Question
In: Statistics and Probability
A random sample of 42 taxpayers claimed an average of $9 comma 786 in medical expenses...
A random sample of 42 taxpayers claimed an average of $9 comma 786 in medical expenses for the year. Assume the population standard deviation for these deductions was $2 comma 387. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below. a. 1 % b. 2 % c. 5 %
Solutions
orchestra answered 1 year agoRelated Solutions
A random sample of 42 taxpayers claimed an average of $9,857 in medical expenses for the...
A random sample of 42 taxpayers claimed an average of $9,857 in
medical expenses for the year. Assume the population standard
deviation for these deductions was $2,409. Construct confidence
intervals to estimate the average deduction for the population with
the levels of significance shown below.
a)1%
b)2%
c)20%
A random sample of 33 taxpayers claimed an average of $9,668 in medical expenses for the...
A random sample of 33 taxpayers claimed an average of $9,668 in
medical expenses for the year. Assume the population standard
deviation for these deductions was $2,415.
Construct confidence intervals to estimate the average deduction
for the population with the levels of significance shown below.
a.
1 %
b.
5 %
c.
20 %
a. The confidence interval with a 1 % level of significance has
a lower limit of $ and an upper limit of $
A random sample of 33 taxpayers claimed an average of $9,668 in medical expenses for the...
A random sample of 33 taxpayers claimed an average of $9,668 in
medical expenses for the year. Assume the population standard
deviation for these deductions was $2,415.
Construct confidence intervals to estimate the average deduction
for the population with the levels of significance shown below.
a.
1 %
b.
5 %
c.
20 %
a. The confidence interval with a 1 % level of significance has
a lower limit of $ and an upper limit of $
A random sample of 37 taxpayers claimed an average of $9,527 in medical expenses for the...
A random sample of 37 taxpayers claimed an average of $9,527 in
medical expenses for the year. Assume the population standard
deviation for these deductions was $2,380.
Construct confidence intervals to estimate the average deduction
for the population with the levels of significance shown below.
a.
5%
b.
10%
c.
20%
The confidence interval with a 5 % level of significance has a
lower limit of _and an upper limit of _
The confidence interval with a 10% level of...
A random sample of 43 taxpayers claimed an average of $9,853 in medical expenses for the...
A random sample of 43 taxpayers claimed an average of $9,853 in
medical expenses for the year. Assume the population standard
deviation for these deductions was 2,418. Construct confidence
intervals to estimate the average deduction for the population with
the levels of significance shown below.
a.1%
b.5%
c.20%
a. The confidence interval with a 1% level of
significance has a lower limit of _____ and an upper limit of
______.
b. The confidence interval with a 5% level of
significance...
It is claimed that the mean of a population is 67. A random sample was obtained...
It is claimed that the mean of a population is 67. A
random sample was obtained and is represented by data labelled
“problem 9”. Complete the hypothesis test with α=5%. Use p-value
method.
49.5
69.3
80.5
81.9
70
62.2
76.7
92.4
92.7
70.9
95.5
62.5
69.1
70.9
79.6
68.4
88
52.3
95.5
85.8
61.2
98.1
100.1
68.8
76.3
66.9
103.2
79.8
60.1
37.3
55.3
83.6
83.7
76.9
71.6
67.6
69.9
80.9
102.2
85.1
Suppose we collect a random sample of n = 9 and find an average income of...
Suppose we collect a random sample of n = 9 and find an average
income of $49,000 with a sample standard deviation s = $12,000.
Provide each of the following using this information.
A 95% confidence interval estimate of the population mean µ.
What is the value for the margin of error? Interpret your
results.
A 90% confidence interval estimate of the population mean
µ.
A 99% confidence interval estimate of the population mean
µ.
A random sample of 42 observations is used to estimate the population variance. The sample mean...
A random sample of 42 observations is used to estimate the
population variance. The sample mean and sample standard deviation
are calculated as 74.5 and 5.6, respectively. Assume that the
population is normally distributed.
a. Construct the 90% interval estimate for the
population variance. (Round intermediate calculations to at
least 4 decimal places and final answers to 2
decimal places.)
b. Construct the 99% interval estimate for the
population variance. (Round intermediate
calculations to at least 4 decimal places and...
4) A random sample of 42 observations was made on a variable XX. The sample mean...
4) A random sample of 42 observations was made on a variable XX.
The sample mean was found to be 7.37 and the sample variance was
found to be 5.53.
Construct a 99% confidence interval for the expected
value (μμ) of XX.
Interpret the confidence interval.
Test the null hypothesis H0:μ=14.5H0:μ=14.5 against
the alternative hypothesis H1:μ>14.5H1:μ>14.5 using a 10%
significance level.
State any assumptions that you needed to make in order
for your answers to parts a. and c. to...
1. Over 9 months, a random sample of 100 women were asked to record their average...
1. Over 9 months, a random sample of 100 women were asked to
record their average menstrual cycle length (in days).
The sample average was 28.86 days, with a sample standard
deviation of 4.24 days.
(a) Calculate the lower bound for the 90% confidence interval
for the true average menstrual cycle length.
(b) Calculate the upper bound for the 90% confidence interval
for the true average menstrual cycle length.
(c) Interpret the confidence interval found in (a,b) in terms of...
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