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Q. 11

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The Practice of Statistics for AP
Found in: Page 797
The Practice of Statistics for AP

The Practice of Statistics for AP

Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339

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Short Answer

Growth hormones are often used to increase the weight gain of chickens. In an experiment using 15 chickens, five different doses of growth hormone (0, 0.2, 0.4, 0.8, and 1.0 milligrams) were injected into chickens (3 chickens were randomly assigned to each dose), and the subsequent weight gain (in ounces) was recorded. A researcher plots the data and finds that a linear relationship appears to hold. Computer output from a least-squares regression analysis for these data is shown below.

role="math" localid="1652870715985" Predictor Coef SE Coef T P Constant 4.54590.61667.37<0.0001 Dose 4.83231.01644.750.0004S=3.135 R-Sq=38.4% R-Sq( adj )=37.7%

(a) What is the equation of the least-squares regression line for these data? Define any variables you use.

(b) Interpret each of the following in context:

(i) The slope

(ii) The y intercept

(iii) s

(iv) The standard error of the slope

(v) r2

(c) Assume that the conditions for performing inference about the slope B of the true regression line are met. Do the data provide convincing evidence of a linear relationship between dose and weight gain? Carry out a significance test at the A α=0.05 level.

(d) Construct and interpret a 95% confidence interval for the slope parameter.

(a) The equation of the least-squares regression line is y^=4.5459+4.8323x.

(b) The values are

(i) b=4

(ii) a=4.5450

(iii) s=3.135

(iv) SEb=1.0164

(v) role="math" localid="1652871593895" r2=38.4%

(c) There's sufficient convincing evidence to justify the argument.

(d) There is a 95% chance that the weight gain will be between 2.636876 and 7.027724. While the does, the ounces climb by 1 milligram.

See the step by step solution

Step by Step Solution

Part(a) Step 1: Given Information

Predictor Coef SE Coef T P Constant 4.54590.61667.37<0.0001 Dose 4.83231.01644.750.0004S=3.135 R-Sq=38.4% R-Sq( adj )=37.7%

Part(a) Step 2: Explanation

a=4.5459

b=4.8323

The general regression line equation

y^=a+bx

By inserting values the regression line becomes:

role="math" localid="1652870965008" y^=4.5459+4.8323x

With x Dose and y weight gain.

Part(b) Step 1: Given Information

Predictor Coef SE Coef T P Constant 4.54590.61667.37<0.0001 Dose 4.83231.01644.750.0004S=3.135 R-Sq=38.4% R-Sq( adj )=37.7%

Part(b) Step 2: Explanation

(i) b=4 is the first output. The weight per milligram might increase by 4.8323 ounces.

(ii). In the result a=4.5450, the y-intercept is mentioned.

This means that the weight will be 4.5459 ounces if the dosage is 0 milligrams.

(iii) s=3.135 is the output. This means that the average prediction error is 3.1350 ounces.

(iv) SEb=1.0164 This suggests that the population's true slope is 1.016-1 on average over all feasible samples.

(v) r2 is written as r2=R-Sq=38.4%. This means that the least-square regression line explains 38.4% of the variance in the variables.

Part(c) Step 1: Given Information

Predictor Coef SE Coef T P Constant 4.54590.61667.37<0.0001 Dose 4.83231.01644.750.0004S=3.135 R-Sq=38.4% R-Sq( adj )=37.7%

Part(c) Step 2: Explanation

Define hypothesis

H0:β=0

H1:β0

The test statistic is

t=b-β0SEb =4.8323-01.0164 =4.754

The P-value is the probability of having the test numbers' value or a more dramatic value. The t-value in the row role="math" localid="1652872078633" d f=n-2 =15-2 =13 is represented by a number (or interval) in Table B column title:

P<0.0005

The null hypothesis is rejected if the p-value is less than or equal to the degree of significance.

P<0.05 Reject H0

Part(d) Step 1: Given Information

Predictor Coef SE Coef T P Constant 4.54590.61667.37<0.0001 Dose 4.83231.01644.750.0004S=3.135 R-Sq=38.4% R-Sq( adj )=37.7%

Part(d) Step 2: Explanation

Degrees of freedom

d f=n-2 =15-2 =13

Table B, in the row of d f=13 and the column of c=95%, has the important t-value.

The essential t-value may be found in table B in the c=95% column and in the d f=13 row.

t*=2.160

The boundaries are

role="math" localid="1652872255479" b-t*×SEb=4.8323-2.160×1.0164 =2.636876

b+t*×SEb=4.8323+2.160×1.0164 =7.027724

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