A random sample of the amount paid (in dollars) for taxi fare from downtown to t
ID: 3217645 • Letter: A
Question
A random sample of the amount paid (in dollars) for taxi fare from downtown to the airport was obtained. (Give your answers correct to two decimal places.)
22
17
19
13
22
16
16
23
20
22
15
21
16
21
14
Use the data to find a point estimate for the mean.
$
Use the data to find a point estimate for the variance.
$
(c) Use the data to find a point estimate for the standard deviation.
$
The number of engines owned per fire department was obtained from a random sample taken from profiles of fire departments from across the United States. (Give your answers correct to two decimal places.)
49 6 7 39 27 43 51 52 49 42 13 52
(a) Use the data to find a point estimate for the mean.
(b) Use the data to find a point estimate for the variance.
(c) Use the data to find a point estimate for the standard deviation.
3. Consider the following. (Enter your answers to four decimal places.)
(a) Find the level of confidence assigned to an interval estimate of the mean formed using the interval x - 1.07 · x to x + 1.07 · x.
(b) Find the level of confidence assigned to an interval estimate of the mean formed using the interval x - 1.57 · x to x + 1.57 · x.
(c) Find the level of confidence assigned to an interval estimate of the mean formed using the interval x - 1.73 · x to x + 1.73 · x.
(d) Find the level of confidence assigned to an interval estimate of the mean formed using the interval x - 2.02 · x to x + 2.02 · x
4. The sampled population is normally distributed, with the given information. (Give your answers correct to two decimal places.)
n = 19, x = 29.9, and = 7
(a) Find the 0.99 confidence interval for .
A sample of 69 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 24.9 years. The population variance is 15.
Give a point estimate for . (Give your answer correct to one decimal place.)
(b) Find the 95% confidence interval for . (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
(c) Find the 99% confidence interval for . (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
22
17
19
13
22
16
16
23
20
22
15
21
16
21
14
Explanation / Answer
Here, n = 69 , mean = 24.9 , population variance= 15 , population std.deviation = sqrt(15) = 3.87
a) point estimate for mean = 24.9
b) z value at 95% confidence interval = 1.96
CI = mean + / - z * (s / sqrt(n))
= 24.9 + / - 1.96 ( 3.87 / sqrt(69))
= (23.98 , 25.81)
Lower limit = 23.98
Upper limit = 25.81
c)
z value at 99% confidence interval = 2.576
CI = mean + / - z * (s / sqrt(n))
= 24.9 + / - 2.576 ( 3.87 / sqrt(69))
= (23.69 , 26.10)
Lower limit = 23.69
Upper limit = 26.10
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