Question

a college student was interested in the average amount college students spend on entertainment each week. he randomly sampled 200 students and found the following 95% confidence interval (24,28) in dollars per week if this student randomly sampled 100 students instead, what would happen to the margin of error (assuming everything else remained the same)? choose the correct answer below. o a. the margin of error would increase. o b. the margin of error would remain the same. o c. the margin of error would decrease. o d. more information would be needed to determine the effect on the margin of error.

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E = z\times\frac{\sigma}{\sqrt{n}}\), where \(z\) is the z - score (depends on confidence level), \(\sigma\) is the standard deviation and \(n\) is the sample size. Here, confidence level is fixed (95% so \(z\) is fixed) and assume \(\sigma\) (population standard deviation) is constant (since everything else remains the same).

Step2: Analyze the effect of sample size change on margin of error

We have \(n_1 = 200\) and \(n_2=100\). The relationship between margin of error and sample size is \(E\propto\frac{1}{\sqrt{n}}\). When \(n\) decreases from \(n_1 = 200\) to \(n_2 = 100\) (i.e., \(n\) is halved). Let \(E_1=z\times\frac{\sigma}{\sqrt{200}}\) and \(E_2 = z\times\frac{\sigma}{\sqrt{100}}\). Then \(\frac{E_2}{E_1}=\sqrt{\frac{200}{100}}=\sqrt{2}\approx1.414\). So \(E_2>E_1\)

Answer:

A. The margin of error would increase.