Question

1. adult educational activities a local county has a very active adult education venue. a random sample of the population showed that 189 out of 400 persons 16 years old or older participated in some type of formal adult education activities, such as basic skills training, apprenticeships, personal interest courses, and part - time college or university degree programs. estimate the true proportion of adults participating in some kind of formal education program with 98% confidence. assume the variable is normally distributed.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is calculated as the number of successes (persons participating) divided by the sample size $n$. Here, the number of successes $x = 189$ and sample size $n = 400$. So, $\hat{p}=\frac{x}{n}=\frac{189}{400} = 0.4725$.

Step2: Determine z-score for 98% confidence

For a 98% confidence interval, the significance level $\alpha=1 - 0.98 = 0.02$. The critical value $z_{\alpha/2}$ is the z-score such that the area in the two tails is $\alpha$. So, $\alpha/2 = 0.01$, and we look for the z-score corresponding to an area of $1 - 0.01 = 0.99$ in the standard normal distribution. From z-tables or using a calculator, $z_{0.01}\approx 2.33$.

Step3: Calculate standard error

The standard error (SE) for a proportion is given by the formula $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting $\hat{p} = 0.4725$ and $n = 400$:
\[

$$\begin{align*} SE&=\sqrt{\frac{0.4725\times(1 - 0.4725)}{400}}\\ &=\sqrt{\frac{0.4725\times0.5275}{400}}\\ &=\sqrt{\frac{0.2492}{400}}\\ &=\sqrt{0.000623}\\ &\approx 0.02496 \end{align*}$$

\]

Step4: Calculate margin of error (ME)

The margin of error is calculated as $ME = z_{\alpha/2}\times SE$. Substituting $z_{\alpha/2}=2.33$ and $SE\approx 0.02496$:
\[
ME = 2.33\times0.02496\approx 0.05816
\]

Step5: Calculate confidence interval

The confidence interval for the population proportion $p$ is given by $\hat{p}\pm ME$. So, the lower limit is $0.4725 - 0.05816 = 0.41434$ and the upper limit is $0.4725 + 0.05816 = 0.53066$.

Answer:

The 98% confidence interval for the true proportion of adults participating in some kind of formal education program is approximately $(0.414, 0.531)$ (or in more precise terms, $(0.4143, 0.5307)$).