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# Hypothesis Tests: Molar Heat of Fusion of Water; Difference in Diesel Truck Fuel Economy

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Exercise 1

Two methods for measuring the molar heat of fusion of water are being compared. Ten measurements made by method A have a mean of 6.02 kilojoules per mole (kJ/mol) with a standard deviation of 0.02 (kJ/mol). Five measurements made by method B have a mean of 6.00 kJ/mol and a standard deviation of 0.01 kJ/mol.

a) Can you conclude that the mean measurements differ between the two methods? Carry out the appropriate hypothesis test.
b) Verify your result in part (a) using the appropriate confidence interval.
c) Assuming the population variances were equal, can you conclude that the mean measurements from Method A exceed that of Method B by more than .001 kJ/mol? Carry out the appropriate hypothesis test.
d) Find a 99.8% two sided confidence interval based on the knowledge of "equal variances".

Exercise 2

A sample of 10 diesel trucks were run both hot and cold to estimate the difference in fuel economy. The results, in mi/gal, are presented in the following table. (From "in-sue Emissions from Heavy-Duty Diesel Vehicles," J. Yanowitz, Ph.D. thesis, Colorado School of Mines, 2001.)
_________________________________
Truck Hot Cold
1 4.56 4.26
2 4.46 4.08
3 6.49 5.83
4 5.37 4.96
5 6.25 5.87
6 5.90 5.32
7 4.12 3.92
8 3.85 3.69
9 4.15 3.74
10 4.69 4.19

a) Can you conclude that the mean fuel mileage of Cold engines is less than that of the hot engines? Carry out the appropriate hypothesis test.

b) Find a 98% confidence interval for the difference in mean fuel mileage between hot and cold engines.

##### Solution Summary

The solution is attached in a 3-paged Word document containing both step-by-step calculation and written explanation for each part of the questions on hypothesis testing.

##### Solution Preview

!.
(a) We can use t-test
H0:
Ha:
Test statistics: T= = 2.581989
We reject the null hypothesis that the two means are equal if or , where is the critical value of the t-distribution with v degree of freedom.

So v=12.96=13
Here we use significance level .
By t-table .

So we reject H0 if T<-2.16 or T>2.16. As the calculated T=2.58>2.16, we reject H0 and claim that the mean measurements differ between the two methods.

(b) The 95% confidence interval is the ...

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