Question

Best known for its testing program, ACT, Inc., also compiles data on a variety of issues in education. In 2004 the company reported that the national college freshman-to-sophomore retention rate held steady at 74% over the previous four years.
1. Consider random samples of 400 freshmen who took the ACT. Use the 68-95-99.7 Rule to describe the sampling distribution model for the percentage of those students we expect to return to that school for their sophomore years. Do you think the appropriate conditions are met?

Answer 1

Answer:

$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

$p \sim N(p, \sqrt{\frac{p(1-p)}{n}})$

So then:

$\mu_p = 0.74$

$\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806