Question

12.6 Can a pretest on mathematics skills predict success in a statistics course?  Assume students in an introductory stats class can be considered a random sample from a larger population of statistics students.  These 62 students took a pretest at the beginning of the semester, and their score from the pre-test and their score from the final exam in the course were recorded.  The least squares regression line was found to be y = 13.8 + 0.81x, and the standard error of b₁ was 0.43.

(a) (10m) Would you reject this null hypothesis versus the one-sided alternative that the slope is positive? Fill in the table to find your answer.

Hypotheses	Test statistic and degrees of freedom	Pvalue statement and Pvalue	Decision: Reject or Fail to Reject Ho and why or why not	English conclusion regarding significance or non significance.
Ho: β1 = 0 HA: β1 ≠ 0	t* = b1/SEb1 = (0.81 – 0)/ (0.43) = 1.88372093 =1.884 df = n-2 = 62 - 2 = 60	1. P(t>1.884) = 0.032 2. 0.025 < P(t>1.884) <0.05 3.                 0.08 -1.8------- 0.0301	Reject Ho since P value < alpha (0.05) (5%) 0.032 < 0.05	And we conclude that at the 5% significance level there is sufficient evidence that slope is positive

(b) (10m) Use the hypothesis test approach to test the null hypothesis that there is no linear relationship between the pretest score and the score on the final exam against the two-sided alternative.   Use a significance level of 5%.

Hypotheses	Test statistic and degrees of freedom	Pvalue statement and Pvalue	Decision: Reject or Fail to Reject Ho and why or why not	English conclusion regarding significance or non-significance.
Ho: β1 = 0 HA: β1 ≠ 0	t* = b1/SEb1 = (0.81 – 0)/ (0.43) = 1.88372093 =1.884 df = n-2 = 62 - 2 = 60	1. 2P(t>1.884) = 0.064 2. 0.05 < 2P(t>1.884) < 0.1 3.                   0.02   -1.5 -------- 0.0643	Fail to reject Hosince P value <= Alpha (0.064 <= 0.05) (p value is larger than prechosen alpha)	And we conclude that at the 5% significance level there is not sufficient evidence of linear relationship

(START PAGE 3 HERE c) (2m) Why were you able to use a z approximation to calculate your pvalue above?

Answer 1

a)
null hypothesis always contains equality
since we want to test if slope is positive
hypotheses are: -
Ho:b = 0
Ha: b > 0

TS = (b^ b under null hypothesis)/(se(b^))
= (0.81 - 0)/0.43
= 1.8837

df =n-2 = 62 - 2 = 60

p-value = =t.dist.rt(1.8837,60) {using Excel}
= 0.0322

if p-value < alpha, we reject the null hypothesis
if p-value > alpha, we fail to reject the null hypothesis

here p-value < alpha (0.05)
we reject the null hypothesis
we conclude that there is sufficient evidence that slope is positive

b)
this is two-sided alternative
hence hypothesis are
Ho:b = 0
Ha: b ≠ 0

TS = (b^ b under null hypothesis)/(se(b^))
= (0.81 - 0)/0.43
= 1.8837

df =n-2 = 62 - 2 = 60

p-value = =T.DIST.2T(1.8837,60)
=0.0645

since p-value < alpha (0.05)
we fail to reject the null hypothesis

we conclude that there is not sufficient evidence of linear relationship