Question

A random sample of ?=80n=80 in-coming college freshmen showed that 44 were planning to bring a car with them to campus next fall. Suppose that we are interested in forming a confidence interval at 90 percent for the proportion of all in-coming college freshmen who are planning to bring a car to campus next fall.

Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places.

(a) The estimate is:  .

(b) The standard error is:

(c) The multiplier is:

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 = 44 Sample Size N = 80 The sample proportion is computed as follows, based on the sample size N = 80 and the number of favorable cases X = 44: X 44 р 0.55 N 80 The critical value for a = 0.1 is c = 21-a/2 1.645. The corresponding confidence interval is computed as shown below: CI(Proportion) P(1-P) p(1 - 9) Ôn – %c V p+2c n n 0.55 - 1.645 x 0.55(1 – 0.55) 80 0.55 +1.645 x 0.55(1 – 0.55) 80 = (0.459,0.641) Therefore, based on the data provided, the 90% confidence interval for the population proportion is 0.459 < p < 0.641, which indicates that we are 90% confident that the true population proportion p is contained by the interval (0.459,0.641).

(a) The estimate is: .

0.55

(b) The standard error is:

1.645*√(0.55*0.45/80)

0.0915

(c) The multiplier is:

1.645

please like ??