Question

Consider the following data between number of visitors x and the amount of wildlife seen in Cheaha State Park y,

visitors	wildlife
500	25
450	35
475	30
420	32
555	28
600	22
375	19

(a) Find the mean of x and the mean of y.

(b) Find the standard deviation of x and the standard deviation of y. Use Excel.

(c) Find the correlation coefficient. Use Excel. (d) Find the slope and intercept of the linear regression line. Then write down the line.

(e) Make predictions for all the values of x in the table. Then calculate the residuals.

(f) Calculate the sum of squared residuals. (g) Calculate the standard deviation of the regression.

(h) Using the mean of x and your data for x, calculate the sum of squared deviations.

(i) Write down an appropriate null and alternative hypothesis.

(j) Calculate the test-statistic.

(k) Make a prediction for when x is 470, and calculate (xp − ¯ x)2. Then make a confidence interval around your prediction.

Answer 1

Solution-: Let, x=Number of visitors and y= The amount of wildlife

Given data:

F Column G Column

x	y
500	25
450	35
475	30
420	32
555	28
600	22
375	19

(a)

Mean (x) =482.14 ( By using MS-Excel function "=AVERAGE(F2:F8)" )

Mean (y) =27.29     ( By using MS-Excel function "=AVERAGE(G2:G8)" )

(b)

Var (x) = 5134.69 ( By using MS-Excel function "=VARP(F2:F8)" )

SD (x) = 76.66   ( By using MS-Excel function "SQRT()" )

Var (y) = 27.35 ( By using MS-Excel function "=VARP(G2:G8)" )

SD (y) = 5.23 ( By using MS-Excel function "SQRT()" )

(c)

r=Corr(x,y)=-0.11   ( By using MS-Excel function "=CORREL(F2:F8,G2:G8)" )

(d) Find the slope and intercept of the linear regression line by using Data Data Analysis Regresssion in MS-Excel

We get output

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.108379
R Square	0.011746
Adjusted R Square	-0.1859
Standard Error	6.151098
Observations	7
ANOVA
	df	SS	MS	F	Significance F
Regression	1	2.248524	2.248524	0.059428	0.817089
Residual	5	189.18	37.83601
Total	6	191.4286
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	31.099	15.8149	1.966448	0.106406	-9.55432	71.75265	-9.55432	71.75265
x	-0.008	0.032445	-0.24378	0.817089	-0.09131	0.075493	-0.09131	0.075493

From this output we get regression line,

y=31.099-0.008*x