The following data on price ($) and the overall score for 6 stereo headphones th

Question

The following data on price ($) and the overall score for 6 stereo headphones that were tested by Consumer Reports were as follows Brand Bose Scullcandy Koss Phillips/O'Neill Denon JVC Price 180 160 95 80 80 35 Score 76 79 61 58 50 27 a. Does the t test indicate a significant relationship between price and the overall score? The test t-Conclusion at a = .05 (to 2 decimal places.) p-value is C - Select your answer - what is your conclusion? Use a = .05 Select your answer b. Test for a significant relationship using the F test. p-value is- Select your answer What is your conclusion? Use a = .05 Because p-value is -select your answer- .05, we -select your answer- Ho: 1 is -select your answer- c. Show the ANOVA table for these data. Round your answers to three decimal places, if necessary Sum of Degrees of Source of Variation Regression Squares Freedom Mean Square p-value - Select your answer Error Total

Explanation / Answer

> price <- c(180,160,95,80,80,35)

> score <- c(76,79,61,58,50,27)

> relation <- lm(price ~ score)

> summary(relation)

Call:

lm(formula = price ~ score)

Residuals:

1 2 3 4 5 6

27.8303 -0.2559 -16.7385 -23.6523 -2.0890 14.9054

Coefficients:

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

(Intercept) -52.681 30.868 -1.707 0.16308

score 2.695 0.506 5.327 0.00598 **

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Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 21.45 on 4 degrees of freedom

Multiple R-squared: 0.8765, Adjusted R-squared: 0.8456

F-statistic: 28.38 on 1 and 4 DF, p-value: 0.005977

Hence,

(a)

t = 5.327 ~ 5.33 (round to 2 decimal places)

p-value = 0.00598

At 0.05 significance level,

0.05 > p-value

so we reject the null hypothesis that the score and price are not related.

Hence, we conclude that the score and the price are related significantly.

(b)

Using the F test,

p-value = 0.00598

At 0.05 significance level,

0.05 > p-value

so we reject the null hypothesis that the score and price are not related.

Hence, we conclude that the score and the price are related significantly.

(c)

> anova(relation)

Analysis of Variance Table

Response: price

Df Sum Sq Mean Sq F value Pr(>F)

score 1 13059.3 13059.3 28.378 0.005977 **

Residuals 4 1840.7 460.2   

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Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

THe df column is the degrees of freedom.

The Sum Sq column is the sum of sqaures.

The Mean Sq column is the Mean square

The F value column is the F

The Pr(>F) column is the p-value.

The score row stands for the regression source of variation.

The Residuals row stands for the error source of variation.

For the total source of variation,

sum of squares = 13059.3 + 1840.7 = 14900

Degrees of freedom = 1 + 4 = 5

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