The mileage (the distance a car can travel on one liter of gasoline) of a car is

Question

The mileage (the distance a car can travel on one liter of gasoline) of a car is measured. Historically, it is known that the measurement technique has a standard deviation of 1 kilometer (the result is random due to the influence of air temperature and wind, tire types, ete.) and produces results that are normally distributed with zero mean error. Twenty five measurements were made and the average value of 12 km 1 was obtained. 1. Find the 90% and 95% confidence intervals for the mileage. 2. What is the confidence level of the confidence interval 11.5, 12.51 km/ 3. How many measurements should be made in order to obtain the margin of error no more than 0.2 km at the 99% confidence level?

Explanation / Answer

1) At 90% confidence interval the critical value is z0.05 = 1.645

The interval is

Mean +/- z0.05 * SD/Sqrt (n)

= 12 +/- 1.645 * 1/sqrt(25)

= 12 +/- 0.329

= -11.671, 12.329

At 95% confidence interval the critical value is z0.025 = 1.96

Mean +/- z0.025 * SD/Sqrt (n)

= 12 +/- 1.96 * 1/sqrt(25)

= 12 +/- 0.392

= 11.608, 12.392

2) mean - z* * SD/sqrsqrt(n ) = 11.5

Or, 12 - z* * 1/sqrt (25) = 11.5

Or, z* * 1/sqrt (25) = 0.5

Or, z* = 0.5 * sqrt(25)

Or, z* = 2.5

The confidence interval is 98.76%

3) At 99% confidence interval the critical value is z0.005 = 2.58

Margin of error = 0.2

Or, z0.005 * sd/sqrt (n ) = 0.2

Or, 2.58 * 1/sqrt(n) = 0.2

Or, sqrt(n) = 2.58/0.2

Or, sqrt(n) = 12.9

Or, n = 166.41 = 166

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