Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Car Booster Seats The National Highway Traffic 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?

774 649 1210 546 431 612

Short answer

There is not sufficient evidence to accept the claim that the sample is from a population with a mean less than 1000 hic.

The results of the crash test of child booster seats suggest that all child booster seats does not meet the specified requirement.

Step-by-step solution

Given information

The results of the crash test are 774, 649,1210, 546, 431, 612.

The safety requirement is that the hic measurement should be less than 1000 hic.

The significance level is 0.01.

Check the requirements

Assume that the population follows the normal distribution and the samples are randomly selected.

The sample size (n) of the crash test of child booster seats is 16.

The t-distribution would be used here.

Describe the hypothesis

Null hypothesis, $\left({\mathbf{H}}_{\mathbf{0}}\right)$is a statement of the claim thatsample is from a population with a mean is equal to 1000 hic.

Alternate hypothesis,$\left({\mathbf{H}}_{\mathbf{1}}\right)$is a statement of the claim thatsample is from a population with a mean less than 1000 hic.

Let $\mu$be the true population mean.

Mathematically, it can be expressed as,

$$\begin{gathered} H_0:\mu =1000 \\ {H}_1:\mu <1000 \end{gathered}$$

The hypothesis is left-tailed.

Calculate the test statistic

Formula for test statistic is given by,

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Where,$\overline{x}$ is the sample mean and s is the standard deviation of sample.

The sample mean is computed as,

$$\begin{gathered} \overline{x}=\frac{\sum x_i}{n} \\ =\frac{774+649+...+612}{6} \\ =703.67 \end{gathered}$$

The sample standard deviation is,

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(774-703.67\right)}^2+{\left(649-703.67\right)}^2+...+{\left(612-703.67\right)}^2}{6-1}} \\ =272.72 \end{gathered}$$

By substituting these values, test statistics is given by,

$$\begin{gathered} t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{703.67-1000}{\frac{272.73}{\sqrt{6}}} \\ =-2.661 \end{gathered}$$

Calculate the critical value

The significance level is 0.01.

Sample size (n) is 6.

The degree of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =6-1 \\ =5 \end{gathered}$$

In the t-distribution table, find the value corresponding to the row value of degree of freedom 5 and column value of area in one tail 0.01 is 3.365 which is critical value role="math" localid="1649063658875" ${\text{t}}_{\text{0.01}}$; but the given test is left tailed therefore use -3.365 as a critical value.

Thus, the critical valuerole="math" localid="1649063667947" ${\text{t}}_{0.01}$is -3.365.

The rejection region is $\left(t:t<-3.365\right)$.

Compare test statistic and critical value

Test statistic is -2.661 and the critical value is -3.365.

According to this, we can conclude that the test statistic -2.661 will not fall in the rejection region.

Therefore, we failed to reject the null hypothesis.

Conclusion

There is not sufficient evidence to accept the claim that the sample is from a population with a mean less than 1000 hic.

The observation value is 1210 which is too high. Therefore, the results suggest that all child booster seats does not meet the specified requirement.