Question

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.

Answer #1

#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

Suppose the local Best Buy store averages 522 customers every
day entering the facility with a standard deviation of 124
customers. A random sample of 40 business days was selected. What
is the probability that the average number of customers in the
sample is between 530 and 540?
0.2387
0.0357
0.1621
0.0572
(no excel work)

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