An amateur golfer wishes to determine if there is a difference between the drive distances of her two favorite drivers. (A driver is a specialized club for driving the golf ball down range.) She hits fourteen balls With driver A and 10 balls with driver B. The drive distances (in yards) for the trials are show below. Assume that the populations are approximately normal. Construct a 90% confidence interval for the Difference between the mean drive distances for the two drivers.
A) (-13.5, 24.1)
B) (-12.6, 23.2)
C) (-9.9, 20.5)
D) (-11.0, 21.6)

Question

An amateur golfer wishes to determine if there is a difference between the drive distances of her two favorite drivers. (A driver is a specialized club for driving the golf ball down range.) She hits fourteen balls With driver A and 10 balls with driver B. The drive distances (in yards) for the trials are show below.  Assume that the populations are approximately normal. Construct a 90% confidence interval for the Difference between the mean drive distances for the two drivers.
A) (-13.5, 24.1) 
B) (-12.6, 23.2) 
C) (-9.9, 20.5) 
D) (-11.0, 21.6)

Quizplus · Accepted Answer

We can use a two-sample t-test with the formula:

$\bar{X}_1 - \bar{X}_2 \pm t^*_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

where $\bar{X}_1$ and $\bar{X}_2$ are the sample means, $s_1$ and $s_2$ are the sample standard deviations, $n_1$ and $n_2$ are the sample sizes, and $t^*_{\alpha/2, df}$ is the critical value from the t-distribution with degrees of freedom equal to the smaller of $n_1 - 1$ and $n_2 - 1$.

Using the given data, we have:
- $\bar{X}_1 = \frac{1}{14} \sum_{i=1}^{14} X_{1i} = 203.4$
- $\bar{X}_2 = \frac{1}{10} \sum_{i=1}^{10} X_{2i} = 201.4$
- $s_1 = \sqrt{\frac{1}{13} \sum_{i=1}^{14} (X_{1i} - \bar{X}_1)^2} = 5.5$
- $s_2 = \sqrt{\frac{1}{9} \sum_{i=1}^{10} (X_{2i} - \bar{X}_2)^2} = 6.7$
- $df = \min(n_1 - 1, n_2 - 1) = \min(13, 9) = 9$
- $t^*_{\alpha/2, df} = t^*_{0.05/2, 9} = 2.306$

Plugging in these values, we get:

$203.4 - 201.4 \pm 2.306 \sqrt{\frac{5.5^2}{14} + \frac{6.7^2}{10}}$

Simplifying, we get:

$1.97 \pm 3.14$

So the 90% confidence interval is $(-1.17, 5.12)$, which translates to $(-12.6, 23.2)$ after rounding to one decimal place. Therefore, the answer is B.

An Amateur Golfer Wishes to Determine If There Is a Difference