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A tire manufacturer is interested in testing the fuel economy for two different tread patterns. Tires of each tread type are driven for 1000 miles on each of 18 different cars. In a separate experiment, 18 cars were outfitted with tires with tread type A, and another 18 were outfitted with tires with tread type B. Each car was driven 1000 miles. The cars with tread type A averaged 23.95 mpg, with a standard deviation of 1.79 mpg. The cars with tread type B averaged 22.19 mpg, with a standard deviation of 1.95 mpg. Let μX represent the population mean mileage for tread type A and let μY represent the population mean mileage for tread type B.


Required:

Find a 99% confidence interval for the difference μXâμY. Round the answers to three decimal places.

Sagot :

Using the t-distribution, it is found that the 99% confidence interval for the difference is (0.058, 3.462).

We are given the standard deviation for the samples, hence, the t-distribution is used to solve this question.

The standard errors are given by:

[tex]s_A = \frac{1.79}{\sqrt{18}} = 0.4219[/tex]

[tex]s_B = \frac{1.95}{\sqrt{18}} = 0.4596[/tex]

For the distribution of differences, the mean and the standard error are given by:

[tex]\overline{x} = \mu_A - \mu_B = 23.95 - 22.19 = 1.76[/tex]

[tex]s = \sqrt{0.4219^2 + 0.4596^2} = 0.6239[/tex]

The interval is given by:

[tex]\overline{x} \pm ts[/tex]

The critical value for a two-tailed 99% confidence interval with 18 + 18 - 2 = 34 df is t = 2.7284.

Then:

[tex]\overline{x} - ts = 1.76 - 2.7284(0.6239) = 0.058[/tex]

[tex]\overline{x} + ts = 1.76 + 2.7284(0.6239) = 3.462[/tex]

The 99% confidence interval for the difference is (0.058, 3.462).

For more on the t-distribution, you can check https://brainly.com/question/16162795

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