In: Statistics and Probability
PART I
Researchers in the corporate office of an airline wonder if there is a significant difference between the cost of a flight on Priceline.com vs. the airline's own website. A random sample of 16 flights were tracked on Priceline and the airline's website and the average difference between the vendors was $4.251 with a standard deviation of $9.2763. The 99% confidence paired-t interval for price (Priceline - Airline Site) was (-2.5826, 11.0846). Which of the following is the appropriate interpretation?
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PART II
A survey conducted by General Motors of 38 drivers in America, 33 indicated that they would prefer a car with a sunroof over one without. When estimating the proportion of all Americans who would prefer a car with a sunroof over one without with 90% confidence, what is the margin of error?
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PART III
A U.S. census bureau pollster noted that in 379 random households surveyed, 218 occupants owned their own home. What is the 99% confidence interval estimate of the proportion of American households who own their own home?
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PART I
99% confidence paired-t interval for price (Priceline - Airline Site) was (-2.5826, 11.0846)
2) We are 99% confident that the average difference in price between the two vendors for all flights is between -2.5826 and 11.0846.
PART II
n = number of drivers in America = 38
x = number of drivers who prefer a car with a sunroof over one = 33
(Round to 4 decimal)
Confidence level = c = 0.90
Margin of error (E):
Where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95
zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65)
E = 1.645 * 0.05484
E = 0.0902 (Round to 4 decimal)
Margin of error = 0.0902
PART III
n = number of households randomly selected = 379
x = number of occupants who owned their own home = 218
(Round to 4 decimal)
Confidence level = c = 0.99
Margin of error (E):
Where zc is z critical value for (1+c)/2 = (1+0.99)/2 = 0.995
zc = 2.58 (From statistical table of z values)
E = 2.58 * 0.025391
E = 0.0655
Margin of error = 0.0655091
99% confidence interval estimate of the proportion of American households who own their own home is
0.5752 - 0.0655091 < p< 0.5752 + 0.0655091
0.50969 < p < 0.64071 (Round to 4 decimal)
99% confidence interval estimate of the proportion of American households who own their own home is (0.50969, 0.64071)