Question
A recently televised broadcast of a popular television show had a 15 share, meaning that among 5000 monitored households with TV sets in use, 15% of them were tuned to the show. A 0.01 significance level is used to test an advertiser’s claim that among the households with TV sets in use, less than 20% were tuned in to the show. Find the P-value.
A) 0.0002
B) 0.0001
C) 0.9999
D) 1.9998
A) 0.0002
B) 0.0001
C) 0.9999
D) 1.9998
Answers
GPT-4o mini
To find the P-value for the test of the advertiser's claim, we can set up the hypothesis test as follows:
- **Null Hypothesis (H0)**: The proportion of households tuned to the show is \( p \geq 0.20 \).
- **Alternative Hypothesis (H1)**: The proportion of households tuned to the show is \( p < 0.20 \).
Given that the sample size \( n = 5000 \) and the sample proportion \( \hat{p} = 0.15 \), we can calculate the test statistic using the formula for the z-test for proportions:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}
\]
where \( p_0 = 0.20 \).
Now, substituting the values we have:
\[
z = \frac{0.15 - 0.20}{\sqrt{\frac{0.20 (1 - 0.20)}{5000}}}
\]
Calculating the denominator:
\[
\sqrt{\frac{0.20 \times 0.80}{5000}} = \sqrt{\frac{0.16}{5000}} = \sqrt{0.000032} \approx 0.005656854
\]
Now substitute this back into the z-score formula:
\[
z = \frac{-0.05}{0.005656854} \approx -8.83
\]
Next, we need to find the P-value associated with this z-score. Since this is a one-tailed test (we are testing if \( p < 0.20 \)), we will find the probability that corresponds to \( z = -8.83 \).
Using a standard normal distribution table or a calculator for the z-score of -8.83, we find that the P-value is extremely low, typically less than 0.0001.
Thus, based on the options given, the closest answer for the P-value is:
**B) 0.0001**
- **Null Hypothesis (H0)**: The proportion of households tuned to the show is \( p \geq 0.20 \).
- **Alternative Hypothesis (H1)**: The proportion of households tuned to the show is \( p < 0.20 \).
Given that the sample size \( n = 5000 \) and the sample proportion \( \hat{p} = 0.15 \), we can calculate the test statistic using the formula for the z-test for proportions:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}
\]
where \( p_0 = 0.20 \).
Now, substituting the values we have:
\[
z = \frac{0.15 - 0.20}{\sqrt{\frac{0.20 (1 - 0.20)}{5000}}}
\]
Calculating the denominator:
\[
\sqrt{\frac{0.20 \times 0.80}{5000}} = \sqrt{\frac{0.16}{5000}} = \sqrt{0.000032} \approx 0.005656854
\]
Now substitute this back into the z-score formula:
\[
z = \frac{-0.05}{0.005656854} \approx -8.83
\]
Next, we need to find the P-value associated with this z-score. Since this is a one-tailed test (we are testing if \( p < 0.20 \)), we will find the probability that corresponds to \( z = -8.83 \).
Using a standard normal distribution table or a calculator for the z-score of -8.83, we find that the P-value is extremely low, typically less than 0.0001.
Thus, based on the options given, the closest answer for the P-value is:
**B) 0.0001**