Among the senior class at a high school, 55% of Ms. Keating’s students plan on majoring in a branch of STEM, while 49% of Ms. Lewis’s students plan on majoring in a branch of STEM. Suppose Ms. Keating chooses 25 of her students at random and Ms. Lewis chooses 23 of her students at random. Since nKpK, nK (1 – pK) and nLpL, nL (1 – pL) are all greater than 10, the Normal condition is met. Let K = the proportion of Ms. Keating’s students from the sample who plan on majoring in a branch of STEM, and let L = the proportion of Ms. Lewis’s students from the sample who plan on majoring in a branch of STEM. What is the probability that the proportion of students who plan on majoring in a branch of STEM is greater for Ms. Keating?Find the z-table here.
The difference K − L is approximately Normal, with mean 0.55 − 0.49 = 0.06 and standard deviation √[(0.55)(0.45)/25 + (0.49)(0.51)/23] ≈ 0.144. The probability that Ms. Keating’s sample proportion is greater is P(K − L > 0), which corresponds to z = (0 − 0.06)/0.144 ≈ −0.42. The area to the right is about 0.662, so the probability is 0.662.
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