Chapter 9.3, Problem 78E

To determine

To Explain: these data provide convincing evidence at the $\alpha =0.10$ vertical jump of students at this school differs from 15 inches.

Answer to Problem 78E

There is no enough convincing proof that the average vertical jump of students at this school differs from 15 inches.

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.01\\ n=10\end{array}$

Formula used:

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

Calculation:

Conditions

The three conditions are: Random, independent.(10% condition), Normal/Large sample.

Random sample: satisfied, because the sample is a random sample.

Independent: satisfied, because the sample of 20 students is less than 10% of the population of all students.

Normal/ Large sample: satisfied, because the pattern in the normal quintile plot is roughly linear, this indicates that the distribution is about Normal.

Since all conditions are satisfied, it is appropriate to perform a hypothesis test for the population mean.

The mean is

  $\begin{array}{c}\overline{x}=\frac{\begin{array}{l}11.0+11.5+12.5+26.5+15.0\\ +12.5+22.0+15.0+13.5+12.0\\ +23.0+19.0+15.5+21.0+12.5\\ +23.0+20.0+8.5+25.5+20.5\end{array}}{20}\\ =\frac{340}{20}\\ =17\end{array}$

The variance is

  $\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(11.0-17\right)}^2+{\left(11.5-17\right)}^2+{\left(12.5-17\right)}^2+{\left(26.5-17\right)}^2+{\left(15.0-17\right)}^2\\ +{\left(12.5-17\right)}^2+{\left(22.0-17\right)}^2+{\left(15.0-17\right)}^2+{\left(13.5-17\right)}^2+{\left(12.0-17\right)}^2\\ +{\left(23.0-17\right)}^2+{\left(19.0-17\right)}^2+{\left(15.5-17\right)}^2+{\left(21.0-17\right)}^2+{\left(12.5-17\right)}^2\\ +{\left(23.0-17\right)}^2+{\left(20.0-17\right)}^2+{\left(8.5-17\right)}^2+{\left(25.5-17\right)}^2+{\left(20.5-17\right)}^2\end{array}}{20-1}}\\ =5.3680\end{array}$

Hypothesis test

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is the population mean is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  $\begin{array}{c}{H}_0:\mu =0\\ H_1:\mu >0\end{array}$

The t-test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{17-15}{5.3680/\sqrt{20}}\\ =1.666\end{array}$

The P-value is the probability of getting the value of the test static, or a value more extreme, assuming that the null hypothesis is true. $df=n-1=20-1=19$ .Note: it is require doubling the boundaries of the value of the test statistic, because the test is two-tailed $0.10=2\left(0.05\right)<P<2\left(0.10\right)=0.20$

Command Ti83/84- calculator: 2*tcdf (-1E99,-1.666,1 9) which will return a P-value of 0.2742. Note: it could replace 1E99 by any other very large positive number.

If the P-value is lesser than the significance level $\alpha$ , then the null hypothesis is rejected.

  $P>0.10\Rightarrow \text{Fail to reject }{H}_0$

There is no enough convincing proof that the average vertical jump of students at this school differs from 15 inches.