Question

10-According to a survey in a country, 47 % of adults do not have any credit cards. Suppose a simple random sample of 400 adults is obtained.

-What is the probability that less than 40% have no credit cards?

-What is the probability that 195 or more have no credit cards?

-If a random sample with 500 adults was collected and results in 200 or fewer who do not own a credit card, with a 95% confidence, what conclusion can be made?

The sample proportion is too low! The sample should come from a population with a lower proportion.

The evidence is enough to show that the sample is coming from the described population.

The sample proportion is too high! The sample should come from a population with a higher proportion.

The conclusion cannot be determined.

The evidence is not enough to show that the sample is coming from other population.

Answer 1

n = 400

p = 0.47

= p = 0.47

= sqrt(p(1 - p)/n)

     = sqrt(0.47 * (1 - 0.47)/400)

     = 0.025

a) P( < 0.4)

= P(( - )/ < (0.4 - )/)

= P(Z < (0.4 - 0.47)/0.025)

= P(Z < -2.8)

= 0.0026

b) P(> 0.4875)

= P(( - )/> (0.4875 - )/)

= P(Z > (0.4875 - 0.47)/0.025)

= P(Z > 0.7)

= 1 - P(Z < 0.7)

= 1 - 0.7580

= 0.2420

c) = 200/500 = 0.4

At 95% confidence interval the critical value is z^* = 1.96

The 95% confidence interval for population proportion is

+/- z^* * sqrt((1 - )/n)

= 0.4 +/- 1.96 * sqrt(0.4 * 0.6/500)

= 0.4 +/- 0.0429

= 0.3571, 0.4429

The population proportion 0.47 does not lie in the confidence interval.

Option - A) The sample proportion is too low. The sample should come from a population with a lower proportion.