Question

1)A statistics practitioner took a random sample of 47 observations from a population whose standard deviation is 35 and computed the sample mean to be 102.

Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.

A. Estimate the population mean with 95% confidence.

Confidence Interval =

B. Estimate the population mean with 90% confidence.

Confidence Interval =

C. Estimate the population mean with 99% confidence.

Confidence Interval =

2) For each situation, state the null and alternative hypotheses: (Type "mu" for the symbol ?μ , e.g.  mu > 1 for the mean is greater than 1, mu < 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1. Please do not include units such as "mm" or "$" in your answer.)

(a) The diameter of a spindle in a small motor is supposed to be 4.5 millimeters (mm) with a standard deviation of 0.1 mm. If the spindle is either too small or too large, the motor will not work properly. The manufacturer measures the diameter in a sample of 9 spindles to determine whether the mean diameter has moved away from the required measurement. Suppose the sample has an average diameter of 4.6 mm.
?0 :  
Ha :

(b) Harry thinks that prices in Caldwell are higher than the rest of the country. He reads that the nationwide average price of a certain brand of laundry detergent is $23.05 with standard deviation $1.79. He takes a sample from 4 local Caldwell stores and finds the average price for this same brand of detergent is $20.25.
?0:
??:

Answer 1

1.
a.
TRADITIONAL METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 35/ sqrt ( 47) )
= 5.105
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 5.105
= 10.006
III.
CI = x ± margin of error
confidence interval = [ 102 ± 10.006 ]
= [ 91.994,112.006 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 102 ± Z a/2 ( 35/ Sqrt ( 47) ) ]
= [ 102 - 1.96 * (5.105) , 102 + 1.96 * (5.105) ]
= [ 91.994,112.006 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [91.994 , 112.006 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 35/ sqrt ( 47) )
= 5.105
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 5.105
= 8.398
III.
CI = x ± margin of error
confidence interval = [ 102 ± 8.398 ]
= [ 93.602,110.398 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 102 ± Z a/2 ( 35/ Sqrt ( 47) ) ]
= [ 102 - 1.645 * (5.105) , 102 + 1.645 * (5.105) ]
= [ 93.602,110.398 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [93.602 , 110.398 ] 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
c.
TRADITIONAL METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 35/ sqrt ( 47) )
= 5.105
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 5.105
= 13.151
III.
CI = x ± margin of error
confidence interval = [ 102 ± 13.151 ]
= [ 88.849,115.151 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =35
sample mean, x =102
population size (n)=47
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 102 ± Z a/2 ( 35/ Sqrt ( 47) ) ]
= [ 102 - 2.576 * (5.105) , 102 + 2.576 * (5.105) ]
= [ 88.849,115.151 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [88.849 , 115.151 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
2.
a.
Given that,
population mean(u)=4.5
standard deviation, σ =0.1
sample mean, x =4.6
number (n)=9
null, Ho: μ=4.5
alternate, H1: μ!=4.5
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 4.6-4.5/(0.1/sqrt(9)
zo = 3
| zo | = 3
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =3 & | z α | = 1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 3 ) = 0.003
hence value of p0.05 > 0.003, here we reject Ho
ANSWERS
---------------
null, Ho: μ=4.5
alternate, H1: μ!=4.5
test statistic: 3
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.003
we have enough evidence to support the claim that whether the mean diameter has moved away from the required measurement.
b.
Given that,
population mean(u)=23.05
standard deviation, σ =1.79
sample mean, x =20.25
number (n)=4
null, Ho: μ=23.05
alternate, H1: μ!=23.05
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 20.25-23.05/(1.79/sqrt(4)
zo = -3.128
| zo | = 3.128
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =3.128 & | z α | = 1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -3.128 ) = 0.002
hence value of p0.05 > 0.002, here we reject Ho
ANSWERS
---------------
null, Ho: μ=23.05
alternate, H1: μ!=23.05
test statistic: -3.128
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.002
we have enough evidence to support the claim that prices in Caldwell are higher than the rest of the country