An inventor has developed a new spray coating that is designed to improve the wear of bicycle tires. To test the new coating, the inventor randomly selects one of the two tires on each of 50 bicycles to be coated with the new spray. The bicycle is then driven for 100 miles and the amount of the depth of the tread left on the two bicycle tires is measured (in millimeters). It is desired to determine whether the new spray coating improves the wear of the bicycle tires. The data and summary information is shown below:
Coated Non-Coated Difference
Mean 1.38 0.85 0.53
Std. Dev. 0.12 0.11 0.06
Sample Size 50 50 50
Use the summary data to construct a 90% confidence interval for the difference between the means.
| 0.53 ± 0.01396 | ||
| 0.53 ± 0.03787 | ||
| 0.53 ± 0.01663 | ||
| 0.53 ± 0.04512 |
Answer : 0.53 ± 0.01396
Working notes for the above answer is as follow
we need to construct a 90% confidence interval for the difference between the means.
A confidence interval on the difference between means is computed using the following formula:
Lower Limit = M1 – M2 -(tCL)(
)
Upper Limit = M1 – M2 +(tCL)()
where M1 – M2 is the difference between sample means,
Coated Non-Coated Difference
Mean 1.38 0.85 0.53
so the diifrance between mean is 0.53
We know the standard error of the difference between means is
= 0.53
From either the calculator or a t table, you can find that the t for a 90% confidence interval for 32 df is 0.01396
So we can say that.,confidence interval for the difference between the means.= 0.53 ± 0.01396
