Question

# The horsepower (Y, in bhp) of a motor car engine was measured at a chosen set...

The horsepower (Y, in bhp) of a motor car engine was measured at a chosen set of values of running speed (X, in rpm). The data are given below (the first row is the running speed in rpm and the second row is the horsepower in bhp): rpm 1100 1400 1700 2300 2700 3200 3500 4000 4600 5200 5600 6100 Horsepower (bhp) 50.26 63.89 77.51 126.43 131.03 154.26 176.92 195.02 225.47 240.79 275.9 312.5

The mean and sum of squares of the rpm are 3450.0000 rpm and 173900000.0000 rpm 2 respectively; the mean of the horsepower values is 169.1650 bhp and the sum of the products of the two variables is 8554561.0000 rpm bhp.

Please provide answers to the following to 3 decimals places where appropriate:

Part a) Compute the regression line for these data, and provide your estimates of the slope and intercept parameters. Please round intermediate results to 6 decimal places. Slope: Intercept: Note: For sub-parts below, use the slope and intercept values in Part a, corrected to 3 decimal places to calculate answers by hand using a scientific calculator.

Part b) Based on the regression model, what level of horsepower would you expect the engine to produce if running at 2400 rpm? Answer:

Part c) Assuming the model you have fitted, if increase the running speed by 100 rpm, what would you expect the change in horsepower to be? Answer:

Part d) The standard error of the estimate of the slope coefficient was found to be 0.001449. Provide a 95% confidence interval for the true underlying slope. Confidence interval: ( , )

Part e) Without extending beyond the existing range of speed values or changing the number of observations, we would expect that increasing the variance of the rpm speeds at which the horsepower levels were found would make the confidence interval in (d)

A. either wider or narrower depending on the values chosen.

B. unchanged.

C. narrower.

D. wider.

Part f) If testing the null hypothesis that horsepower does not depend linearly on rpm, what would be your test statistic? (For this part, you are to calculate the test statistic by hand using appropriate values from the answers you provided in part (a) accurate to 3 decimal places, and values given to you in part (d).) Answer:

Part g) Assuming the test is at the 1% significance level, what would you conclude from the above hypothesis test?

A. Since the observed test statistic falls in either the upper or lower 1/2 percentiles of the t distribution with 10 degrees of freedom, we cannot reject the null hypothesis that the horsepower does not depend linearly on rpm.

B. Since the observed test statistic does not fall in either the upper or lower 1 percentiles of the t distribution with 10 degrees of freedom, we cannot reject the null hypothesis that the horsepower does not depend linearly on rpm.

C. Since the observed test statistic falls in either the upper or lower 1/2 percentiles of the t distribution with 10 degrees of freedom, we can reject the null hypothesis that the horsepower does not depend linearly on rpm.

D. Since the observed test statistic does not fall in either the upper or lower 1/2 percentiles of the t distribution with 10 degrees of freedom, we can reject the null hypothesis that the horsepower does not depend linearly on rpm.

E. Since the observed test statistic does not fall in either the upper or lower 1/2 percentiles of the t distribution with 10 degrees of freedom, we cannot reject the null hypothesis that the horsepower does not depend linearly on rpm.

a)

 ΣX ΣY Σ(x-x̅)² Σ(y-ȳ)² Σ(x-x̅)(y-ȳ) total sum 41400.00 2029.98 31070000.00 78090.88 1551130.00 mean 3450.00 169.17 SSxx SSyy SSxy

Sample size,   n =   12
here, x̅ = Σx / n=   3450.000
ȳ = Σy/n =   169.165
SSxx =    Σ(x-x̅)² =    31070000.0000
SSxy=   Σ(x-x̅)(y-ȳ) =   1551130.0

estimated slope , ß1 = SSxy/SSxx =   1551130/31070000=   0.0499
intercept,ß0 = y̅-ß1* x̄ =   169.165- (0.0499 )*3450=   -3.0718

Regression line is, Ŷ=   -3.072   + (   0.050   )*x

b)

Predicted Y at X=   2400   is
Ŷ=   -3.07184   +   0.04992   *2400=   116.745

c)

Increase by 0.05 * 100

= 5

d)

α=   0.05
t critical value=   t α/2 =    2.228   [excel function: =t.inv.2t(α/2,df) ]
estimated std error of slope = Se/√Sxx =    8.079/√31070000=   0.00145

margin of error ,E= t*std error =    2.228   *   0.001   =   0.003229
estimated slope , ß^ =    0.0499

lower confidence limit = estimated slope - margin of error =   0.0499   -   0.003   =   0.0467
upper confidence limit=estimated slope + margin of error =   0.0499   +   0.003   =   0.0532

e)

D. wider.

Please let me know in case of any doubt.

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