According to researchers, the mean length of imprisonment for motor-vehicle-theft offenders in a nation is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in a city in the nation had a mean length of imprisonment of 17.9 months. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in the city differs from the national mean? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in the city is 8.0 months. Set up the hypotheses for the one-mean z-test. Họ: H = 16.7 Ha:H * 16.7 (Type integers or decimals. Do not round.) The test statistic is z = 1.50 . (Round to two decimal places as needed.) The P-value is. (Round to three decimal places as needed.)
According to researchers, the mean length of imprisonment for motor-vehicle-theft offenders in a nation is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in a city in the nation had a mean length of imprisonment of 17.9 months. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in the city differs from the national mean? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in the city is 8.0 months. Set up the hypotheses for the one-mean z-test. Họ: H = 16.7 Ha:H * 16.7 (Type integers or decimals. Do not round.) The test statistic is z = 1.50 . (Round to two decimal places as needed.) The P-value is. (Round to three decimal places as needed.)
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Chapter1: Starting With Matlab
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Question Help
According to researchers, the mean length of imprisonment for motor-vehicle-theft offenders in a nation is 16.7 months. One hundred randomly selected
motor-vehicle-theft offenders in a city in the nation had a mean length of imprisonment of 17.9 months. At the 1% significance level, do the data provide sufficient
evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in the city differs from the national mean? Assume that the population
standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in the city is 8.0 months.
Set up the hypotheses for the one-mean z-test.
Ho: µ =
16.7
Ha: µ +
16.7
(Type integers or decimals. Do not round.)
The test statistic is z = 1.50 .
(Round to two decimal places as needed.)
The P-value is
(Round to three decimal places as needed.)"
Transcribed Image Text:9.4.87-P
Question Help
According to researchers, the mean length of imprisonment for motor-vehicle-theft offenders in a nation is 16.7 months. One hundred randomly selected
motor-vehicle-theft offenders in a city in the nation had a mean length of imprisonment of 17.9 months. At the 1% significance level, do the data provide sufficient
evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in the city differs from the national mean? Assume that the population
standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in the city is 8.0 months.
Set up the hypotheses for the one-mean z-test.
Ho: µ =
16.7
Ha: µ +
16.7
(Type integers or decimals. Do not round.)
The test statistic is z = 1.50 .
(Round to two decimal places as needed.)
The P-value is
(Round to three decimal places as needed.)

Transcribed Image Text:8.2.81
Question Help
In a study, the mean number of days that 124 adolescents in substance abuse treatment used medical marijuana in the last 5 months was 119.17. Assuming the
population standard deviation is 34 days, a 95% confidence interval for the mean number of days, µ, of medical marijuana use in the last 5 months of all adolescents in
substance abuse treatment is from 113.19 days to 125.15 days; this interval's margin of error is 5.98 days. OComplete parts (a) through (d) below.
Click here to view page 1 of the table of areas under the standard normal curve.
Click here to view page 2 of the table of areas under the standard normal curve.
The 95% confidence interval is from 103.57 day(s) to 127.51 day(s).
(Round to two decimal places as needed.)
b. Compare the 95% confidence intervals obtained in part (a) and in the original study by drawing a graph. Choose the correct graph below.
O A.
В.
95% CI for u
95% Cl for u
(n = 124)
(n = 124)
95% Cl for u
95% Cl for u
(n = 31)
(n = 31)
108
118
128
104
118
132
D.
95% Cl for u
95% CI for u
(n = 124)
(n = 124)
95% CI for u
95% Cl for u
(n = 31)
(n = 31)
100
114
128
112
120
128
c. Compare the margins of error for the two 95% confidence intervals.
The margin of error for the interval obtained in part (a) is
day(s), which is
the margin of error of the interval obtained in the original study.
(Round to two decimal places as needed.)
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