Question

1. error analysis in a random sample of 2500 consumers, 61% prefer game a over game b. describe and correct the error in giving an interval that is likely to contain the exact percent of all consumers who prefer game a over game b. x margin of error = ±1/√n = ±1/√2500 = ±0.02 it is likely that the exact percent of all consumers who prefer game a over game b is between 60.96% and 61.02%. 26. error analysis in a survey, 8% of adult internet users said they participate in sports fantasy leagues online. the margin of error is ±4%. describe and correct the error in approximating the sample size. x ±0.08 = ±1/√n 0.0064 = 1/n n ≈ 156

Explanation:

Step1: Identify the error in 25

The formula $\pm\frac{1}{\sqrt{n}}$ is a rough - estimate for proportion close to 50%. For a proportion $p = 0.61$, the correct margin - of - error formula is $E=\pm z\sqrt{\frac{p(1 - p)}{n}}$. For a 95% confidence interval, $z\approx1.96$. First, calculate the correct margin of error.
$E = 1.96\sqrt{\frac{0.61\times(1 - 0.61)}{2500}}$
$=1.96\sqrt{\frac{0.61\times0.39}{2500}}$
$=1.96\sqrt{\frac{0.2379}{2500}}$
$=1.96\sqrt{0.00009516}$
$=1.96\times0.00975$
$\approx\pm0.0191$
The interval is $0.61\pm0.0191$, or between $59.09\%$ and $62.91\%$.

Step2: Identify the error in 26

The margin - of - error formula is $E = z\sqrt{\frac{p(1 - p)}{n}}$. For a 95% confidence interval, $z\approx1.96$. Given $E = 0.04$ and $p = 0.08$, we solve for $n$.
$0.04=1.96\sqrt{\frac{0.08\times(1 - 0.08)}{n}}$
$\frac{0.04}{1.96}=\sqrt{\frac{0.08\times0.92}{n}}$
$(\frac{0.04}{1.96})^2=\frac{0.08\times0.92}{n}$
$n=\frac{0.08\times0.92\times(1.96)^2}{(0.04)^2}$
$n=\frac{0.0736\times3.8416}{0.0016}$
$n=\frac{0.28374176}{0.0016}$
$n\approx177.34$
Since the sample size $n$ must be an integer, $n = 178$.

Answer:

For 25: The error was using the wrong margin - of - error formula. The correct interval is between $59.09\%$ and $62.91\%$.
For 26: The error was using the wrong margin - of - error formula. The correct sample size $n\approx178$.