Question

A researcher collected data from a small random sample of ten students by asking them individually how much time (y) they spent studying and how much time (x) they spent on social media, on one day during an exam week. Both times were given in hours, rounded to the nearest half hour. They shown in the table below.

x	5	2.5	3	4.5	2.5	2	3	2	6.5	6
y	2	4.5	2	1	4	3.5	2	4	0	0.5

The researcher wishes to find a simple linear regression model Yi ∼ N(a + bxi , σ2 ).

(i) Find data summaries for this data and use them to calculate estimates for the slope and intercept parameters (b and a) for this linear regression. (ii) Construct the ANOVA table for this regression. (iii) Calculate the coefficient of Determination, R2 and comment on how well the regression line fits the data. (iv) Find a 90% confidence interval for the mean expected number of hours of studying in a day, for an individual who spends 1.5 hours on social media.

Answer 1

(i) Find data summaries for this data and use them to calculate estimates for the slope and intercept parameters (b and a) for this linear regression.

Regression eq of Y on X

Where 'a' = Intercept

= 2.35 - 0.844 * 3.7

= 5.475

'b' =

=

= -0.844

   = 0.892

Therefore the regression eq is

(ii) Construct the ANOVA table for this regression.

n = 10 k = 2

	df	SS (Formula is mentioned in the 1st column)	MS (SS/df)
Explained Variation ()	1 (k-1)	17.91	17.91
Error Variation	8 (n - k)	4.62	0.58
Total  (Syy)	9 (n-1)	22.53

Syy = 22.53

Sxx = 25.1

= 21.2

F statistic = = 31.01257

(iii) Calculate the coefficient of Determination, R2 and comment on how well the regression line fits the data.

= 0.79

= 79%

Since 'R' > 70% we can say that the model is good fit for the data.

(iv) Find a 90% confidence interval for the mean expected number of hours of studying in a day, for an individual who spends 1.5 hours on social media.

We need mean predicted so we will subsitute in the regression eq

= 4.208

Where

= 4.62

= 1.353

t-score Using t-distribution tables and

= 2.306

90% confidence interval for mean predicted response is

Ans: