2. A researcher compared 2 experimental engines for gas consumption for a couple

Question

2. A researcher compared 2 experimental engines for gas consumption for a couple of times. Series of trials had the following results for gas consumption (L/100 km). Engine A Engine B Sample Mean 5.4 4.9 Sample Standard Deviation 0.26 0.4 9 9 Sample Size Analysis indicates both engines' data are normally distributed. (a) Based on the common rule of thumb by comparing sample standard deviations are the variances for gas consumption of engines A and B considered to be equal? (b) Assume the variances are unequal. Construct 95% confidence interval for AuA-HB. Using the degrees of freedom of the smaller of ni 1 and n2 1. (c) Assume the variances are equal. We would like to see whether Engine B has lower level of gas consumption on average than Engine A, at 5% level of significance. State the null hypothesis and the alternative hypothesis. Perform an appropriate hypothesis test using the critical value method, and p-value method. Make your conclusion

Explanation / Answer

Part a

As per the thumb rule, the variances for the gas consumption of engines A and B are not same because for the both samples we are given same sample size but significantly different sample standard deviations. We assume approximately near or equal sample standard deviations for the equality of the population variances.

Part b

We are given

N1 = 9 and N2 = 9

N1 - 1 = 8 and N2 – 2 = 8

Required degrees of freedom = 8

Confidence level = 95%

Confidence interval = (X1bar – X2bar) -/+ t*sqrt[(S1^2/N1)+(S2^2/N2)]

Confidence interval = (5.4 – 4.9) -/+ 1.8595*sqrt[(0.26^2/9)+(0.4^2/9)]

Confidence interval = 0.5 -/+ 1.8595* 0.159025

Confidence interval = 0.5 -/+ 0.29571

Lower limit = 0.5 – 0.29571 = 0.20429

Upper limit = 0.5 + 0.29571 = 0.79571

Part c

Here, we have to use two sample t test for population mean.

H0: µ1 = µ2

Ha: µ1 > µ2

This is a one tailed test. This is a lower tailed test.

Test statistic formula is given as below:

t = = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

t = (5.4 – 4.9) / sqrt[(0.26^2/9)+(0.4^2/9)]

t = 0.5 / 0.159025

t = 3.14416

Degrees of freedom = 8

= 0.05

Critical value = t = 1.8595

P-value = 0.0069

P-value <

So, we reject the null hypothesis

This means we conclude that there is sufficient evidence that engine B has lower level of gas consumption on average than Engine A.

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