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Consider the following Excel regression of perceived sound quality as a function of price for 27 stereo speakers. Regression Statistics R Square 0.01104 Standard Error 4.02545 Observations 27 Statistic Coefficients Std Error t Stat P-value Lower 95% Upper 95% Intercept 88.4902 1.67814 52.731 0.0000 85.0340 91.9464 Price –0.00239 0.00453 –0.528 0.6019 –0.01172 0.00693 Picture Click here for the Excel Data File(a) Is the coefficient of Price significantly different from zero at α = .05?YesNo(b) Calculate the correlation coefficient. At α = .05, does the correlation differ from zero? (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.) r tcalc t.025 ± Since tcalc (Click to select)>< t.025, we (Click to select)rejectfail to reject the hypothesis of no correlation.(c) Given these results, is the difference in quality associated with a higher price likely to be apparent to an average listener?YesNo

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Answer:

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Step-by-step explanation:

Given the following :

Regression

Statistics

R Square 0.01104

Standard Error 4.02545

Observations 27

Stat - - - - Coef - - Std Error - - t-Stat - - P-value

Int. - - 88.4902 - 1.67814 - - 52.731 - - 0.0000

Price (–0.00239) - 0.00453 (–0.528) - - 0.6019

Stat - - - - lower 95% - - - upper 95%

Int. - - - - - 85.0340 - - - - - - 91.9464

Price - - (–0.01172) - - - - - - 0.00693

(A) No, At α = .05, the Coefficient of price is not significantly different from zero, because :

The p-value (0.6019) obtained is greater than 0.05, Hence we fail to reject the null hypothesis.

B) the correlation Coefficient (r) :

Since ; r^2 = 0.01104

r = √0.01104

r = 0.1050714

The p-value obtained at α = .05 for r value of 0.1051 using the online p-value calculator is .602, this shows that correlation does not differ from 0.

Using :

r(√n-2) / √1 - r^2

0.1051(√27-2) / √1 - 0.01104

0.5255 / 0.9944646

= 0.5284250

For a 2 tailed test: α = .025

t0. 025 = 2.060 and - 2.060

Obtained t value is less than the critical, hence the null hypothesis stands

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