Question

In: Statistics and Probability

Medical literature says that about 8% of males are color-blind. A university’s introductory psychology course is...

  1. Medical literature says that about 8% of males are color-blind. A university’s introductory psychology course is taught in a large lecture hall. Among the students, there are 320 males. Each semester when the professor discusses visual perception, he shows the class a test for color blindness. The percentage of males who are color-blind varies from semester to semester. Let p = proportion of males that are color blind. Consider the sampling distribution of for a random sample of n = 320 males.a)Using the appropriate conditions, clearly list and show a check for each of the necessary assumptions for using the Central Limit Theorem to describe the sampling distribution of .

b)Clearly state what the Central Limit Theorem says about the sampling distribution of for a random sample of n=320 males, if the medical literature is correct that about 8% of males are color-blind.

c)Sketch the sampling distribution of that you gave in part (b). Make sure to include all appropriate labels.

d)When the 320 psychology class students took the test for color blindness, only 11 of them tested as being color-blind. Do you think this provides evidence that the medical literature isn’t correct? Explain in one clear sentence and support your answer with a probability. Hint: Use your sketch from part c!

Solutions

Expert Solution

(a) The central limit theorem states that the sampling distribution of the mean of any independent, random variable will be normal or nearly normal, if the sample size is large enough.

(b) The theorem says that for large sample sizes, the sampling distribution of proportion is also approximately normal. This is especially true if both np and nq are greater than 10. In our case, np = 320 * 0.08 = 25.6 (> 10) and nq = 294.4 (> 10)

(c) μ = np = 25.6, σ = √(npq) = √(320 * 0.08 * 0.92) = 4.853

(d) x1 = 10.5 and x2 = 11.5

z1 = (x - μ)/σ = (10.5 - 25.6)/4.853 = -3.111 and z2 = (11.5 - 25.6)/4.853 = -2.9054

P(x = 11) = P(-3.111 < z < -2.9054) = 0.0009

The result means the probability of 11 in the sample being colorblind is 0.0009

Either the figure quoted in the medical literature is not correct or this sample is unusually different.

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