Chapter 9.2, Problem 57E

(a)

To determine

To Explain: the parameter of interest.

Answer to Problem 57E

Population proportion $p$ of all students who are willing to report cheating by other students

Explanation of Solution

The parameter of interest is the population value of which require to know the value or other information.

PARAMETER OF INTEREST = Population proportion $p$ of all students who are willing to report cheating by other students

(b)

To determine

To Explain: that the conditions for performing the significance test are met in this case.

Answer to Problem 57E

All conditions are satisfied.

Explanation of Solution

Given:

  $\begin{array}{c}n=172\\ p=0.15\end{array}$

Conditions for carrying out a one-sample $z$ test: Random, Normal and independent

Random: satisfied, the reason is that the sample is a simple random sample.

Normal: The distribution can be assumed to be normal if the number of failures $n\left(1-p\right)$ and the number of successes $np$ are both greater than 10.

  $\begin{array}{c}np=172\times 0.15=25.8\\ n\left(1-p\right)=172\times \left(1-0.15\right)=146.2\end{array}$

Both are greater than 10, therefore the normal requirement is satisfied.

Independent: can be assumed, because the sample size (172) is less than 10% of the population size.

Thus all conditions are satisfied.

(c)

To determine

To Explain: the P-value.

Answer to Problem 57E

  $P=0.146=14.6\%$

P-value is the probability of getting a sample that contains a more extreme proportion than the give sample proportion is 0.146 or 14.6%.

Explanation of Solution

P-value is given in the output as:

  $P=0.146=14.6\%$

P-value is the probability of getting a sample that contains a more extreme proportion than the give sample proportion is 0.146 or 14.6%.

(d)

To determine

To Explain: that the data give convincing evidence that the population proportion differs from 0.15, justify answer with appropriate evidence.

Answer to Problem 57E

No

Explanation of Solution

  $\begin{array}{c}{H}_0:p=0.15\\ H_1:p\ne 0.15\end{array}$

The significance level is 1 decreased by the confidence level:

  $\alpha =1-95\%=1-0.95=0.05$

P-value given in the output:

  $P=0.146$

If the P-value is lesser than the significance level, then reject the null hypothesis.

  $P=0.146>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is not enough proof to help the claim.