Question

Consider the following data on price ($) and the overall score for six stereo headphones tested by a certain magazine. The overall score is based on sound quality and effectiveness of ambient noise reduction. Scores range from 0 (lowest) to 100 (highest).

Brand	Price ($)	Score
A	180	74
B	150	73
C	95	63
D	70	58
E	70	40
F	35	28

Find the value of the test statistic. (Round your answer to three decimal places.)_____

Find the p-value. (Round your answer to four decimal places.)

p-value = ____

2.-Test for a significant relationship using the F test. Use α = 0.05.

Find the value of the test statistic. (Round your answer to two decimal places.)_____

Find the p-value. (Round your answer to three decimal places.)

p-value = ____

What is your conclusion?

Reject H₀. We conclude that the relationship between price ($) and overall score is significant.

Do not reject H₀. We conclude that the relationship between price ($) and overall score is significant.    

Reject H₀. We cannot conclude that the relationship between price ($) and overall score is significant.

Do not reject H₀. We cannot conclude that the relationship between price ($) and overall score is significant.

(c)Show the ANOVA table for these data. (Round your p-value to three decimal places and all other values to two decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Regression
Error
Total

Answer 1

In order to solve this question I used R software.

R codes and output:

> price=c(180,150,95,70,70,35)
> score=c(74,73,63,58,40,28)
> fit=lm(price~score)
> summary(fit)

Call:
lm(formula = price ~ score)

Residuals:
1 2 3 4 5 6
32.573 5.208 -23.444 -35.270 12.157 8.775

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -47.5498 39.4593 -1.205 0.2946
score 2.6348 0.6747 3.905 0.0175 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 27.87 on 4 degrees of freedom
Multiple R-squared: 0.7922, Adjusted R-squared: 0.7403
F-statistic: 15.25 on 1 and 4 DF, p-value: 0.01747

> anova(lm(price~score))
Analysis of Variance Table

Response: price
Df Sum Sq Mean Sq F value Pr(>F)
score 1 11843.5 11843.5 15.25 0.01747 *
Residuals 4 3106.5 776.6
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>

Que.1

Test statistic for testing slope coefficient is t = 3.905

and p value = 0.0175

Since p-value is less than 0.05, slope coefficient is statistically significant.

Que.2

F test = 15.25

p-value= 0.0175
Since p-value is less than 0.05, we reject H₀ and conclude that relationship between price and overall score is significant.

c. ANOVA table:

Source of variation	SS	DF	MS	F	p-value
Regression	11843.5	1	11843.5	15.25	0.0175
Error	3106.5	4	776.625
Total	14950	5