Question

In: Statistics and Probability

you may need to use the appropriate technology to answer this question. Consider the following data...

you may need to use the appropriate technology to answer this 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 61
D 70 58
E 70 42
F 35 28

(a) The estimated regression equation for this data is  ŷ = 26.268 + 0.297x,  where x = price ($) and y = overall score. Does the t test indicate a significant relationship between price and the overall score? Use α = 0.05.

State the null and alternative hypotheses.

State the null and alternative hypotheses.

H0: β0 = 0
Ha: β0 ≠ 0H0: β1 = 0
Ha: β1 ≠ 0    H0: β1 ≥ 0
Ha: β1 < 0H0: β1 ≠ 0
Ha: β1 = 0H0: β0 ≠ 0
Ha: β0 = 0

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 =

What is your conclusion?

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 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.

(b)

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

State the null and alternative hypotheses.

H0: β0 ≠ 0
Ha: β0 = 0H0: β0 = 0
Ha: β0 ≠ 0    H0: β1 = 0
Ha: β1 ≠ 0H0: β1 ≥ 0
Ha: β1 < 0H0: β1 ≠ 0
Ha: β1 = 0

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?

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

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

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

D) Reject H0. We 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

Solutions

Expert Solution

We can perform this analysis in MS Excel as follows

Regression in excel

I performed the analysis using basic commands in MS Excel and I am giving the steps to be followed-

  1. Select the Data tab and choose the Data Analysis in the top right-hand corner
  2. In the Data Analysis menu choose Regression: Single Factor and click OK
  3. In the ‘Input Range’ box, select all the data in the columns you created, including the variable names
  4. Check the ‘Labels in First Row’ box
  5. In the ‘Output Range’ box, enter a cell range where Excel will place the output and click OK
  6. If the p-value were less than 0.05, you would reject the null hypothesis that says the means of all categories are equal. If the p-value were greater than 0.05, then you would fail to reject the null.

A) The Null hypothesis can be given as

H0: β0 ≠ 0

and alternative hypothesis

Ha: β0 ≠ 0

Now the results from Excel is as follows

As mentioned in question we get the same regression equation

ŷ = 26.268 + 0.297x,  where x = price ($) and y = overall score

test statisitc: t= 4.195

p-value = 0.0137 which is less than 0.05

Conclusion:   we reject the null hypothesis in favor of the alternative hypothesis

B) & C) Test for a significant relationship using the F test and ANOVA table

The Null hypothesis can be given as

H0: β0 ≠ 0

and alternative hypothesis

Ha: β0 ≠ 0

test statistic: F= 17.598

p-value = 0.0137 which is less than 0.05

Conclusion:   we reject the null hypothesis in favor of the alternative hypothesis


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