Just need help with c,d, and e! The general manager of an engineering firm wants

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Just need help with c,d, and e!

The general manager of an engineering firm wants to know whether a technical artist's experience influences the quality of his or her work. A random sample of 24 artists is selected and their years of work experience and quality rating (as assessed by their supervisors) recorded. Work experience (EXPER) is measured in years and quality rating (RATING) takes a value of 1 through 7, with 7 = excellent and 1 = poor. The simple regression model RATING = beta_1 + beta_2 EXPER + e is proposed. The least squares estimates of the model, and the standard errors of the estimates, are RATING (se) = 3.204 (0.709) + 0.076 EXPER (0.044) (a) Sketch the estimated regression function. Interpret the coefficient of EXPER. (b) Construct a 95% confidence interval for beta_2, the slope of the relationship between quality rating and experience. In what are you 95% confident? (c) Test the null hypothesis that beta_2 is zero against the alternative that it is not using a two-tail test and the alpha = 0.05 level of significance. What do you conclude? (d) Test the null hypothesis that beta_2 is zero against the one-tail alternative that it is positive at the alpha = 0.05 level of significance. What do you conclude? (e) For the test in part (c), the p-value is 0.0982. If we choose the probability of a Type I error to be alpha = 0.05, do we reject the null hypothesis, or not, just based on an inspection of the p-value? Show, in a diagram, how this p-value is computed.

Explanation / Answer

c) TS = (b1^ - b)/se (b1^)

here Ho : b =0

hence TS = (0.076 -0)/0.044 = 76/44 = 1.727272

df = n-2 = 22

t-citical = 2.074

as P(T > t*) = 0.025

since TS   < t-critical

we fail to reject the null

d) here TS remains same = 1.7272

but

t-critical = P(t >t*) = 0.05

= 1.717

since TS > t-critical     {1.727272 > 1.717}

hence we reject the null hypothesis

e)

we fail to reject ,as p-value > 0.05

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