Question

A Minnesota school district has a novel idea about a factor influencing school attendance: they think the more the school cafeteria serves “junkie” kid preferred food, the more they want to come to school. To test this, they take a sample of eight school days and measure the % of saturated fat in the school lunch and the % school attendance. The students are told in advance what the school lunch will be.

Evaluate the data and then explain if serving tasty, yet lousy, food is related to attendance.   
                        % Saturated Fat
                        in School Lunch           % Attendance

13.3                             35

24.9                             80

  9.0                             10

34.5                             75

36.1                             85

22.1                             75

22.7                             70

24.5                             80

1.State your null formally and in lay terms for a simple regression

2. Calculate r and the regression line (y = a + bx) and reject/accept at a=.05.

3. Explain your findings in lay terms using r-square, r, b

4. Calculate a 95% confidence interval for the slope and explain in layterms.

5. Calculate a 95% confidence interval for Y when X is “25” and explain in layterms.

Answer 1

1. Null Hypothesis , H₀ : % Saturated Fat in School Lunch doesnt affect Attendance in school.

Null Hypothesis is the default Hypothesis , that is arrived without any tests. In this case, the null hypothesis is that % Saturated Fat in School Lunch doesnt affect Attendance in school. We have to do a regression test to check, if it actually affects or not.

Alternate Hypothesis, H_a : % Saturated Fat in School Lunch affects Attendance in school.

Alternate Hypothesis is the Hypothesis , that we want to prove in the test.

2.

% Saturated Fat in School Lunch	% Attendance
13.3	35
24.9	80
9.0	10
34.5	75
36.1	85
22.1	75
22.7	70
24.5	80

correlation coefficient, r = 0.8572

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.857272
R Square	0.734916
Adjusted R Square	0.690735
Standard Error	14.84423
Observations	8
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3665.394	3665.394	16.63434	0.006513
Residual	6	1322.106	220.351
Total	7	4987.5
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	5.92091	15.11906	0.391619	0.70887	-31.0741	42.91592	-31.0741	42.91592
% Saturated Fat in School Lunch	2.472649	0.606261	4.078522	0.006513	0.989182	3.956117	0.989182	3.956117

The regression equation is given as ,

Y (% Attendance) = 5.92091 + 2.472649 * X (% Saturated Fat in School Lunch).

Since the p value of the variable % Saturated Fat in School Lunch is 0.0065, i.e less than 0.05., the alternate hypothesis is accepted and it is established that the % Saturated Fat in School Lunch affects % Attendance.

3. In this case, r² = 0.7349 = 73.49%.

It means that 73.49% of the variation of % Attendance is explained by the variation in % Saturated Fat in School Lunch, which is a good amount of explanation.

Also, slope coefficient of % Saturated Fat in School Lunch is 2.472649. It implies two things -

1. The slope is positive, which means % Saturated Fat in School Lunch affects school attendance in a positive way as assumed in the hypothesis.

2. It implies that for every 1% increase in saturated fats in school lunch, there is a corresponding 2.472649% change in the attendance

The correlation between the two variables is 0.8572, which says that there is a high positive correlation between these two variables.

4. Mean of Slope ,b = 2.47265

SE of slope = 0.60626.

For a 95 % confidence, Z score = 1.96.

Therefore, Upper interval of confidence interval of slope = Mean + 1.96 * SE of slope = 2.47265 + 1.96 * 0.60626. = 3.6609

Lower interval of confidence interval of slope = Mean - 1.96 * SE of slope = 2.47265 - 1.96 * 0.60626. = 1.284

These upper and lower bounds imply that slope lies in the range of (1.284, 3.6609) with a 95% probability .

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