Question

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	2000	2300	2700	3200	3500	4000	4300	4600	5200	6100
Horsepower (bhp)	54.08	50.42	75.76	89.24	109.65	115.47	159.63	165.24	200.73	206.22	223.33	258.43	306.4

The mean and sum of squares of the rpm are 3238.46153238.4615 rpm and 165030000.0000165030000.0000 rpm 22 respectively; the mean of the horsepower values is 154.9692154.9692 bhp and the sum of the products of the two variables is 8020328.00008020328.0000 rpm bhp. A scatterplot displaying the data is shown below:



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 24002400 rpm?

Answer:

Part c)

Assuming the model you have fitted, if increase the running speed by 100100 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 0.0012030.001203. 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. wider.
C. unchanged.
D. narrower.


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 does not fall in either the upper or lower 1/2 percentiles of the t distribution with 1111 degrees of freedom, we cannot reject the null hypothesis that the horsepower does not depend linearly on rpm.
B. Since the observed test statistic falls in either the upper or lower 1/2 percentiles of the t distribution with 1111 degrees of freedom, we can reject the null hypothesis that the horsepower does not depend linearly on rpm.
C. Since the observed test statistic does not fall in either the upper or lower 1/2 percentiles of the t distribution with 1111 degrees of freedom, we can reject the null hypothesis that the horsepower does not depend linearly on rpm.
D. Since the observed test statistic falls in either the upper or lower 1/2 percentiles of the t distribution with 1111 degrees of freedom, we cannot 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 percentiles of the t distribution with 1111 degrees of freedom, we cannot reject the null hypothesis that the horsepower does not depend linearly on rpm.

Answer 1

(A)

	rpm	horsepower	rpm*horsepower	rpm2	horsepower2
	1100	54.08	59488	1210000	2924.6464
	1400	50.42	70588	1960000	2542.1764
	1700	75.76	128792	2890000	5739.5776
	2000	89.24	178480	4000000	7963.7776
	2300	109.65	252195	5290000	12023.1225
	2700	115.47	311769	7290000	13333.3209
	3200	159.63	510816	10240000	25481.7369
	3500	165.24	578340	12250000	27304.2576
	4000	200.73	802920	16000000	40292.5329
	4300	206.22	886746	18490000	42526.6884
	4600	223.33	1027318	21160000	49876.2889
	5200	258.43	1343836	27040000	66786.0649
	6100	306.4	1869040	37210000	93880.96
Sum =	42100	2014.6	8020328	165030000	390675.151

Based on the above table, the following is calculated:

Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n) are obtained as follows:

m(slope)=0.0521

intercept= -13.9052

Therefore, we find that the regression equation is:

horsepower = -13.9052 + 0.0521 rpm

2.We have

horsepower = -13.9052 + 0.0521 rpm

rpm=2400

horsepower= -13.9052+0.0521*2400

horsepower=111.1348

3.We have

horsepower = -13.9052 + 0.0521 rpm

rpm=100

horsepower= -13.9052+0.0521*100

horsepower= -8.6952

4.

95% confidence interval of slope:

CI=

Now,

lower limit=0.0521-2.200985*0.001203=0.04945

upper limit=0.0521+2.200985*0.001203=0.05474

CI=(0.04945,0.05474)

please rate my answer and comment for doubts.