Moore PSAlaignment Computers in some vehicles calculate various quantities relat

Question

Moore PSAlaignment Computers in some vehicles calculate various quantities related to performance One of these is the fuel efficiency, or gas mieage, usually expressed as miles per gallon (mpg). For one vehicle equioped in this way the miles per gallon were time the tank and the then cakulating mies per gaton, the ver also recorded the miles per gallon by dividing the miles driven by the number of galons at n-up. The driver wants to determine if these calculations are dimerent. 416 50.7 36.6 373 34.2 44.7 47.8 43.2 474 42.0 365 4.2 37.2 35.6 30.5 40,5 40.0 41.0 42.8 39.2 13 14 19 Computer 43 1 44.4 483 4S 393 32.5 43.5 444 38.8 44 s 45.4 45.3 45.7 14.2 Js 2 39.8 449 47 s (a state the appropriate Ho. State the appropriate H, (b) Carry out the test. (Round your answer for t to four decimal places Give the degrees of freedom. Give the value. (Round your ansa er to four decimal places)

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 0
Alternative hypothesis: 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(4.3442/20) + (4.59752/20)] = 1.414

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (4.3442/20 + 4.59752/20)2 / { [ (4.3442 / 20)2 / (19) ] + [ (4.59752 / 20)2 / (19) ] }
DF = 4.00146858477 / (0.0468538922 +0.05878592542) = 38 (approx.)

t = [ (x1 - x2) - d ] / SE = [ (43.1 - 40.44) - 0 ] / 1.414 = 1.88

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 38 degrees of freedom.

We use the t Distribution Calculator to find P(t < 1.88)

The P-Value is 0.067791.
The result is not significant at p < .05.

Interpret results. Since the P-value (0.068) is more than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. We do not have sufficient evidence that the results of the two computations are different.

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