isds 361a exam 2 practice chpt 7-9 fall 2020
Study online at https://quizlet.com/_914zcx
1. the standard deviation of the sampling standard error
distribution of the sample mean
2. the mean of the sampling distribution of equal to the population mean
the sample mean is
3. a random sample of size 49 is taken from 300 and 3
a population whose mean is 300 and the why: sample mu = pop mu, so 300 = 300
standard deviation is 21. the mean and
the standard error of the samplig distri- why: sigma of x bar = sigma / square root of
bution of the sample mean, respectively "n" (formula for the standard deviation of the
are: sampling distribution of the sample mean)
- since standard deviation = 21 and n = 49,
21/sqrt (49)
= 21/7
=3
4. given an infinite population with the 1.5
mean of 40 and a standard deviation of why: use standard deviation of x bar formula
15, a sample size of 100 is taken from it. = sigma / square root of n
the standard error of the sampling distri- = 15 / sqrt (100)
bution of the sample mean is: =
1.5
5. if a random sample of size n is drawn normal for all values of n
from a normal population, then the sam-
pling distribution of the sample mean will
be:
6. for a random sample of size 50 drawn central limit theorem
from a population, what theorem or rule
is used to determine the type of distrib-
1/8
, isds 361a exam 2 practice chpt 7-9 fall 2020
Study online at https://quizlet.com/_914zcx
ution for the sampling distribution of the
sample mean?
7. given an infinite population with a mean 0.98332
of 75 and a standard deviation of 12, the why: mean = 75 and s.d = 12
probability that the mean of a sample of - P(x<78) = P (z<(78-75)/12/sqrt(36))
36 observations, taken at random from = P (z<3x6/12)
this population is less than 78 is: = P (z<1.5)
= 0.98332
8. a sample of size 40 is taken from an in- P(Z>1.05)
finite population whose mean and stan- why: P(x>70) = P(z<70-68)
dard deviation are 68 and 12 respectively. = P(z>2/12/sqrt(40))
the probability that the sample mean is = P(z>2x2rt(10)/12)
larger than 70 equals: = P (z>4rt(10)/12)
= P(z>1.05)
9. a sample of 250 observations is selected 0.0274
at random from an n infinite population. = sqrt (0.25 x 0.75 x 1/250)
given that the population proportion is = 0.0274
0.25, the standard error of the sampling
distribution of the sample proportion will
be:
10. a sample of size 200 is taken at random 0.281
from an infinite population. given that = why: std dev = sqrt [(pop proportion x (1-
the population proportion is 0.60, the pop proportion))/n)]
probability that the sample proportion is = sqrt[(0.6)(1-0.6)/200)]
less than 0.58 is = 0.0346
USE EXCEL
= norm.dist(0.58,0.6,0.0346, true)
= 0.2816
2/8
Study online at https://quizlet.com/_914zcx
1. the standard deviation of the sampling standard error
distribution of the sample mean
2. the mean of the sampling distribution of equal to the population mean
the sample mean is
3. a random sample of size 49 is taken from 300 and 3
a population whose mean is 300 and the why: sample mu = pop mu, so 300 = 300
standard deviation is 21. the mean and
the standard error of the samplig distri- why: sigma of x bar = sigma / square root of
bution of the sample mean, respectively "n" (formula for the standard deviation of the
are: sampling distribution of the sample mean)
- since standard deviation = 21 and n = 49,
21/sqrt (49)
= 21/7
=3
4. given an infinite population with the 1.5
mean of 40 and a standard deviation of why: use standard deviation of x bar formula
15, a sample size of 100 is taken from it. = sigma / square root of n
the standard error of the sampling distri- = 15 / sqrt (100)
bution of the sample mean is: =
1.5
5. if a random sample of size n is drawn normal for all values of n
from a normal population, then the sam-
pling distribution of the sample mean will
be:
6. for a random sample of size 50 drawn central limit theorem
from a population, what theorem or rule
is used to determine the type of distrib-
1/8
, isds 361a exam 2 practice chpt 7-9 fall 2020
Study online at https://quizlet.com/_914zcx
ution for the sampling distribution of the
sample mean?
7. given an infinite population with a mean 0.98332
of 75 and a standard deviation of 12, the why: mean = 75 and s.d = 12
probability that the mean of a sample of - P(x<78) = P (z<(78-75)/12/sqrt(36))
36 observations, taken at random from = P (z<3x6/12)
this population is less than 78 is: = P (z<1.5)
= 0.98332
8. a sample of size 40 is taken from an in- P(Z>1.05)
finite population whose mean and stan- why: P(x>70) = P(z<70-68)
dard deviation are 68 and 12 respectively. = P(z>2/12/sqrt(40))
the probability that the sample mean is = P(z>2x2rt(10)/12)
larger than 70 equals: = P (z>4rt(10)/12)
= P(z>1.05)
9. a sample of 250 observations is selected 0.0274
at random from an n infinite population. = sqrt (0.25 x 0.75 x 1/250)
given that the population proportion is = 0.0274
0.25, the standard error of the sampling
distribution of the sample proportion will
be:
10. a sample of size 200 is taken at random 0.281
from an infinite population. given that = why: std dev = sqrt [(pop proportion x (1-
the population proportion is 0.60, the pop proportion))/n)]
probability that the sample proportion is = sqrt[(0.6)(1-0.6)/200)]
less than 0.58 is = 0.0346
USE EXCEL
= norm.dist(0.58,0.6,0.0346, true)
= 0.2816
2/8
