To determine the [tex]\( t \)[/tex]-test statistic for Adam's road trip average speeds, we need to utilize the [tex]\( t \)[/tex]-test formula:
[tex]\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\][/tex]
where:
- [tex]\(\bar{x}\)[/tex] is the sample mean
- [tex]\(\mu\)[/tex] is the population mean
- [tex]\(s\)[/tex] is the sample standard deviation
- [tex]\(n\)[/tex] is the sample size
Given the values:
- Sample speeds: [tex]\(60.5, 63.2, 54.7, 51.6, 72.3, 70.7, 67.2, 65.4 \, \text{mph}\)[/tex]
- Sample standard deviation, [tex]\(s = 7.309\)[/tex]
- Population mean, [tex]\(\mu = 65 \, \text{mph}\)[/tex]
- Sample size, [tex]\(n = 8\)[/tex]
First, we calculate the sample mean, [tex]\(\bar{x}\)[/tex]:
[tex]\[
\bar{x} = \frac{60.5 + 63.2 + 54.7 + 51.6 + 72.3 + 70.7 + 67.2 + 65.4}{8}
\][/tex]
The sum of the speeds is:
[tex]\[
60.5 + 63.2 + 54.7 + 51.6 + 72.3 + 70.7 + 67.2 + 65.4 = 505.6
\][/tex]
Therefore, the sample mean, [tex]\(\bar{x}\)[/tex]:
[tex]\[
\bar{x} = \frac{505.6}{8} = 63.2
\][/tex]
Now, substitute the values into the formula:
[tex]\[
t = \frac{63.2 - 65}{7.309 / \sqrt{8}}
\][/tex]
First, compute the denominator:
[tex]\[
s / \sqrt{n} = 7.309 / \sqrt{8}
\][/tex]
[tex]\[
\sqrt{8} \approx 2.828
\][/tex]
[tex]\[
7.309 / 2.828 \approx 2.585
\][/tex]
Now, compute the [tex]\( t \)[/tex]-value:
[tex]\[
t = \frac{63.2 - 65}{2.585}
\][/tex]
[tex]\[
t = \frac{-1.8}{2.585}
\][/tex]
[tex]\[
t \approx -0.6965616123331693
\][/tex]
Finally, rounded to the hundredths place, the [tex]\( t \)[/tex]-statistic is approximately [tex]\( -0.70 \)[/tex]. Given the answer choices, none match exactly the computed [tex]\( t \)[/tex]-statistic. Instead, the closest rounded value (though note there are no close matches to [tex]\( -0.70\)[/tex]) provided is not represented in the options [tex]\( -2.87 \)[/tex] or [tex]\( -1.44 \)[/tex].
