Question

Jump around Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was $15$ inches. They wondered if the average vertical jump of students at their school differed from $15$ inches, so they obtained a list of student names and selected a random sample of $20$ students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches):

Do these data provide convincing evidence at the $\alpha =0.10$ level that the average vertical jump of students at this school differs from 15 inches?

Short answer

There is not enough convincing proof that the average vertical jump of students at this school differs from $15$inches.

Step-by-step solution

Given information

$$\begin{gathered} \alpha =0.01 \\ n=10 \end{gathered}$$

Concept

$t=\frac{\overline{x}-{\mu }_0}{{\displaystyle \frac{s}{\sqrt{n}}}}$

Calculation

Conditions

Random and independent are the three requirements.

Normal/Large sample ($10$ percent condition).

Because the sample is a random sample, the satisfaction level is high.

Independent: pleased, because the sample of $20$ students represents less than 10% of the total student population.

Because the pattern in the normal quantile plot is generally linear, this implies that the distribution is around Normal. Normal/ Large sample: satisfied

Because all of the prerequisites have been met, a hypothesis test for the population mean is appropriate.

The mean is

$\overline{x}=17$

The variance is

$s=5.3680$

Calculation

Hypothesis test:

The assertion is either the null hypothesis or the alternative hypothesis. The null hypothesis asserts that the claim value and the population mean are the same. The alternative hypothesis statement is the polar opposite of the claim if the claim is the null hypothesis.

$$\begin{gathered} H_0:\mu =0 \\ {H}_1:\mu >0 \end{gathered}$$

The $t-$test statistic is

$$\begin{gathered} t=\frac{\overline{x}-\mu 0}{{\displaystyle \frac{s}{\sqrt{n}}}} \\ =\frac{17-15}{{\displaystyle \frac{5.3680}{\sqrt{20}}}} \\ =1.666 \end{gathered}$$

If the null hypothesis is true, the P-value is the likelihood of getting the test result static, or a value more extreme.

$df=n-1=20-1=19$

Because the test is two-tailed, it is necessary to double the test statistic's value boundaries. $0.10=2(0.05)<P<2(0.10)=0.20$

Command $Ti83/84-$calculator$:2*tcdf(-1E99,-1.666,19)$which will return a P-value of $0.2742$Note: it could replace $1E99$by any other very large positive number.

The null hypothesis is rejected if the $P-$value is less than the significance level.

$P>0.10\Rightarrow Failtoreject{H}_0$

There is not enough convincing proof that the average vertical jump of students at this school differs from $15$inches.