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Question 23, 8.3.23-T
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Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement? $\begin{array}{llllll}626 & 655 & 1017 & 562 & 516 & 584\end{array}$
What are the hypotheses?
A. $H_{0}: \mu> 1000$ hic
$H_{1}: \mu< 1000$ hic
C. $H_{0}: \mu< 1000$ hic
$H_{1}: \mu \geq 1000$ hic
B.
\[
\begin{array}{l}
H_{0}: \mu=1000 \text { hic } \\
H_{1}: \mu< 1000 \text { hic }
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu=1000 \mathrm{hic} \\
H_{1}: \mu \geq 1000 \mathrm{hic}
\end{array}
\]
Identify the test statistic.
$t=\square$ (Round to three decimal places as needed.)
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Answer
Rounding to three decimal places, the final answer is \(\boxed{-4.588}\).
Step 1 ：The question is asking to test the claim that the sample is from a population with a mean less than 1000 hic. This is a one-tailed t-test because we are testing if the mean is less than a certain value.
Step 2 ：The null hypothesis is always a statement of no effect or no difference. In this case, the null hypothesis would be that the mean is equal to 1000 hic. The alternative hypothesis is what we are testing for, which is that the mean is less than 1000 hic.
Step 3 ：The test statistic for a one-sample t-test is calculated as: \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\) where: \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 4 ：Given the data [626, 655, 1017, 562, 516, 584], we calculate the sample mean \(\bar{x}\) as 660.0, the sample standard deviation \(s\) as 181.51914499578274, and the sample size \(n\) as 6.
Step 5 ：Substituting these values into the formula, we get the test statistic \(t = \frac{660.0 - 1000}{181.51914499578274 / \sqrt{6}} = -4.588091865272006\).
Step 6 ：Rounding to three decimal places, the final answer is \(\boxed{-4.588}\).
