Chapter 9, Problem 10CRP

(a)

(i)

To determine

The level of significance, null and alternative hypothesis.

Answer to Problem 10CRP

Solution:

The level of significance is α = 0.05. The null hypothesis is $H_0:\mu =0.10$ and alternative hypothesis $H_1:\mu \ne 0.10$.

Explanation of Solution

The level of significance is defined as the probability of rejecting the null hypothesis when it is true, it is denoted by $\alpha =0.05$.

Null hypothesis $H_0:\mu =0.10$

Alternative hypothesis $H_1:\mu \ne 0.10$

(ii)

To determine

To find:

The sampling distribution that should be used and compute the value of the sample test statistic.

Answer to Problem 10CRP

Solution:

The normal distribution should be used. The sample test statisticis 1.29.

Explanation of Solution

Calculation:

We will use the normal distribution to binomial, if the condition np > 5 and nq > 5 are satisfied.

$\begin{array}{l}np=68*0.10\\ =6.8\\ nq=68*0.90\\ =61.2\end{array}$

Both conditions are met, so we can use normal distribution.

Using $r=10,n=68$

$\begin{array}{l}\widehat{p}=\frac{r}{n}\\ =\frac{10}{68}\\ =0.147\end{array}$

The sample test statistic z is

$\begin{array}{l}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0{q}_0}{n}}}\\ z=\frac{0.147-0.10}{\sqrt{\frac{0.10*0.90}{68}}}\\ z=1.29\end{array}$

(iii)

To determine

To find:

The P-value of the test statistic and sketch the sampling distribution showing the area corresponding to the P-value.

Answer to Problem 10CRP

Solution:

The P-value of the test statistic is 0.1970.

Explanation of Solution

Calculation:

We have z = 1.29

$\begin{array}{l}P-value=P(z>1.29)\\ =1-P(z\le 1.29)\\ \text{Using Table 3 from the Appendix to find the specified area:}\\ =1-0.9015\\ =0.0985\end{array}$

For a two tailed test, the P-value is 0.1970.

Graph:

To draw the required graphs using the Minitab, follow the below instructions:

Step 1: Go to the Minitab software.

Step 2: Go to Graph > Probability distribution plot > View probability.

Step 3: Select ‘Normal’ and Mean = 0, Standard deviation = 1.

Step 4: Click on the Shaded area > X value.

Step 5: Enter X-value as 1.29 and select ‘Both Tail’.

Step 6: Click on OK.

The obtained distribution graph is:

$\begin{array}{l}P-\text{value}\text{=}2(0.0985)\\ P-\text{value}\text{=}0.1970\end{array}$

(iv)

To determine

Whether we reject or fail to reject the null hypothesisand whether the data is statistically significant for a level of significance of 0.05.

Answer to Problem 10CRP

Solution:

The P-value > α, hence we fail to reject the $H_0$. The data is not statistically significant for a level of significance of 0.05.

Explanation of Solution

The P-value of 0.1970 is greater than the level of significance (α) of 0.05. Therefore we don't have enough evidence to reject the null hypothesis $H_0$. Hence, the data is not statistically significant for a level of significance of 0.05.

(v)

To determine

The interpretation for the conclusion.

Answer to Problem 10CRP

Solution:

There is not enough evidence to conclude that the students at this school are significantly different from the overall proportion of 16-19 year older who is victims of crime.

Explanation of Solution

The P-value of 0.1970 is greater than the level of significance ($\alpha$) of 0.05. Therefore we don't have enough evidence to reject the null hypothesis $H_0$. There is not enough evidence to conclude that the students at this school are significantly different from the overall proportion of 16-19 year older who is victims of crime.

(b)

To determine

To find:

The 90% confidence interval for the proportion of students in the school who have been victims of a crime.

Answer to Problem 10CRP

Solution:

The 90% confidence interval for p is (0.079, 0.221).

Explanation of Solution

Calculation:

We have to find 90% confidence interval. Using,

$\begin{array}{l}\widehat{p}=0.147\approx 0.15\\ n=68\\ \text{Using}\text{Table}3\text{of}\text{Appendix}\\ z_c=1.64\end{array}$

90% confidence interval is

$\begin{array}{l}\widehat{p}-z_c\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}<p<\widehat{p}+z_c\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ 0.15-1.645\sqrt{\frac{0.15*(1-0.15)}{68}}<p<0.15+1.645\sqrt{\frac{0.15*(1-0.15)}{68}}\\ 0.079<\mu <0.22\end{array}$

The 90% confidence interval for p is (0.079, 0.221).

(c)

To determine

To find: The minimum sample size so that the margin of error is not more than 0.05.

Answer to Problem 10CRP

Solution:

The sample size is 193 so that the margin of error is not more than 0.05

Explanation of Solution

Calculation:

We have the sample size formula as follows

$\begin{array}{l}n=\widehat{p}(1-\widehat{p}){\left(\frac{z_c}{E}\right)}^2\\ \widehat{p}=0.147\\ E=0.05\\ \text{Using}\text{Table}3\text{of}\text{Appendix}\\ z_c=1.96\end{array}$

$\begin{array}{l}n=\widehat{p}(1-\widehat{p}){\left(\frac{z_c}{E}\right)}^2\\ n=\left(0.147\right)\left(0.853\right){\left[\frac{1.96}{0.05}\right]}^2\\ n=192.68\\ n\approx 193\end{array}$

Therefore n = 193 is the minimum sample size to ensure that the margin of error E is not more than 0.05.