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Sagot :
To solve this problem, let's go through the hypothesis testing steps:
Step 1: State the null and alternative hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The proportion of households tuned in is 20%, i.e., [tex]\( p = 0.20 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The proportion of households tuned in is not 20%, i.e., [tex]\( p \neq 0.20 \)[/tex].
Step 2: Calculate the sample proportion ([tex]\( \hat{p} \)[/tex])
The sample proportion is given by:
[tex]\[ \hat{p} = \frac{750}{5000} = 0.15 \][/tex]
Step 3: Calculate the standard error of the proportion
The standard error (SE) is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} \][/tex]
where [tex]\( p_0 = 0.20 \)[/tex] and [tex]\( n = 5000 \)[/tex].
From the calculation, we get:
[tex]\[ SE = \sqrt{\frac{0.20 \times (1 - 0.20)}{5000}} = \sqrt{\frac{0.20 \times 0.80}{5000}} = 0.005656854249492381 \][/tex]
Step 4: Calculate the z-score
The test statistic (z-score) is calculated using the following formula:
[tex]\[ z = \frac{\hat{p} - p_0}{SE} \][/tex]
[tex]\[ z = \frac{0.15 - 0.20}{0.005656854249492381} = -8.838834764831846 \][/tex]
Step 5: Determine the p-value
The p-value for a two-tailed test is calculated as follows:
[tex]\[ p\text{-value} = 2 \times (1 - \Phi(|z|)) \][/tex]
where [tex]\( \Phi \)[/tex] is the cumulative distribution function of the standard normal distribution.
Given the [tex]\( z \)[/tex]-score of [tex]\( -8.838834764831846 \)[/tex], the p-value is extremely small, essentially [tex]\( 0.0 \)[/tex].
Step 6: Make a decision
Compare the p-value with the significance level [tex]\( \alpha = 0.01 \)[/tex]:
- If the p-value < [tex]\( \alpha \)[/tex], reject the null hypothesis ([tex]\( H_0 \)[/tex]).
- If the p-value > [tex]\( \alpha \)[/tex], do not reject the null hypothesis ([tex]\( H_0 \)[/tex]).
Here the p-value [tex]\( 0.0 \)[/tex] is less than the significance level [tex]\( \alpha = 0.01 \)[/tex].
Conclusion
Since the p-value is less than the significance level:
[tex]\[ \text{My P-value less than a Alpha, so I Reject Null Hypothesis} \][/tex]
Therefore, we have enough evidence to reject the null hypothesis and conclude that the proportion of households tuned in is significantly different from 20%.
Step 1: State the null and alternative hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The proportion of households tuned in is 20%, i.e., [tex]\( p = 0.20 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The proportion of households tuned in is not 20%, i.e., [tex]\( p \neq 0.20 \)[/tex].
Step 2: Calculate the sample proportion ([tex]\( \hat{p} \)[/tex])
The sample proportion is given by:
[tex]\[ \hat{p} = \frac{750}{5000} = 0.15 \][/tex]
Step 3: Calculate the standard error of the proportion
The standard error (SE) is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} \][/tex]
where [tex]\( p_0 = 0.20 \)[/tex] and [tex]\( n = 5000 \)[/tex].
From the calculation, we get:
[tex]\[ SE = \sqrt{\frac{0.20 \times (1 - 0.20)}{5000}} = \sqrt{\frac{0.20 \times 0.80}{5000}} = 0.005656854249492381 \][/tex]
Step 4: Calculate the z-score
The test statistic (z-score) is calculated using the following formula:
[tex]\[ z = \frac{\hat{p} - p_0}{SE} \][/tex]
[tex]\[ z = \frac{0.15 - 0.20}{0.005656854249492381} = -8.838834764831846 \][/tex]
Step 5: Determine the p-value
The p-value for a two-tailed test is calculated as follows:
[tex]\[ p\text{-value} = 2 \times (1 - \Phi(|z|)) \][/tex]
where [tex]\( \Phi \)[/tex] is the cumulative distribution function of the standard normal distribution.
Given the [tex]\( z \)[/tex]-score of [tex]\( -8.838834764831846 \)[/tex], the p-value is extremely small, essentially [tex]\( 0.0 \)[/tex].
Step 6: Make a decision
Compare the p-value with the significance level [tex]\( \alpha = 0.01 \)[/tex]:
- If the p-value < [tex]\( \alpha \)[/tex], reject the null hypothesis ([tex]\( H_0 \)[/tex]).
- If the p-value > [tex]\( \alpha \)[/tex], do not reject the null hypothesis ([tex]\( H_0 \)[/tex]).
Here the p-value [tex]\( 0.0 \)[/tex] is less than the significance level [tex]\( \alpha = 0.01 \)[/tex].
Conclusion
Since the p-value is less than the significance level:
[tex]\[ \text{My P-value less than a Alpha, so I Reject Null Hypothesis} \][/tex]
Therefore, we have enough evidence to reject the null hypothesis and conclude that the proportion of households tuned in is significantly different from 20%.
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