Chapter 10, Problem 92CR

According to a large national survey conducted by the Pew Research Center (“What Americans Think About NSA Surveillance, National Security and Privacy,” May 2, 2015, pewresearch.org, retrieved December 1, 2016), 54% of adult Americans disapprove of the National Security Agency collecting records of phone and Internet data. Suppose that this estimate was based on a random sample of 1000 adult Americans.

1. a. Is there convincing evidence that a majority of adult Americans feel this way? Test the relevant hypotheses using a 0.05 significance level.
2. b. The actual sample size was much larger than 1000. If you had used the actual sample size when doing the calculations for the test in Part (a), would the P-value have been larger than, the same as, or smaller than the P-value you obtained in Part (a)? Provide a justification for your answer.

To determine

Check whether it can be concluded that the proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data is greater than 0.5.

Answer to Problem 92CR

Yes, the sample data provide the evidence that the proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data is greater than 0.5.

Explanation of Solution

Calculation:

In a sample of 1,000 adults of Country A, 54% of adults were founded to disapprove the National Security Agency collecting records of phone and internet data.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data.

Step 2:

Null hypothesis: $H_0:p=0.5$

That is, the proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data is 0.5.

Step 3:

Alternative hypothesis: $H_a:p>0.5$

That is, the the proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data is greater than 0.5.

Step 4:

Significance level, $\alpha$:

It is assumed that the significance level, $\alpha =0.05$.

Step 5:

Test statistic, z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$,

Where, $\widehat{p}$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$, in the test statistic.

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion from a random sample.
• The large sample z test can be used if the sample size n satisfies the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is given that the sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =1,000\left(0.5\right)\\ =500\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =1,000\left(1-0.5\right)\\ =1,000\left(0.5\right)\\ =500\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although, the population size is not known, it is reasonable to assume that the sample size of 1,000 acts as the representative for all adults of Country A. It is also definite that the sample size is less than 10% of the population of the adults of Country A.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$\begin{array}{c}z=\frac{0.54-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{1,000}}}\\ =\frac{0.04}{0.0158}\\ \approx 2.53\end{array}$

Thus, the value of test statistic is 2.53.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is right-tailed test. Therefore, the P-value is the area under the z curve and to the right of the calculated z value.

The P-value for the test statistic value of 2.53 is obtained as follows:

$\begin{array}{c}P\text{-value}=\text{Area to the Right of }2.53\\ =P\left(z>2.53\right)\\ =1-P\left(z<2.53\right)\end{array}$

Use Standard Normal Probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 2.53:

• Locate 2.5 in the left column of the table.
• Obtain the value in the corresponding row below .03.

That is, $P\left(z<2.53\right)=0.9943$.

The probability of the event is as given below:

$\begin{array}{c}P\left(z>2.53\right)=1-P\left(z<2.53\right)\\ =1-0.9943\\ =0.0057\end{array}$

Thus, the P-value for the test statistic of 2.53 is 0.0057.

Step 9:

Decision rule:

If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ of 0.0057 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.0057\right)<\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, the sample data provide the evidence that the proportion of adults of Country A who disapprove the National Security Agency collecting records of phone and internet data is greater than 0.5.

To determine

Justify whether the P-value is larger, same as, or smaller than the P-value obtained in Part (a) when the actual sample is used.

Answer to Problem 92CR

The P-value would be smaller than that of the P-value obtained in Part (a).

Explanation of Solution

Greater sample size has the chance to increase the test statistic and also reduces the P-value.

If the actual sample size is much higher than that used in Part (a), then there is more chance to have greater test statistic. In such a way, the P-value would also have been less than that obtained in Part (a).