Question

CAR 1	MILEAGE	CAR 2	MILEAGE	CAR 3	MILEAGE	CAR 4	MILEAGE
1	14.9	2	10.8	3	19	4	18.9
1	17.7	2	10.7	3	13.8	4	19.2
1	17.7	2	11	3	20.1	4	19.4
1	18.7	2	12	3	19.8	4	21
1	19.8	2	7.5	3	12.2	4	13.5
1	21.1	2	10.5	3	24.3	4	17.2
1	17.3	2	9.1	3	21.8	4	12.7
1	19.8	2	10.7	3	20.7
1	16.3	2	7.5	3	16.4
1	17.8	2	12.1	3	25.4
COUNT	10	COUNT	10	COUNT	10	COUNT	7
MEAN	18.11	MEAN	10.2285714	MEAN	19.35	MEAN	17.4142857

With the above data, could you please help me answer questions 10-13:

Calculate a 95% confidence interval for the mean mileage of make 2. Use the method for single means when σ is not known, but use the Error Mean Square as the estimate of the variance. The degrees of freedom will be the Error DF, not n-1!

Reminders: Confidence Interval = mean ± margin of error Margin of error = critical value * standard error Use critical value for T at α/2 = 0.025 and df = error df (t table or EXCEL T.INV function) Use standard error = √(error mean square/number of observations of that make of car)

10. What was the margin of error for the confidence interval for gasoline mileage of make 2?

11. What was the lower 95% confidence limit for make 2 mileage?

12. What was the upper 95% confidence limit for make 2 mileage?

Conduct a test of the hypothesis that the mean mileage of makes 2 and 3 do not differ. Use the method for single means when σ is not known with the Error MS serving as the pooled variance.

Reminders: Test statistic t = difference of means / standard error of difference of means. The standard error of the difference equals square root of the sum of variances of the two means. The variance of each mean is estimated by the error mean square/number of observations in that mean.

13. What is the value of the t test statistic for testing the hypothesis that makes 2 and 3 do not differ in mileage?

Answer 1

sample Mean = 10.19
t critical = 2.26
sM = √(1.6412/10) = 0.52
μ = M ± t(sM)
μ = 10.19 ± 2.26*0.52

10] MARGIN OF ERROR= 2.26*0.52= 1.175
μ = 10.19 ± 1.175

11] LOWER LIMIT= 10.19-1.175= 9.015

12] UPPER LIMIT= 10.19+1.175=11.365

95% CI [9.015, 11.365].

You can be 95% confident that the population mean (μ) falls between 9.015 and 11.365

13-14]

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=9.

Hence, it is found that the critical value for this two-tailed test is tc​=2.262, for α=0.05 and df=9.

The rejection region for this two-tailed test is R={t:∣t∣>2.262}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=8.235>tc​=2.262, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.000, and since p=0.000<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.