Questions: A company that manufactures baseball bats believes that its new bat will allow players to hit the ball 30 feet farther than its current model. The owner hires a professional baseball player known for hitting home runs to hit ten balls with each bat and he measures the 90% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the old model. Round the endpoints of the interval to one decimal place, if necessary. Hitting Distance (in Feet) New Model Old Model --- --- 278 244 264 235 253 257 261 283 259 270 243 211 239 230 267 237 263 298 248 274

Solution Steps

A baseball bat company believes that its new bat will allow players to hit the ball 30 feet further than its current model. To test this claim, the company hired a professional baseball player to hit ten balls with each bat. The table shows the hitting distances for each hit. We will construct a 90% confidence interval for the true difference between the mean hitting distance with the new bat and the mean hitting distance with the old bat. The population variances are assumed to be equal. Population 1 represents the distances of balls hit with the new model and Population 2 represents the distances of balls hit with the old model. Round the endpoints of the interval to one decimal place.

First, calculate the difference in hitting distances between the new and old models for each hit. This is done by subtracting the old model distance from the new model distance. Here are the differences:

34, 32, 34, 36, 30, 29, 31, 24, 32, 38

Next, calculate the mean ($\bar{d}$) and standard deviation ($s_d$) of these differences.

$\bar{d} = \frac{34+32+34+36+30+29+31+24+32+38}{10} = 32$

$s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{(2^2 + 0 + 2^2 + 4^2 + (-2)^2 + (-3)^2 + (-1)^2 + (-8)^2 + 0 + 6^2)}{9}} \approx 4.17$

For a 90% confidence interval with 9 degrees of freedom ($n-1=10-1=9$), the $t$-critical value ($t_{\alpha/2}$) is approximately 1.833 (using a t-table or calculator).

The margin of error (ME) is calculated as:

$ME = t_{\alpha/2} \cdot \frac{s_d}{\sqrt{n}} = 1.833 \cdot \frac{4.17}{\sqrt{10}} \approx 2.4$

Finally, construct the 90% confidence interval by adding and subtracting the margin of error from the mean difference:

Lower bound: $\bar{d} - ME = 32 - 2.4 = 29.6$

Upper bound: $\bar{d} + ME = 32 + 2.4 = 34.4$

Final Answer: