Question

In order to test the city efficiency of the 10 Toyota Prius vehicles, the following data were collected (in mpg units):

1	45.3
2	45.2
3	45.1
4	44.8
5	45.2
6	45.5
7	44.9
8	44.8
9	45.1
10	44.8

a) Compute a sample mean and a sample standard deviation for data collected.

b) Suppose that the data set in part a) is used to determine that the Toyota hybrid is more efficient than a Honda hybrid. For a random sample of 12 Honda hybrid vehicles the average consumption was estimated 47.6 mpg with the sample standard deviation s of 0.3 mpg. Use α = 0.05

I. List data assumptions.

II. State H0 and Ha.

III. Calculate the test statistic.

IV. Make decision using the p-value approach.

V. Draw conclusion.

c) Compute 95% confidence interval for the difference in mean mpgs between a Toyota hybrid and a Honda hybrid.

Answer 1

(a) Sample mean = (45.3+45.2+45.1+44.8+45.2+45.5+44.9+44.8+45.1+44.8)/10 = 45.07

Sample standard deviation = (45.3-45.07)^2+45.2-45.07)^2+45.1-45.07)^2+44.8-45.07)^2+45.2-45.07)^2+45.5-45.07)^2+44.9-45.07)^2+44.8-45.07)^2+45.1-45.07)^2+44.8-45.07)^2)/10 - 1 = 0.241

(b) The common assumptions made when doing a t-test include those regarding the scale of measurement, random sampling, normality of data distribution, adequacy of sample size and equality of variance in standard deviation.

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

s2p = (10 - 1)*0.241^2 + (12 - 1)*0.3^2/(10 + 12 - 2) = 0.076

se = 0.076*(1/10 + 1/15) = 0.118

t = (45.07 - 47.6)/0.118 = -21.485

The p-value is 0.000.

Since the p-value (0.000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the Toyota hybrid is more efficient than a Honda hybrid.

(c) The 95% confidence interval for the difference in mean mpgs between a Toyota hybrid and a Honda hybrid is between -2.77564 and -2.28436.

= (45.07 - 47.6) 2.09*0.118

= -2.77564, -2.28436

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