Question

Suppose a teacher recorded the attendance of her students in a recent statistics class because she wanted to investigate the linear relationship between the number of classes they missed and their final grades. The accompanying table shows these data for a random sample of nine students. Complete parts a through c. Click the icon to view the table showing missed classes and final grade. a. Using a =0.02, test the significance of the regression slope. The hypotheses for this test are The test statistic is (Type an integer or decimal rounded to two decimal places as needed.) The p-value is (Type an integer or decimal rounded to three decimal places as needed.) Therefore, at the 0.02 level of significance, the null hypothesis. There sufficient evidence to conclude that the slope is zero b. Construct a 98% confidence interval for the population slope. The confidence interval is ( OD). (Type integers or decimals rounded to three decimal places as needed.) c. What conclusions about missed classes and final grade can be drawn from these results? The conclusion is that there a linear relationship between missed classes and final grade because the null hypothesis rejected and because the confidence interval for the population slope 0.

Answer 1

S.NO х 4 1 2 3 4 5 6 6 1 4 Y 75 80 93 70 97 83 91 85 93 767 85.2222 (x-7) 1.7778 11.1111 2.7778 1.7778 7.1111 0.4444 7.1111 5.4444 0.4444 38.0000 SSX 0 2 0 5 2 24 2.667 (7-7) 104.49 27.27 60.49 231.72 138.72 4.94 33.38 0.05 60.49 661.5556 SSY (X-X)(y-7) -13.6296 -17.4074 -12.9630 -20.2963 -31.4074 1.4815 -15.4074 -0.5185 -5.1852 - 115.3333 7 8 9 Total Mean SXY sample size nu 9 2.6667 38.000 85.2222 Son (X-X) = Sw (X-X)(y-7)= slope B. S./S.- intercept= . = V-B,x= -115.333 = -3.0351 Cou 93.3158

SSE =Syy-(Sxy)2/Sxx=

311.508772

	s2 =SSE/(n-2)=	44.5013
std error σ              =	=se =√s2=	6.6709

std error of slope =se(β1) =s/√Sxx=

1.082167

a)

null hypothesis: Ho:           β1=		0
Alternate Hypothesis: Ha: β1≠		0

test stat t =(b1-β1)/se(β1)=		-2.80
p value =	0.026	(from excel:tdist(-2.805,7,2)

fail to reject the null hypothesis , there is not sufficient evidence,,,,, slope is not different from zero,

b)

for 98 % CI value of t=		2.998	(from excel:tinv(0.02,7)
margin of error E=t*std error =		3.2443
lower bound=b1+E=		-6.279
Upper bound=b1+E=		0.209

98% confidence interval =(-6.279 , 0.209)

c)

there is not a linear ,,,,,,,,,,,,null hypothesis is not rejected,.,,,,,contains 0.