Question

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” The Minitab output shows the results of a significance test and a 95% confidence interval based on the survey data.13

a. Define the parameter of interest.

b. Check that the conditions for performing the significance test are met in this case.

c. Interpret the P-value.

d. Do these data give convincing evidence that the population proportion differs from $0.15$? Justify your answer with appropriate evidence.

Short answer

Part a) Population proportion$p$of all students who are willing to report cheating by other students.

Part b) All conditions are satisfied.

Part c) $P=0.146=14.6\%$

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

Part d) The required answer is No.

Step-by-step solution

Part a)  Step 1: The objective is to explain the parameter of interest.

The parameter of interest is the population value, for which the value or other information is required. PARAMETER OF INTEREST = Population proportion$p$ of all students willing to report other students' cheating.

Part b) Step 2: Given information

$$\begin{gathered} n=172 \\ p=0.15 \end{gathered}$$

Part b) Step 2: The objective is to explain that the conditions for performing the significance test are met in this case. 

Random, Normal, and Independent conditions for performing a one-sample$z$test.

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

Normal: If the number of failures $n(1-p)$and the number of successes $np$are both greater than$10$, the distribution is assumed to be normal.

$$\begin{gathered} np=172\times 0.15 \\ =25.8 \end{gathered}$$

$$\begin{gathered} n(1-p)=172\times (1-0.15) \\ =146.2 \end{gathered}$$

Both are greater than$10,$indicating that the normal requirement has been met.

Because the sample size$172$is less than$10\%$of the population size, independence can be assumed.

Therefore, all conditions are satisfied.

Part c) Step 1: The objective is to explain theP value.

The P-value is displayed in the output as:

$P=0.146=14.6\%$

The probability of receiving a sample with a more extreme proportion than the given sample proportion is $0.146$ or $14.6\%$

Part d) Step 1: The objective is to explain that the data give convincing evidence that the population proportion differs from0.15 and justify the answer with appropriate evidence.

$$\begin{gathered} H_0:p=0.15 \\ {H}_1:pnotequalto0.15 \end{gathered}$$

The confidence level reduces the significance level to $1$:

$$\begin{gathered} \alpha =1-95\% \\ =1-0.95 \\ =0.05 \end{gathered}$$

P-value is given in the output:

$P=0.146$

If the P-value is less than the significance level, the null hypothesis should be rejected.

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

There is insufficient evidence to support the claim.