Please Answer each question item (a through i) in the space provided below. A st
ID: 3295525 • Letter: P
Question
Please Answer each question item (a through i) in the space provided below.
A statistics teacher collected the following data to determine if the number of hours a student studied during the semester could be used to predict the final grade for the course. The Excel output follows the data.
Student
Hours Studying
Final Grade
1
42
92
2
58
95
3
32
81
4
39
78
5
37
75
6
51
88
7
49
85
8
45
85
Summary Output
Regression Statistics
Multiple R 0.752344
R-square 0.566022
Standard error 4.832541
Observations 8
ANOVA Df SS MS F p-value
Regression 1 182.7543 182.7543 7.825578 0.03127
Residual 6 140.1207 23.35345
Total 7 322.875
Coefficient Standard Error t-stat p-value
Intercept 58.00609 9.755659 5.945891 .001011
Hours 0.608927 0.217674 2.797424 0.03127
a. Determine the least-squares regression line. _______________________
b. Interpret the value of the slope.
c. Determine the standard error of estimate ________________
d. Construct a 95% confidence interval for the average final grade when hours spent studying = 50
Assume that x 2 = 16069 and x = 353
e. Construct a 95% prediction interval for an individual y value when x = 6.5.
f. What percentage of the variation in y is explained by the regression line?
g. In testing the hypotheses H0: 1 = 0 and H1 : 1 0 what is the value of the calculated test statistic, t ?
h. In testing the hypotheses H0: 1 = 0 and H1 : 1 0 what is the decision rule at the 0.05 significance level?
i. In testing the hypotheses, H0: 1 = 0 and H1 : 1 0 what is the conclusion at the 0.05 significance level? And interpretation?
Student
Hours Studying
Final Grade
1
42
92
2
58
95
3
32
81
4
39
78
5
37
75
6
51
88
7
49
85
8
45
85
Explanation / Answer
a. Least Square Regression Line
y = 58.00 + 0.609 x
y = Final Grade and x = Hours Studying
b. Interpret the value of the slope = 0.609 and if we increase 1 hours of studying, then it will increase in final grade by the value of 0.61.
c. Standard error of the estimate se = 4.8325
d. 95% Confidence Interval when hours spent studying = 50
so Final Grade = 58.00 + 0.609 x 50 = 88.45
x 2 = 16069 and x = 353
95% Confidence Interval = y50 +- tn-2,0.05 sy * sqrt [1/n + (x* -xbar)2 / (n-1)sx2]
= 88.45 +- 2.447 * 4.8325 * sqrt [1/8 + (50 - 44.125)2/ 492.875 ]
= 88.45 +- 5.222
= (83.23, 93.67)
f. What percentage of the variation in y is explained by the regression line?
56.66% of the variation in y is explained by the regression line.
g. Test statistic : t = 2.797 as given in regression.
h. Decision Rule at the 0.05 significance level, so, here p - value is 0.03 so it is under significance level so we shall reject the null hypothesis. So, 1 0 .
i. We can conclude that the linear relationship is significant in nature.