Question 1170393: A researcher suspects that the actual prevalence among children and adolescents is higher than the previously reported prevalence of GAD among children and adolescents. The previously reported prevalence is 3.9% and the researcher conducts a study to test the accuracy of the previous reported prevalence of GAD by recruiting 98 children and adolescents from various pediatrician's offices and tests them for GAD using the DSM-5. The researcher determines that the prevalence of GAD is 6.1%. What should the researchers conclusion be for 5% significance level?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this hypothesis test:
**1. Define the Hypotheses:**
* **Null Hypothesis (H0):** The actual prevalence of GAD among children and adolescents is 3.9% (p = 0.039).
* **Alternative Hypothesis (H1):** The actual prevalence of GAD among children and adolescents is higher than 3.9% (p > 0.039).
**2. Set the Significance Level:**
* α = 0.05 (5% significance level)
**3. Calculate the Sample Proportion:**
* Sample size (n) = 98
* Number of children with GAD = 98 * 0.061 = 5.978. Since you cant have a fraction of a child, we will round to 6.
* Sample proportion (p̂) = 6 / 98 ≈ 0.0612
**4. Perform the Hypothesis Test (One-Proportion Z-Test):**
* We'll use a one-proportion z-test because we're dealing with proportions.
* The test statistic (z) is calculated as:
$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$
Where:
* p̂ = sample proportion (0.0612)
* p = population proportion (0.039)
* n = sample size (98)
* Now we plug in the values:
$z = \frac{0.0612 - 0.039}{\sqrt{\frac{0.039(1-0.039)}{98}}}$
$z = \frac{0.0222}{\sqrt{\frac{0.039(0.961)}{98}}}$
$z = \frac{0.0222}{\sqrt{\frac{0.03748}{98}}}$
$z = \frac{0.0222}{\sqrt{0.0003824}}$
$z = \frac{0.0222}{0.01955}$
$z ≈ 1.135$
**5. Determine the Critical Value or P-value:**
* Since this is a right-tailed test (H1: p > 0.039), we need to find the critical z-value for α = 0.05.
* Using a standard normal distribution table or calculator, the critical z-value is approximately 1.645.
* Alternativly we can calculate the p value.
* Using a z value of 1.135, and a right tailed test, the p value is .128.
**6. Make a Decision:**
* **Critical Value Method:**
* Our calculated z-value (1.135) is less than the critical z-value (1.645).
* Therefore, we fail to reject the null hypothesis.
* **P-value method:**
* The p value of .128 is larger than the significance level of .05.
* Therefore, we fail to reject the null hypothesis.
**7. Draw a Conclusion:**
* At the 5% significance level, there is not enough evidence to support the researcher's claim that the actual prevalence of GAD among children and adolescents is higher than 3.9%.
* The researcher should conclude that the difference between the sample proportion and the previously reported proportion is not statistically significant.
**In summary:** The researcher does not have sufficient evidence to reject the previously reported prevalence.
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