Answer :
We are given a claim that [tex]$25\%$[/tex] of adults would describe themselves as organized. A random sample of [tex]$100$[/tex] adults shows that [tex]$42$[/tex] of them describe themselves as organized. We will test whether the true proportion of organized adults is greater than [tex]$25\%$[/tex] using a significance level of [tex]$\alpha=0.01$[/tex].
Below is a detailed step-by-step solution.
──────────────────────────────
Step 1. Write Down the Hypotheses
We set up our null and alternative hypotheses as follows:
[tex]\[
H_0: p = 0.25 \quad \text{and} \quad H_a: p > 0.25.
\][/tex]
This is a one-tailed test using the upper tail.
──────────────────────────────
Step 2. Check the Conditions for Inference
1. Random Condition:
The problem states we have a random sample of adults.
2. 10% Condition:
The sample size is [tex]$100$[/tex]. Assuming that the adult population is very large, [tex]$100$[/tex] is less than [tex]$10\%$[/tex] of the population.
3. Large Counts Condition:
Under the null hypothesis, the expected number of successes and failures are calculated as:
[tex]\[
np_0 = 100 \times 0.25 = 25 \quad \text{and} \quad n(1-p_0) = 100 \times 0.75 = 75.
\][/tex]
Both [tex]$25$[/tex] and [tex]$75$[/tex] are greater than or equal to [tex]$10$[/tex]. Hence, the condition is met.
──────────────────────────────
Step 3. Compute the Test Statistic
1. Observed Proportion:
The sample proportion is:
[tex]\[
\hat{p} = \frac{42}{100} = 0.42.
\][/tex]
2. Standard Error (SE):
Under the null hypothesis, the standard error for the sample proportion is:
[tex]\[
SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.25 \times 0.75}{100}}.
\][/tex]
Simplifying:
[tex]\[
SE = \sqrt{\frac{0.1875}{100}} \approx 0.0433.
\][/tex]
3. Z-Statistic:
The test statistic is:
[tex]\[
z = \frac{\hat{p} - p_0}{SE} = \frac{0.42 - 0.25}{0.0433} \approx 3.93.
\][/tex]
──────────────────────────────
Step 4. Find the P-value
Since the alternative hypothesis is [tex]$p > 0.25$[/tex], we look at the upper tail of the standard normal distribution. The p-value is the probability of observing a [tex]$z$[/tex]-statistic as extreme or more extreme than [tex]$3.93$[/tex]. This p-value is approximately:
[tex]\[
p\text{-value} \approx 4.32 \times 10^{-5}.
\][/tex]
──────────────────────────────
Step 5. Conclusion
We compare the p-value with [tex]$\alpha = 0.01$[/tex]. Since
[tex]\[
4.32 \times 10^{-5} < 0.01,
\][/tex]
we reject the null hypothesis.
Thus, there is convincing evidence at the [tex]$1\%$[/tex] significance level that more than [tex]$25\%$[/tex] of adults would describe themselves as organized.
──────────────────────────────
Summary of Conditions for Inference:
- Random: The sample of adults is random.
- 10% Condition: [tex]$100$[/tex] adults is less than [tex]$10\%$[/tex] of the full adult population.
- Large Counts:
[tex]\[
np_0 = 25, \quad n(1-p_0) = 75, \quad \text{and both are at least } 10.
\][/tex]
This completes the solution.
Below is a detailed step-by-step solution.
──────────────────────────────
Step 1. Write Down the Hypotheses
We set up our null and alternative hypotheses as follows:
[tex]\[
H_0: p = 0.25 \quad \text{and} \quad H_a: p > 0.25.
\][/tex]
This is a one-tailed test using the upper tail.
──────────────────────────────
Step 2. Check the Conditions for Inference
1. Random Condition:
The problem states we have a random sample of adults.
2. 10% Condition:
The sample size is [tex]$100$[/tex]. Assuming that the adult population is very large, [tex]$100$[/tex] is less than [tex]$10\%$[/tex] of the population.
3. Large Counts Condition:
Under the null hypothesis, the expected number of successes and failures are calculated as:
[tex]\[
np_0 = 100 \times 0.25 = 25 \quad \text{and} \quad n(1-p_0) = 100 \times 0.75 = 75.
\][/tex]
Both [tex]$25$[/tex] and [tex]$75$[/tex] are greater than or equal to [tex]$10$[/tex]. Hence, the condition is met.
──────────────────────────────
Step 3. Compute the Test Statistic
1. Observed Proportion:
The sample proportion is:
[tex]\[
\hat{p} = \frac{42}{100} = 0.42.
\][/tex]
2. Standard Error (SE):
Under the null hypothesis, the standard error for the sample proportion is:
[tex]\[
SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.25 \times 0.75}{100}}.
\][/tex]
Simplifying:
[tex]\[
SE = \sqrt{\frac{0.1875}{100}} \approx 0.0433.
\][/tex]
3. Z-Statistic:
The test statistic is:
[tex]\[
z = \frac{\hat{p} - p_0}{SE} = \frac{0.42 - 0.25}{0.0433} \approx 3.93.
\][/tex]
──────────────────────────────
Step 4. Find the P-value
Since the alternative hypothesis is [tex]$p > 0.25$[/tex], we look at the upper tail of the standard normal distribution. The p-value is the probability of observing a [tex]$z$[/tex]-statistic as extreme or more extreme than [tex]$3.93$[/tex]. This p-value is approximately:
[tex]\[
p\text{-value} \approx 4.32 \times 10^{-5}.
\][/tex]
──────────────────────────────
Step 5. Conclusion
We compare the p-value with [tex]$\alpha = 0.01$[/tex]. Since
[tex]\[
4.32 \times 10^{-5} < 0.01,
\][/tex]
we reject the null hypothesis.
Thus, there is convincing evidence at the [tex]$1\%$[/tex] significance level that more than [tex]$25\%$[/tex] of adults would describe themselves as organized.
──────────────────────────────
Summary of Conditions for Inference:
- Random: The sample of adults is random.
- 10% Condition: [tex]$100$[/tex] adults is less than [tex]$10\%$[/tex] of the full adult population.
- Large Counts:
[tex]\[
np_0 = 25, \quad n(1-p_0) = 75, \quad \text{and both are at least } 10.
\][/tex]
This completes the solution.
