A researcher reports that 80% of high school graduates, but only 40% of high school dropouts, would pass a basic literacy test. Assume that the researcher’s claim is true. Suppose we give a basic literacy test to a random sample of 60 high school graduates and a separate random sample of 75 high school dropouts. Let ^ p G and ^ p D be the sample proportions of graduates and dropouts, respectively, who pass the test.
Find the probability that the proportion of graduates who pass the test is at most 0.20 higher than the proportion of dropouts who pass, assuming that the researcher’s report is correct.
Enter the probability here:

Question

katherinestyles katherinestyles

12-01-2020
Mathematics

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A researcher reports that 80% of high school graduates, but only 40% of high school dropouts, would pass a basic literacy test. Assume that the researcher’s claim is true. Suppose we give a basic literacy test to a random sample of 60 high school graduates and a separate random sample of 75 high school dropouts. Let ^ p G and ^ p D be the sample proportions of graduates and dropouts, respectively, who pass the test.
Find the probability that the proportion of graduates who pass the test is at most 0.20 higher than the proportion of dropouts who pass, assuming that the researcher’s report is correct.
Enter the probability here:

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danialqeem danialqeem · Answer 1 · 2020-01-12T13:36:33+01:00

danialqeem danialqeem

12-01-2020

This was what I got.

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MrRoyal MrRoyal · Answer 2 · 2021-12-07T13:19:58+01:00

The probability that at least 20% of the graduates pass is 0.995

The given parameters are:

Graduates

[tex]\mathbf{n = 60}[/tex] -- sample size

[tex]\mathbf{p = 80\%}[/tex] --- proportion that graduates

Dropouts

[tex]\mathbf{n = 75}[/tex] -- sample size

[tex]\mathbf{p = 40\%}[/tex] --- proportion that dropout

Start by calculating the mean and the standard deviation

The mean of the graduates is:

[tex]\mathbf{\mu = np}[/tex]

[tex]\mathbf{\mu = 60 \times 80\%}[/tex]

[tex]\mathbf{\mu_1 = 48}[/tex]

The mean of the dropouts is:

[tex]\mathbf{\mu = 75 \times 40\%}[/tex]

[tex]\mathbf{\mu_2 = 30}[/tex]

The standard deviation is then calculated as:

[tex]\mathbf{\sigma = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}}[/tex]

This gives

[tex]\mathbf{\sigma = \sqrt{\frac{80\% \times (1 - 80\%)}{60} + \frac{40\%(1 - 40\%)}{75}}}[/tex]

[tex]\mathbf{\sigma = \sqrt{0.0027 + 0.0032}}[/tex]

[tex]\mathbf{\sigma = \sqrt{0.0059}}[/tex]

Take square root of 0.0059

[tex]\mathbf{\sigma = 0.07681}[/tex]

The required probability is the probability that at least 20% of the graduates pass.

This is represented as:

[tex]\mathbf{P(p_1 - p_2 \ge 0.20)}[/tex]

Using normal cdf at [tex]\mathbf{\sigma = 0.07681}[/tex] and [tex]\mathbf{mu = 0.4}[/tex], we have:

[tex]\mathbf{P(p_1 - p_2 \ge 0.20) = 0.995}[/tex]

Hence, the probability that at least 20% of the graduates pass is 0.995