Question

DIRECTIONS: Make sure your responses are neat and readable. You must show me your calculation in a separate piece of paper. Homework that is difficult to grade due to messiness will be returned ungraded. If you would like to get full credit, do not forget to attach the Excel output. The number of hours 10 students spent studying for a test and their scores on that test is represented in the table below. 0 2 4 5 5 Hours spent studying, x Test scores, y 5 75 6 93 7 90 8 95 9 95 4051 64 69 73 Excel will be used to answer the following questions related to the data in the table. 1) Graph a scatterplot of the data. (6p) 2) Find the sample correlation coefficient for the data. (Pearson correlation coefficient between hours spent studying, x, and test score, y). (Round off your final answer to three decimal places). (7p) 3) Describe the type of correlation and interpret the correlation in the context of the data. Be specific in describing the magnitude, direction, and strength of the relationship. (12p) 4) Write out the null and alternative hypotheses and conduct a hypothesis testing for the data. At the 1% level of significance, is your test statistic statistically significant? Briefly explain how you reached your conclusion. For this question you must state followings: degrees of freedom, one or two tailed hypothesis testing, the critical value, test statistic, do you reject or fail to reject the null hypothesis. (25p) 5) Find the equation of the regression line for the hours spent studying and test scores of students. What are the intercept (a) and slope (b) values of the regression formula, Y = a +bX for predicting test scores variable from hours spent studying variable. Write the formula replacing a and b by their respective numerical values (round off each element in the final formula to three decimal places). (15p) 6) What is the predicted test score for a student who spent 3 hours for studying for the test. (10p)

Answer 1

(1)

Following is the scatter plot:

100 90 80 70 50 2 6 8 10 х 60 40

Scatter plot shows a strong, positive and linear relationship between the variables.

(2)

Following table shows the calculations:

	X	Y	X^2	Y^2	XY
	0	40	0	1600	0
	2	51	4	2601	102
	4	64	16	4096	256
	5	69	25	4761	345
	5	73	25	5329	365
	5	75	25	5625	375
	6	93	36	8649	558
	7	90	49	8100	630
	8	95	64	9025	760
	9	95	81	9025	855
Total	51	745	325	58811	4246

Sample size: n 10 From table we have (Σ3) (Σχ) -Στ. 325- ((51)A2/10) = 64.9 (Σ) sST Sy y. 58811- ((745)A2 /10) 'YY n = 3308.5 Σ3ΣΥ) y - SP ST 4246- ((745 * 51)/10) = 446.5 The correlation coefficient is: 446.5/ sqrt( 64.9 * 3308.5) r 0.964

(3)

The correlation coefficient is positive. It has magnitude 0.964. It shows a very strong, linear and positive relationship between the variables.

(4)

Here we have r = 0.964, n 10 Hypotheses are: Ho p 0 Ha p 0 Level of significance: 0.01 Test is two tailed df =n-2-8 Degree of freedom: Test Statistics: = 0.964/sqrt ((1 -0.964^2)/(10-2)) (1-r2)/(n-2) = 10.254 Critical value: 3.355 Ift< -3.355 or t > 3.355, Reject HO Rejection Region: Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value 0 Decision: Since p-value is less than level of significance so we reject the null hypothesis =TINV (0.01,8) Excel function for critical value: Excel function for p-value: =TDIST (10.254,8,2)

(5)

The slope is: 446.5/64.9 b =6.88 The intercept is: 1 bo 4-Σ-4Σ:) - = ( 745 - (6.88 * 51))/ 10 bo = 39.412 The regression equation is: y' 6.88 * x 39.412

Slope: It shows that for each unit increase in hours spent studying test scores is increased by 6.880.

Intercept: When studying hours are zero test score will be 39.412.

(6)

The predicted value for x = 6 is y' 39.412 6.88* 6 = 80.692

(7)

The coefficient of determination is: s2 (446.5)^2 / (64.9 * 3308.5) 2- 0.929

It shows that 93.9% of variation in test scores is explained by hours of studying.

(8)

The SSE is SSE y-by-bxy 58811 - (39.412 * 745)- (6.88* 4246) =236.58 The standard error of estimate: SSE sqrt(236.58/(10-2)) SE= n-2 = SE = 5.4381

(9)

Yes it seems it is good predictor. Because r-square is high, SE is small. Test shows that there is a significant relationship between the variables.

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As you need excel output, following is the output generated by excel:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.963570611
R Square	0.928468322
Adjusted R Square	0.919526862
Standard Error	5.439009075
Observations	10
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3071.837442	3071.837442	103.8385614	7.37348E-06
Residual	8	236.6625578	29.58281972
Total	9	3308.5
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	39.41294299	3.848922049	10.23999512	7.10882E-06	30.53731284	48.28857314
X	6.8798151	0.67514559	10.19012077	7.37348E-06	5.32292658	8.436703621