Question

Moondollars is a coffee franchise that has many coffee stores all over the world. MoonDollars coffee is investigating a linear regression model to predict the annual income for a store based on the population of the city where the store is located. In the proposed regression model, annual store income is the response variable and population of the city is the explanatory variable.

A random sample of 20 stores is selected and measurements are observed.

Population ('000s)	Annual income ('000s $)
1,880	1,779
970	341
3,090	2,459
2,370	2,548
1,750	1,611
850	1,052
1,730	1,917
1,290	918
740	569
570	414
2,610	3,017
2,230	1,135
1,630	1,149
1,260	1,174
510	466
1,180	1,698
2,330	2,183
1,360	856
520	907
1,280	1,207

a)Calculate the point prediction for the value x = 1,500. Give your answer as a whole number.

y^ = _________

b)Give the 95% prediction interval for the value x = 1,500. Give your answers as whole numbers.

______ ≤ y^ ≤ _______

Answer 1

These are the data that have been provided for the dependent and independent variables:

Obs.	Population	Income
1	1880	1779
2	970	341
3	3090	2459
4	2370	2548
5	1750	1611
6	850	1052
7	1730	1917
8	1290	918
9	740	569
10	570	414
11	2610	3017
12	2230	1135
13	1630	1149
14	1260	1174
15	510	466
16	1180	1698
17	2330	2183
18	1360	856
19	520	907
20	1280	1207

Now, with the provided sample values of the predictor and the response variable, we need to construct the following table to compute the estimated regression coefficients:

Obs.	Population	Income		​	​
1	1880	1779	3534400	3164841	3344520
2	970	341	940900	116281	330770
3	3090	2459	9548100	6046681	7598310
4	2370	2548	5616900	6492304	6038760
5	1750	1611	3062500	2595321	2819250
6	850	1052	722500	1106704	894200
7	1730	1917	2992900	3674889	3316410
8	1290	918	1664100	842724	1184220
9	740	569	547600	323761	421060
10	570	414	324900	171396	235980
11	2610	3017	6812100	9102289	7874370
12	2230	1135	4972900	1288225	2531050
13	1630	1149	2656900	1320201	1872870
14	1260	1174	1587600	1378276	1479240
15	510	466	260100	217156	237660
16	1180	1698	1392400	2883204	2003640
17	2330	2183	5428900	4765489	5086390
18	1360	856	1849600	732736	1164160
19	520	907	270400	822649	471640
20	1280	1207	1638400	1456849	1544960
Sum =	30150	27400	55824100	48501976	50449460

The sum of squares obtained from the table above are:

The slope and y-intercept coefficients are computed using the following formulas:

Therefore, the regression equation is:

Now that we have the regression equation, we can compute the predicted value for Population = 1500, by simply plugging in the value of Population = 1500 in the regression equation found above:

Now, we are going to compute the 95% prediction interval for the predicted value . The first step consists of finding the standard error of the estimate, for which we need to deal with the regression sum of squares and with the sum of squared errors.

We know that the total sum of squares is:

Also, the regression sum of squares is computed as follows:

Since we know that , then we can compute the sum of squared errors as follows:

​

Hence, the means squared error is :

Now, taking square root, we get the standard error of the estimate:

Since we want to compute a 95% prediction interval for the predicted value , the significance level used is α=0.05. The critical t-value for df=n−2=20−2=18 degrees of freedom, and α=0.05 is tc​=2.101. Now, with all this information, we are ready to compute the margin of error of the prediction interval:

Hence solution in the snapshot is:-

a) Y = 1363 and

CI is (499,2228)

Please let me know in comments if anything is unclear. Will reply ASAP. Please upvote, Thanks!