Question

1. The desired percentage of Silicon Dioxide in a certain type of aluminous cement is 5.5. Sixteen independently obtained samples are analyzed, the sample statistics are: sample mean, x=5.25 and, sample standard deviation, s=0.3. We need to test if the true mean Silicon Dioxide percentage significantly differs from 5.5.

1. Write down the null and alternative hypotheses to be tested. Clearly define the terms used.

2. What is the type of statistical test procedure that should be used to test the hypotheses? Explain.

3. Construct a 95% confidence interval. Test the hypotheses using the confidence interval. Interpret the test result clearly.

4. Test the hypotheses using the Test Statistic / Critical Value method (you must clearly indicate the test statistic, and the critical value(s) use Take α=5%).

5. What is the P value of the test? Test the hypotheses using the P value. Take α=5%.

2. The Izod Impact Test was performed on 20 specimens of PVC pipe. The sample mean is 1.25 and the sample standard deviation is 0.25. We need to test if the true mean Izod impact strength is lesser than 1.5.

1. Write down the null and alternative hypotheses to be tested. Clearly define the terms used.

2. What is the type of statistical test procedure that should be used to test the hypotheses? Explain.

3. Construct a 95% confidence interval. Test the hypotheses using the confidence interval. Interpret the test result clearly

4. Test the hypotheses using the Test Statistic / Critical Value method (you must clearly indicate the test statistic, and the critical value(s) use Take α=5%).

5. What is the P value of the test? Test the hypotheses using the P value. Take α=5%.

Answer 1

1.

a.
Given that,
population mean(u)=5.5
sample mean, x =5.25
standard deviation, s =0.3
number (n)=16
null, Ho: μ=5.5
alternate, H1: μ!=5.5
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.131
since our test is two-tailed
reject Ho, if to < -2.131 OR if to > 2.131
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =5.25-5.5/(0.3/sqrt(16))
to =-3.333
| to | =3.333
critical value
the value of |t α| with n-1 = 15 d.f is 2.131
we got |to| =3.333 & | t α | =2.131
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.3333 ) = 0.0045
hence value of p0.05 > 0.0045,here we reject Ho
ANSWERS
---------------
null, Ho: μ=5.5
alternate, H1: μ!=5.5
test statistic: -3.333
critical value: -2.131 , 2.131
decision: reject Ho
p-value: 0.0045
we have enough evidence to support the claim that if the true mean Silicon Dioxide percentage significantly differs from 5.5
b.
TRADITIONAL METHOD
given that,
sample mean, x =5.25
standard deviation, s =0.3
sample size, n =16
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3/ sqrt ( 16) )
= 0.075
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
margin of error = 2.131 * 0.075
= 0.16
III.
CI = x ± margin of error
confidence interval = [ 5.25 ± 0.16 ]
= [ 5.09 , 5.41 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =5.25
standard deviation, s =0.3
sample size, n =16
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
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 = [ 5.25 ± t a/2 ( 0.3/ Sqrt ( 16) ]
= [ 5.25-(2.131 * 0.075) , 5.25+(2.131 * 0.075) ]
= [ 5.09 , 5.41 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 5.09 , 5.41 ] 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
2.
Given that,
population mean(u)=1.5
sample mean, x =1.25
standard deviation, s =0.25
number (n)=20
null, Ho: μ=1.5
alternate, H1: μ<1.5
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.729
since our test is left-tailed
reject Ho, if to < -1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1.25-1.5/(0.25/sqrt(20))
to =-4.472
| to | =4.472
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =4.472 & | t α | =1.729
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -4.4721 ) = 0.00013
hence value of p0.05 > 0.00013,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=1.5
alternate, H1: μ<1.5
test statistic: -4.472
critical value: -1.729
decision: reject Ho
p-value: 0.00013
we have enough evidence to support the claim that if the true mean Izod impact strength is lesser than 1.5
b.
TRADITIONAL METHOD
given that,
sample mean, x =1.25
standard deviation, s =0.25
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.25/ sqrt ( 20) )
= 0.056
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 0.056
= 0.117
III.
CI = x ± margin of error
confidence interval = [ 1.25 ± 0.117 ]
= [ 1.133 , 1.367 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1.25
standard deviation, s =0.25
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
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 = [ 1.25 ± t a/2 ( 0.25/ Sqrt ( 20) ]
= [ 1.25-(2.093 * 0.056) , 1.25+(2.093 * 0.056) ]
= [ 1.133 , 1.367 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 1.133 , 1.367 ] 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