A random sample of adults was surveyed about whether they shop online. Of the 90 adults surveyed, 72 stated they shop online. What is the 99% confidence interval for p, the proportion of adults who shop online?Find the z-table here.

The sample proportion is 72/90 = 0.80. For a 99% confidence interval, use z ≈ 2.576; the standard error is √[0.80(0.20)/90] ≈ 0.042, giving a margin of error of 2.576(0.042) ≈ 0.109. Thus the interval is 0.80 ± 0.109, or approximately (0.69, 0.91), choice C.

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