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Question

# A. Find the 95%confidence interval forβ1 was(.014,.064), and thus concluded that there is a positive association

between X and Y in the population. To form the confidence interval, you used the R output for the standard error ofˆβ1(.01277). Now, verify this standard error using the formula given in class for SD(ˆβ1).

B Carry out the hypothesis test of the null hypothesis that there is no association between X and Y. State the hypotheses in appropriate notation, report the value of the test statistic and the P-value, and state your conclusion.

C Consider individuals whose ACT test score is 28. (i) Obtain the value ofˆY for such individuals. (ii) Obtain the value of the standard error ofˆY

D Obtain a 95%confidence interval for the mean freshman GPA for students whose ACT test score is 28.Interpret the confidence interval.

E Show the ANOVA table. Report R2; interpret its value.

For this,

Here is summary

Call:

lm(formula = gpa ~ act, data = gpa.df)

Residuals:

Min    1Q  Median    3Q   Max

-2.74004 -0.33827 0.04062 0.44064 1.22737

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 2.11405  0.32089  6.588 1.3e-09 ***

act     0.03883  0.01277  3.040 0.00292 **

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.6231 on 118 degrees of freedom

Multiple R-squared: 0.07262, Adjusted R-squared: 0.06476

F-statistic: 9.24 on 1 and 118 DF, p-value: 0.002917

Mean  24.725

SD  4.472065

\$fit

fit    lwr    upr

1 3.201209 3.061384 3.341033

\$se.fit

 0.07060873

\$df

 118

\$residual.scale

 0.623125

Anova table.

Response: gpa

Df Sum Sq Mean Sq F value   Pr(>F)

act      1  3.588  3.5878  9.2402 0.002917 **

Residuals 118 45.818  0.3883

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

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