Question

ch 22 eggs that are contaminated with salmonella can cause food poisoning among consumers. a large egg producer takes an srs of 200 eggs from all the eggs shipped in one day. the laboratory reports that 16 of these eggs had salmonella contamination. unknown to the producer, 0.2% (two - tenths of 1%) of all eggs shipped had salmonella. in this situation
\\( \frac { 16 } { 200 } \\) is a parameter and 0.2% is a statistic.
both 0.2% and \\( \frac { 16 } { 200 } \\) are statistics.
0.2% is a parameter and \\( \frac { 16 } { 200 } \\) is a statistic.
both 0.2% and \\( \frac { 16 } { 200 } \\) are parameters.
question 8
ch 22* a 2011 nbc news survey found that 80% of a sample of 4500 american teens said they owned an mp3 player such as an ipod. assume that the sample was an srs. based on the sample, the large - sample 90% confidence interval for the proportion of all american teens who own an mp3 player is
0.80 ± 0.0117.
0.80 ± 0.0098.
0.80 ± 0.0060.

Explanation:

Step1: Recall the formula for the confidence interval

The formula for the large - sample confidence interval for a proportion is \(\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)

where \(\hat{p}\) is the sample proportion, \(z\) is the critical value, and \(n\) is the sample size.

Step2: Identify the values

Given \(\hat{p}=0.80\), \(n = 4500\). For a \(90\%\) confidence interval, the critical value \(z\) (from the standard normal distribution) is \(z=1.645\)

Step3: Calculate the margin of error

First, calculate \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.80\times(1 - 0.80)}{4500}}=\sqrt{\frac{0.80\times0.20}{4500}}=\sqrt{\frac{0.16}{4500}}\approx\sqrt{3.556\times10^{-5}}\approx0.00596\)

Then, the margin of error \(E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.645\times0.00596\approx0.0098\)

Answer:

\(0.80\pm0.0098\)