Consider the following Excel regression of perceived sound quality as a function of price for 27 stereo speakers. Regression Statistics R Square0.01104 Standard Error4.02545 Observations27 StatisticCoefficientsStd Error t Stat P-valueLower 95%Upper 95% Intercept88.49021.6781452.7310.0000 85.0340 91.9464 Price–0.002390.00453–0.5280.6019–0. 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

Kindly check explanation

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

step-by-step explanation:

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg