Question

The ages of a group of

125125

randomly selected adult females have a standard deviation of

16.116.1

years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so let

sigmaσequals=16.116.1

years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want

9595%

confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population?

The required sample size is

Answer #1

Solution :

Given that,

Population standard deviation = = 16.1

Margin of error = E = 0.5

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = [1.96 * 16.1 /0.5 ]2

n = 3983.12

Sample size = n = 3984

Yes,because statistic students are typically younger than people in the general population.

The ages of a group of 142 randomly selected adult females have
a standard deviation of 18.1 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigma=18.1 years for the sample size
calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 90% confidence that the
sample mean is within one-half...

The ages of a group of 130 randomly selected adult females have
a standard deviation of 16.2 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let σ=16.2 years for the sample size
calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 95% confidence that the
sample mean is within one-half...

The ages of a group of 135 randomly selected adult females have
a standard deviation of 17.9 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigmaequals17.9 years for the sample
size calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 95% confidence that the
sample mean is within one-half...

The ages of a group of 131 randomly selected adult females have
a standard deviation of 18.9 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigmaequals18.9 years for the sample
size calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 99% confidence that the
sample mean is within one-half...

The ages of a group of 154 randomly selected adult females have
a standard deviation of 17.9 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigmaequals17.9 years for the sample
size calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 95% confidence that the
sample mean is within one-half...

The ages of a group of 124 randomly selected adult females have
a standard deviation of 17.6 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigmaσequals=17.6 years for the sample
size calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 99% confidence that the
sample mean is within one-half...

The ages of a group of 124 randomly selected adult females have
a standard deviation of 18.5 years. Assume that the ages of female
statistics students have less variation than ages of females in the
general population, so let sigma equals18.5 years for the sample
size calculation. How many female statistics student ages must be
obtained in order to estimate the mean age of all female
statistics students? Assume that we want 90 % confidence that the
sample mean is...

An IQ test is designed so that the mean is 100 and the standard
deviation is 21 for the population of normal adults. Find the
sample size necessary to estimate the mean IQ score of statistics
students such that it can be said with 95% confidence that the
sample mean is within 4 IQ points of the true mean. Assume that
sigmaσequals=21 and determine the required sample size using
technology. Then determine if this is a reasonable sample size for...

An IQ test is designed so that the mean is 100 and the standard
deviation is
2525
for the population of normal adults. Find the sample size
necessary to estimate the mean IQ score of statistics students such
that it can be said with
9090%
confidence that the sample mean is within
66
IQ points of the true mean. Assume that
sigmaσequals=2525
and determine the required sample size using technology. Then
determine if this is a reasonable sample size for...

The recommended dietary allowance (RDA) of iron for adult
females is 18milligrams (mg) per day. The given iron intakes (mg)
were obtained for 45 random adult females. At the 1% significance
level, do the data suggest that adult females are, on average,
getting less than the RDA of 18mg of iron? Assume that the
population standard deviation is 4.9mg. Preliminary data analyses
indicate that applying the z-test is reasonable. note (sample
mean = 14.66mg)
state the hypothesis for one-mean z...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 8 minutes ago

asked 10 minutes ago

asked 22 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago