You conducted your own study on Facebook to try to find out the proportion of Fa

Question

You conducted your own study on Facebook to try to find out the proportion of Facebook users in the United States who are afraid of spiders. You recruited 50 of your Facebook friends who live in the United States to answer your question. 36 of them answered "yes I am afraid of spiders". Let's also say that you know that the standard error of the sample proportion is = .06. What is the correct way to construct a 95% CI for this proportion?

a .72 +/- .95*(.06)

b .72 +/- 1.96*(.06)

c .06 +/- .95(36/50)

d .28 +/- 1.96*(.06)

Explanation / Answer

The sampling distribution of the sampling mean follows a normal distribution. The mean of the sampling distribution will be the same as the population mean. As we do not know the population mean, we estimate it to be equal to the sample mean. As 36 out of 50 friends said yes, the sample mean is 36/50 = 0.72.

From this, we conclude that options c) and d) are incorrect as the confidence interval (CI) should be around the mean of the sampling distribution which is estimated to be 0.72.

For a 95% CI, we need probability values between 0.025 and 0.975. From the z-table, the corresponding z-score is +/-1.96.

From this, we conclude that option a) is also incorrect as we need a multiple of 1.96 around the sample mean.

The CI is defined as sample mean +/- z-score * standard error = 0.72 +/- 1.96 * 0.06.

Thus, option b) is correct.

The answer is b) 0.72 +/- 1.96 * 0.06

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