Question

Featured below are data on fuel economy (in miles/gallon) for a random sample of 12 mid-size cars.

X: {25.8, 20.2, 28.7, 24.6, 19.5, 33.1, 30.5, 27.7, 28.4, 30.9, 22.3, 24.2}

Construct a 95% confidence interval for the true mean fuel economy

Conduct a test to determine whether the true mean is higher than 25.0 miles/ gallon.

The null and alternative hypothesis

The test statistic

The p-value of the test,

Your decision and the interpretation in the context of the problem.

Answer 1

Solution:

x	x2
25.8	665.64
20.2	408.04
28.7	823.69
24.6	605.16
19.5	380.25
33.1	1095.61
30.5	930.25
27.7	767.29
28.4	806.56
30.9	954.81
22.3	497.29
24.2	585.64
∑x=315.9	∑x2=8520.23

Mean ˉx=∑xn

=25.8+20.2+28.7+24.6+19.5+33.1+30.5+27.7+28.4+30.9+22.3+24.2/12

=315.9/12

=26.325

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√8520.23-(315.9)212/11

=√8520.23-8316.0675/11

=√204.1625/11

=√18.5602

=4.3082

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,11 =2.201

Margin of error = E = t/2,df * (s /n)

= 2.201 * (4.31 / 12)

= 2.74

Margin of error = 2.74

The 95% confidence interval estimate of the population mean is,

- E < < + E

26.32 - 2.74< < 26.32 + 2.74

23.58 < < 29.06

(23.58 , 29.06 )

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :    = 25.0

Ha : > 25.0

Test statistic = t

= ( - ) / s / n

= (26.32-25.0) / 4.31 / 12

= 1.061

Test statistic = t = 1.061

P-value =0.1557

= 0.05  

P-value >

0.1557 > 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to suggest that