Question

A college wants to decide whether it should keep an ATM on campus. The results of a sample of 100 students show that 24 of them do not use the ATM. In the following four questions, you will generate an interval estimate for the population proportion of students who do not use the ATM, at the 95% Confidence Level. First, give the appropriate value for Z.

Find the estimated standard error of the proportion.

Give the lower limit of the Confidence Interval.

Give the Upper Limit of the Confidence Interval estimate of the true population proportion of students who did not use the ATM.

Answer #1

Solution :

Given that,

n = 100

x = 24

= x / n = 24 / 100 = 0.24

1 - = 1 - 0.24 = 0.76

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2
= Z0.025 = **1.96**

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.24 * 0.76) / 100)

**= 0.084**

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.24 - 0.084 < p < 0.24 + 0.084

0.156 < p < 0.324

(0.156,0.324)

**Lower limit = 0.156**

**Upper limit = 0.324**

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