Question

Question 1 (1 point) Airline companies recognize that empty seats represent lost revenues that can never be recovered. To avoid losing revenues, the companies often book more passengers than there are available seats. Then, when a flight experiences fewer no-shows than expected, some passengers are 'bumped' from their flights (are denied boarding). Incentives are provided to encourage passengers to give up their reserved seat! voluntarily, but occasionally some passengers are involuntarily bumped from the flight. Obviously, these incidents can reſiect poorly on customer satisfaction. Suppose Southwest Airlines would like to estimate the true proportion of involuntarily bumped passengers across all domestic flights in the industry. In a pilot sample of 813 domestic passengers, 352 were involuntarily bumped. What is the estimate of the population proportion and what is the standard error of this estimate? 1) Estimate of proportion: 0.567. Standard error: 0.0174. 2) Estimate of proportion: 0.433, Standard error: 0.0174. 3) Estimate of proportion: 0.567, Standard error: 0.0006. 4) The true population proportion is needed to calculate this. 5) Estimate of proportion: 0.433. Standard error: 0.0006,

Answer 1

Answer Date: 10/11/2019 The mean of the sampling distribution of proportion is, 49 = 352 = 0.433 The standard deviation of the sampling distribution of proportion is, O = P(1-P) PVn (0.433)(1-0.433) V 813 = 0.0174 The answer is 2.

The probability is that between 15.05% and 15.29% of the homes would be foreclosed on is. First, compute the mean and standard deviation of sampling distribution of sample proportion. The mean of the sampling distribution of sample proportion is, Ho = P = 0.1536 The standard deviation of the sampling distribution of sample proportion is, 0; -, Þ(1-P) (0.1536)(1 -0.1536) V 1148 = 0.0107 The probability is that between 15.05% and 15.29% of the homes would be foreclosed on is,

First, compute the z score then find probability based on standard normal table. For p=0.1505 converts to -P-4 0.1505 -0.1536 0.0107 =-0.2900 For p=0.1529 converts to P-up z = 0.1529 -0.1536 0.0107 = -0.0655 From the standard normal distribution table, the associated probability for z values and subtracts the probability is,

p(0.1505 SP < 0.1529)=P(-0.2900 SZS -0.0655) = P(Z 5-0.0655)-p(z 5-0.2900) = 0.4739 -0.3859 = 0.0880 The probability is that between 15.05% and 15.29% of the homes would be foreclosed on is 0.0880. The answer is 3] 3 The average lifetime is, First, compute the mean and standard deviation of sampling distribution of sample mean then find the average lifetime. The mean of the sampling distribution of sample means is, Az = 1 = 53470.11

The standard deviation of the sampling distribution of sample means is, 22 5531.93 67 = 675.83 The average lifetime is, From the standard normal table, the indicated z score for the area of 0.5900 subtract from 1 and indicative area 0.4100 is -0.22754. x = !1+(zxo) = 53470.11+(-0.22754 x 675.83) = 53316.33 The average lifetime is 53316.33. The answer is 2.

With a confidence of 90%, the margin of error for the proportion of all crap that would incorporate the gene into their DNA is, First, compute the proportion, the z-critical value then find confidence interval. The proportion is, P= I - 34 234 = 0.1453 The Z-critical value is, From the standard normal table, the z-critical value at 90% confidence is 1.645.

E = zcx =1.645 x 0.1453 (1 -0.1453) 234 = 0.0378 With a confidence of 90%, the margin of error for the proportion of all crap that would incorporate the gene into their DNA is 0.0378. The answer is 2