Question

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 address the original claim. a safety administration conducted crash tests of child booster seats for cars. listed below are results from those test 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 0.05 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? 632 583 1112 604 512 625 what are the hypotheses? a. ( h_0:mu = 1000 ) hic ( h_1:mugeq1000 ) hic b. ( h_0:mu>1000 ) hic ( h_1:mu<1000 ) hic c. ( h_0:mu<1000 ) hic ( h_1:mugeq1000 ) hic d. ( h_0:mu = 1000 ) hic ( h_1:mu<1000 ) hic identify the test statistic. ( t=square ) (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample mean $\bar{x}$

The formula for sample mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Given data \(x = [632,583,1112,604,512,625]\), \(n = 6\).
\(\sum_{i=1}^{6}x_{i}=632 + 583+1112+604+512+625=4068\)
\(\bar{x}=\frac{4068}{6}=678\)

Step2: Calculate sample standard deviation \(s\)

The formula for sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(632 - 678)^{2}=(- 46)^{2}=2116\)
\((x_{2}-\bar{x})^{2}=(583 - 678)^{2}=(-95)^{2}=9025\)
\((x_{3}-\bar{x})^{2}=(1112 - 678)^{2}=(434)^{2}=188356\)
\((x_{4}-\bar{x})^{2}=(604 - 678)^{2}=(-74)^{2}=5476\)
\((x_{5}-\bar{x})^{2}=(512 - 678)^{2}=(-166)^{2}=27556\)
\((x_{6}-\bar{x})^{2}=(625 - 678)^{2}=(-53)^{2}=2809\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=2116+9025+188356+5476+27556+2809=235338\)
\(s=\sqrt{\frac{235338}{6 - 1}}=\sqrt{\frac{235338}{5}}=\sqrt{47067.6}\approx216.95\)

Step3: Calculate the test statistic \(t\)

The formula for the \(t\) - test statistic in a one - sample \(t\) - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Here, \(\mu = 1000\), \(\bar{x}=678\), \(s\approx216.95\), \(n = 6\)
\(t=\frac{678 - 1000}{216.95/\sqrt{6}}=\frac{-322}{216.95/2.45}\approx\frac{-322}{88.55}\approx - 3.64\)

Answer:

\(t\approx - 3.64\)