Question

# 13. A local store claims that it averages 483 customers per day entering the facility with...

13. A local store claims that it averages 483 customers per day entering the facility with a standard deviation of 112 customers. To test this claim, you select a random sample of 48 business days.

a. What is the standard error of the sample mean?

b. What i s the probability of obtaining a sample mean of 522 or more?

c. What is the probability of obtaining a sample mean within 32 customers of the population mean?

14. According to the American Veterinary Medical Association, 37% of households own a dog. A random sample of 60 households was selected .

a. What is the standard error of the proportion?

b. What is the probability that 18 or fewer of these households own a dog?

c. What is the probability that between 23 and 28 of these households own a dog?

15. The roundtrip airfare between two cities is normally distributed. A random sample of 16 flights has an average airfare of \$422.10 with a standard deviation of \$56.30. a. Find the standard error of the sample mean.

b. What is the critical value for the 90% confidence interval for the mean airfare?

c. What is the margin of error for a 90% confidence interval for the mean airfare?

d. What is the 90% confidence interval for the mean airfare?

e. Interpret the 90% confidence interval from part (d). f. Several years ago, the mean airfare was \$403.80. U sing the confidence interval from part (d), is there evidence that the mean airfare has changed? Explain.

16. A random sample of 150 mortgages was selected. From this sample , 17 were found to be delinquent on their current payment. Find the 98% confidence int erval for the proportion based on this sample.

#13.

Mean = 483 and sd = 112

n = 48

a)

SE = 112/sqrt(48) = 16.1658

b)

P(X > 522)

= P(z > (522 - 483)/16.1658)

= P(z > 2.4125)

= 0.0079

c)

32/16.1658 = 1.9795

P(-1.9795 < z < 1.9795)

= P(z < 1.9795) - P(z < - 1.9795)

= 0.9761 - 0.0239

= 0.9522

#14.

p = 0.37 and n =60

a)

SE = sqrt(0.37 * 0.63 * 60) = 3.7398

b)

mean = 60*0.37 = 22.2

P(X < 18)

= P(z < (18 - 22.2)/3.7398)

= P(z < -1.1231)

= 0.1307

c)

P(23 < X < 28)

= P((23 - 22.2)/3.7398 < z < (28 - 22.2)/3.7398)

= P(0.2139 < z < 1.5509)

= 0.9395 - 0.5847

= 0.3548