Question

In a random sample of 250 students at a university, 207 stated that they were nonsmokers. Based on this sample, compute a 90% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. = What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval? 0 X ?

Answer 1

We need to construct the 90% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases: Favorable Cases X = 207 Sample Size N = 250 The sample proportion is computed as follows, based on the sample size N 250 and the number of favorable cases X = 207: X P 207 250 = 0.828 N The critical value for a = 0.1 is ze = 21-a/2 = 1.645. The corresponding confidence interval is computed as shown below: P(1 - ) CI(Proportion) @(1 - i) P – 20 V p+ZcV n n 0.828 – 1.645 x 0.828(1 – 0.828) 250 0.828 + 1.645 x 0.828(1 – 0.828) 250 = (0.789, 0.867)

lower bound is 0.7 89

upper bound is 0.867

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