Question

1. what is the standard error of the proportion, (sigma_{p}=sqrt{\frac{p(1 - p)}{n}})?

• a. (0.050)
• b. (0.087)
• c. (0.048)

1. for a random sample of 100 bags of popcorn, what interval captures 95% of all sample proportions under the normal curve?

• a. between (25.5%) and (44.5%)
• b. between (15.2%) and (35.1%)
• c. between (7.6%) and (42.4%)

Explanation:

Step1: Calculate the standard error of the proportion

The formula for the standard error of a proportion is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\). Given \(p = 0.35\) and \(n=100\).
Substitute the values into the formula: \(\sigma_{\hat{p}}=\sqrt{\frac{0.35\times(1 - 0.35)}{100}}\)
First, calculate \(1-p=1 - 0.35 = 0.65\). Then \(p(1 - p)=0.35\times0.65=0.2275\).
\(\frac{p(1 - p)}{n}=\frac{0.2275}{100}=0.002275\).
\(\sigma_{\hat{p}}=\sqrt{0.002275}\approx0.048\)

Step2: Analyze the confidence - interval for the normal curve

For a \(95\%\) confidence interval of a proportion \(\hat{p}\) (where \(\hat{p}\) is the sample proportion), the confidence interval is \(\hat{p}\pm z\times\sigma_{\hat{p}}\), and for a \(95\%\) confidence interval, \(z = 1.96\).
The mean of the sampling distribution of the sample proportion \(\mu_{\hat{p}}=p = 0.35\)
The lower limit \(L=\mu_{\hat{p}}-z\times\sigma_{\hat{p}}=0.35-1.96\times0.048\approx0.35 - 0.094=0.256\)
The upper limit \(U=\mu_{\hat{p}}+z\times\sigma_{\hat{p}}=0.35 + 1.96\times0.048\approx0.35+0.094 = 0.444\)

Answer:

1. C. \(0.048\)
2. A. Between \(25.5\%\) and \(44.5\%\)