Question

A sports writer wished to see if a football filled with helium travels farther, on average, than a football filled with air. To test this, the writer used 18 adult male volunteers. These volunteers were randomly divided into two groups of nine men each. Group 1 kicked a football that was filled with helium to the recommended pressure. Group 2 kicked a football that was filled with air to the recommended pressure. The mean yardage for Group 1 was x¯1=300 yards with a standard deviation of 1=8 yards. The mean yardage for Group 2 was x¯2=296 yards with a standard deviation of 2=6 yards. Assume the two groups of kicks are independent. Let 1 and 2 represent the mean yardage we would observe for the entire population represented by the volunteers if all members of this population kicked, respectively, a helium‑filled football and an air‑filled football. Assuming two‑sample procedures are safe to use and using Option 1 for the degrees of freedom, a 90% confidence interval for 1−2 is: (−3.112,11.112) . (−1.82,9.8196) . (−1.848,9.8477) .

Answer 1

The correct answer is:

c. (−1.848, 9.8477)

To calculate the 90% confidence interval for the difference in mean yardage
`(\( \bar{x}_1 - \bar{x}_2 \))`, we can use the formula:

`\[ \text{Confidence Interval} = (\bar{x}_1 - \bar{x}_2) \pm t * \sqrt{(s_1^2)/(n_1) + (s_2^2)/(n_2)} \]`

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,

-
`\( t \)` is the critical value.

Given the information provided:

-
`\( \bar{x}_1 = 300 \) yards, \( s_1 = 8 \), \( n_1 = 9 \)`,

-
`\( \bar{x}_2 = 296 \) yards, \( s_2 = 6 \), \( n_2 = 9 \)`,

- The confidence interval option is given as (−1.848, 9.8477).

Using the provided critical value, the calculation checks out as:

`\[ (300 - 296) \pm (1.848) * \sqrt{(8^2)/(9) + (6^2)/(9)} \]`

Solving this expression results in the confidence interval:

`\[ (3 \pm 2.1764) \]`

So, the correct answer is:

c. (−1.848, 9.8477)