Question

The Greenline Taxi Company in New York City keeps careful records of items left behind by riders. Most items are claimed, but many remain in a lost‑and‑found area at the company’s headquarters. Records indicate that the proportion of riders who leave an item in a taxi is 0.12. Suppose a random sample of 250 riders is obtained.

a) Find the sampling distribution of the sample proportion of riders who leave an item behind. Round the mean to two decimal places and the variance to four decimal places.

b) What is the probability that the sample proportion of riders who leave an item behind is more than 0.15?than 0.15? Round to four decimal places.

c) What is the probability that the sample proportion of riders who leave an item behind is between 0.11 and 0.115?0.11 and 0.115? Round to four decimal places.

d) Some people speculate that in bad economic times, riders (and people in general) are more careful with their belongings. Suppose this sample was obtained during a recession, and the sample proportion of riders who left an item behind was 0.09.0.09. Is there any evidence to suggest that the true proportion is less than 0.12?0.12? Round to four decimal places.

Answer 1

a)

Mean of proportion of riders who leave an item in a taxi = 0.12

Standard error or sample proportion , SE =

= 0.0206

np(1-p) = 250 * 0.12 * (1-0.12) = 26.4

Since np(1-p) > 10, the sample size is large enough to approximate the sampling distribution of proportion as normal distribution.

The sampling distribution of the sample proportion of riders is normal distribution with mean = 0.12 and standard deviation of 0.0206

b)

Probability that the sample proportion of riders who leave an item behind is more than 0.15 = P( > 0.15)

= P[Z > (0.15 - 0.12)/0.0206]

= P[Z > 1.46]

= 0.0721

c)

Probability that the sample proportion of riders who leave an item behind is between 0.11 and 0.115

= P(0.11 < < 0.115)

= P( < 0.115) - P( < 0.11)

= P[Z < (0.115 - 0.12)/0.0206] - P[Z < (0.11 - 0.12)/0.0206]

= P[Z < -0.24] - P[Z < -0.49]

= 0.4052 - 0.3121

= 0.0931

d)

Null Hypothesis H0: p = 0.12

Alternative Hypothesis H1: p < 0.12

Test statistic , Z = (0.09 -  0.12)/0.0206 = -1.46

P-value = P(Z < -1.46) = 0.0721

Probability that the sample proportion of riders who leave an item behind is less than 0.09 when the true proportion is 0.12 is 0.0721.

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no significant evidence to suggest that the true proportion is less than 0.12