A simple random sample of size n is drawn from a population that is normally distributed. The sample mean,
x,
is found to be
115,
and the sample standard deviation, s, is found to be
10.
(a) Construct
a
95%
confidence interval about
μ
if the sample size, n, is
22.
(b) Construct
a
95%
confidence interval about
μ
if the sample size, n, is
12.
(c) Construct
a
90%
confidence interval about
μ
if the sample size, n, is
22.
(d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
(a) Construct
a
95%
confidence interval about
μ
if the sample size, n, is
22.
Lower bound:
nothing;
Upper bound: nothing
(Use ascending order. Round to one decimal place as needed.)
(b) Construct
a
95%
confidence interval about
μ
if the sample size, n, is
12.
Lower bound:
nothing;
Upper bound: nothing
(Use ascending order. Round to one decimal place as needed.)
How does
decreasing
the sample size affect the margin of error, E?
A.As the sample size
decreases,
the margin of error stays the same.
B.As the sample size
decreases,
the margin of error
decreases.
C.As the sample size
decreases,
the margin of error
increases.
(c) Construct
a
90%
confidence interval about
μ
if the sample size, n, is
22.
Lower bound:
nothing;
Upper bound: nothing
(Use ascending order. Round to one decimal place as needed.)
Compare the results to those obtained in part (a). How does
decreasing
the level of confidence affect the size of the margin of error, E?
A.As the level of confidence
decreases,
the size of the interval
decreases.
B.As the level of confidence
decreases,
the size of the interval stays the same.
C.As the level of confidence
decreases,
the size of the interval
increases.
(d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
A.
Yes, the population needs to be normally distributed.
B.
Yes, the population does not need to be normally distributed.
C.
No, the population needs to be normally distributed.
D.
No, the population does not need to be normally distributed.
a)
sample mean 'x̄= | 115.000 | |
sample size n= | 22 | |
std deviation s= | 10.0000 | |
std error ='sx=s/√n=10/√22= | 2.1320 |
for 95% CI; and 21 df, value of t= | 2.080 | from excel: t.inv(0.975,21) | ||
margin of error E=t*std error = | 4.434 | |||
lower bound=sample mean-E = | 110.6 | |||
Upper bound=sample mean+E = | 119.4 | |||
f |
b)
lower bound=sample mean-E = | 108.6 | |
Upper bound=sample mean+E = | 121.4 |
C ).As the sample size decreases, the margin of error increases.
c)
lower bound=sample mean-E = | 111.3 | |
Upper bound=sample mean+E = | 118.7 |
A.As the level of confidence decreases, the size of the interval decreases.
d)
A) Yes, the population needs to be normally distributed.
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