567 | -10.5 | 6.5365 | 26.845 |
27.3 | 131 | 1.8433 | 3.5 | 3.3979 | 6.4517 |
24.2 | 126 | -1.2567 | -1.5 | 1.5792 | 1.885 |
21.6 | 115 | -3.8567 | -12.5 | 14.8739 | 48.2083 |
26.9 | 130 | 1.4433 | 2.5 | 2.0832 | 3.6083 |
29.1 | 139 | 3.6433 | 11.5 | 13.2739 | 41.8983 |
25.3 | 123 | -0.1567 | -4.5 | 0.0245 | 0.705 |
23.1 | 121 | -2.3567 | -6.5 | 5.5539 | 15.3183 |
24.8 | 127 | -0.6567 | -0.5 | 0.4312 | 0.3283 |
28.2 | 134 | 2.7433 | 6.5 | 7.5259 | 17.8317 |
22.8 | 119 | -2.6567 | -8.5 | 7.0579 | 22.5817 |
26.7 | 133 | 1.2433 | 5.5 | 1.5459 | 6.8383 |
30.1 | 142 | 4.6433 | 14.5 | 21.5605 | 67.3283 |
27.4 | 136 | 1.9433 | 8.5 | 3.7765 | 16.5183 |
24.5 | 127 | -0.9567 | -0.5 | 0.9152 | 0.4783 |
25.6 | 125 | 0.1433 | -2.5 | 0.0205 | -0.3583 |
21.9 | 116 | -3.5567 | -11.5 | 12.6499 | 40.9017 |
23.8 | 122 | -1.6567 | -5.5 | 2.7445 | 9.1117 |
28.9 | 138 | 3.4433 | 10.5 | 11.8565 | 36.155 |
M: 25.4567 | M: 127.5 | SSx: 184.7337 | SP: 564.95 |
Sum of squares (SSX) = (X- Mx)2= 184.7337 Sum of products (SP) = (X- Mx)(Y- My) = 564.95 Regression Equation = =bX+a b=SP/SSX= 564.95/184.73 = 3.05819 a= MY-bMX= 127.5 - (3.06*25.46) = 49.64877 = 3.05819X+ 49.64877
Regression equation: = 3.05819X+ 49.64877
Let's say we want to predict the SBP for a person with a BMI of 26.7. We can plug in this value into the regression equation:
= 3.05819(26.7) + 49.64877
= 131.30244
Solution 8 (a)
The predicted SBP for a person with a BMI of 26.7 is 131.30244 mmHg.
The difference between the predicted y (131.30244) and the real y (133) would be the residual or error, which is:
e = y - e = 133 - 131.30244
e = 1.6975
The difference between the predicted y (131.30244 mmHg) and the real y (133 mmHg) is 1.6975 mmHg. This means that the prediction for the SBP of a person with a BMI of 26.7 is slightly lower than the actual SBP value that was observed. In other words, the model underestimated the SBP value for this particular individual. It's important to note that some degree of error is expected in any statistical model, and the goal is to minimize this error as much as possible.
Solution 8 (b)
Using the regression equation, we can make a prediction for the SBP of a person with a BMI of 27.5:
= 3.05819(27.5) + 49.64877
= 133.748995
The predicted SBP for a person with a BMI of 27.5 is 133.748995 mmHg.
Solution 8 (c)
The domain of the data is the range of values for the independent variable (BMI) that were included in the dataset. In this case, the domain is between the minimum value of BMI (21.5 kg/m^2) and the maximum value of BMI (30.1 kg/m^2) for the 30 students in the sample.
Solution 9
The Scatterplot with the line of best fit is given below -
Expert Answer:
