Question

1)

Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 47 with a mean of 73.8 and a standard deviation of 6.8. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

80% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

2) Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for a sample of size 729 with a mean of 48.2 and a standard deviation of 14.8. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

< μ <
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

3) In a survey, 20 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $34 and standard deviation of $12. Estimate how much a typical parent would spend on their child's birthday gift (use a 80% confidence level).

Give your answers to one decimal place. Provide the point estimate and margin or error.

___ ±___

4)

SAT scores are distributed with a mean of 1,500 and a standard deviation of 303. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your confidence interval to 25 points with 90 percent confidence, how many students should you sample? (Round up to a whole number of students.)

5) Express the confidence interval 24.6%±6.5%24.6%±6.5% in the form of a trilinear inequality.

% <p< %

6) Express the confidence interval 18.3%±5.3%18.3%±5.3% in interval form.

Express the answer in decimal format (do not enter as percents).

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =73.8
standard deviation, s =6.8
sample size, n =47
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6.8/ sqrt ( 47) )
= 1
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 46 d.f is 1.3
margin of error = 1.3 * 1
= 1.3
III.
CI = x ± margin of error
confidence interval = [ 73.8 ± 1.3 ]
= [ 72.5 , 75.1 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =73.8
standard deviation, s =6.8
sample size, n =47
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 46 d.f is 1.3
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 73.8 ± t a/2 ( 6.8/ Sqrt ( 47) ]
= [ 73.8-(1.3 * 1) , 73.8+(1.3 * 1) ]
= [ 72.5 , 75.1 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 80% sure that the interval [ 72.5 , 75.1 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population mean
2.
TRADITIONAL METHOD
given that,
sample mean, x =48.2
standard deviation, s =14.8
sample size, n =729
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 14.8/ sqrt ( 729) )
= 0.548
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 728 d.f is 1.647
margin of error = 1.647 * 0.548
= 0.903
III.
CI = x ± margin of error
confidence interval = [ 48.2 ± 0.903 ]
= [ 47.297 , 49.103 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =48.2
standard deviation, s =14.8
sample size, n =729
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 728 d.f is 1.647
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 48.2 ± t a/2 ( 14.8/ Sqrt ( 729) ]
= [ 48.2-(1.647 * 0.548) , 48.2+(1.647 * 0.548) ]
= [ 47.297 , 49.103 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 47.297 , 49.103 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
3.
TRADITIONAL METHOD
given that,
sample mean, x =34
standard deviation, s =12
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 12/ sqrt ( 20) )
= 2.7
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 1.328
margin of error = 1.328 * 2.7
= 3.6
III.
CI = x ± margin of error
confidence interval = [ 34 ± 3.6 ]
= [ 30.4 , 37.6 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =34
standard deviation, s =12
sample size, n =20
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 1.328
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 34 ± t a/2 ( 12/ Sqrt ( 20) ]
= [ 34-(1.328 * 2.7) , 34+(1.328 * 2.7) ]
= [ 30.4 , 37.6 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 80% sure that the interval [ 30.4 , 37.6 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population mean
4.
given data,
standard deviation =303
confidence level is 90%
sample size = number of students
margin of error of your confidence interval to 25 points
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 303
ME =25
n = ( 1.645*303/25) ^2
= (498.44/25 ) ^2
= 397.5 ~ 398          
5.
confidence interval 24.6%±6.5%
confidence interval = 24.6-6.5, 24.6+6.5
confidence interval = 18.1,31.1
confidenceinterval 18.1%<p<31.1%
6.
the confidence interval 18.3%±5.3%
confidence interval = 18.3-5.3,18.3+5.3
confidence interval = (13,23.6)% = 0.13,0.236