Question

1. A cereal company claims that the mean weight of the cereal in its packets is more than 14 oz. The weights (in oz) of the cereal in a random sample of 8 cereal packets are listed below. You may assume the sample data comes from a population that follows a normal distribution. Using a 0.05 significance level, test the companies claim. State the null hypothesis H0. State the alternative hypothesis H1. What is the test statistic? State the alpha level. Determine p value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.

14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2

1. An article in a journal reports that 34% of American fathers take no responsibility for childcare. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 225 fathers from Littleton, yielded 97 who did not help with childcare. Using a 0.05 level of significance test the claim that the proportion is at least 34%. State the null hypothesis H0. State the alternative hypothesis H1. What is the test statistic? State the alpha level. Determine p value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.
2. The average time it takes for a person to experience relief from aspirin is 25 minutes. A new ingredient is added to help speed up relief. A study was conducted wit h35 people that resulted in a mean of 22.1 minutes with a standard deviation of 1.9 minutes. Using a 0.05 significance level, test the claim that the mean amount of time relief will be felt is less than 25 minutes. State the null hypothesis H0. State the alternative hypothesis H1. What is the test statistic? State the alpha level. Determine p value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.
3. A patient medicine company supervisor assumes that the bottling machine is operating properly if only 5% if the processed bottles are not full. A random sample of 100 bottles had 7 bottles that weren’t full. Using a significance level of 0.01, conduct a test to see if the machine is operating properly. State the null hypothesis H0. State the alternative hypothesis H1. What is the test statistic? State the alpha level. Determine p value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.
4. According to management accounting, salary figures for certified management accountants who are in the field less than 1 year are normally distributed with a mean of $31,279, with a standard deviation of $1,797. Test the hypothesis that the mean for all Denver first year CMAs is not equal to $31,129. Use 0.05 level of significance. State the null hypothesis H0. State the alternative hypothesis H1. What is the test statistic? State the alpha level. Determine p value. Do you or do you not reject the null hypothesis? Why? Write a clear conclusion using a complete sentence.

Answer 1

a.
Given that,
population mean(u)=14
sample mean, x =14.05
standard deviation, s =0.3464
number (n)=8
null, Ho: μ=14
alternate, H1: μ>14
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =14.05-14/(0.3464/sqrt(8))
to =0.4083
| to | =0.4083
critical value
the value of |t α| with n-1 = 7 d.f is 1.895
we got |to| =0.4083 & | t α | =1.895
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.4083 ) = 0.34764
hence value of p0.05 < 0.34764,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=14
alternate, H1: μ>14
test statistic: 0.4083
critical value: 1.895
decision: do not reject Ho
p-value: 0.34764
we do not have enough evidence to support the claim that mean weight of the cereal in its packets is more than 14 oz.
b.
Given that,
possible chances (x)=97
sample size(n)=225
success rate ( p )= x/n = 0.4311
success probability,( po )=0.34
failure probability,( qo) = 0.66
null, Ho:p=0.34
alternate, H1: p>=0.34
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.43111-0.34/(sqrt(0.2244)/225)
zo =2.885
| zo | =2.885
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =2.885 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.88504 ) = 0.00196
hence value of p0.05 > 0.00196,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.34
alternate, H1: p>=0.34
test statistic: 2.885
critical value: 1.64
decision: reject Ho
p-value: 0.00196
we have enough evidence to support the claim that the proportion is at least 34%.
c.
Given that,
population mean(u)=25
sample mean, x =22.1
standard deviation, s =1.9
number (n)=35
null, Ho: μ=25
alternate, H1: μ<25
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.691
since our test is left-tailed
reject Ho, if to < -1.691
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =22.1-25/(1.9/sqrt(35))
to =-9.0298
| to | =9.0298
critical value
the value of |t α| with n-1 = 34 d.f is 1.691
we got |to| =9.0298 & | t α | =1.691
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -9.0298 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=25
alternate, H1: μ<25
test statistic: -9.0298
critical value: -1.691
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the mean amount of time relief will be felt is less than 25 minutes.
d.
Given that,
possible chances (x)=7
sample size(n)=100
success rate ( p )= x/n = 0.07
success probability,( po )=0.05
failure probability,( qo) = 0.95
null, Ho:p=0.05
alternate, H1: p!=0.05
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.07-0.05/(sqrt(0.0475)/100)
zo =0.9177
| zo | =0.9177
critical value
the value of |z α| at los 0.01% is 2.58
we got |zo| =0.918 & | z α | =2.58
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.91766 ) = 0.3588
hence value of p0.01 < 0.3588,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.05
alternate, H1: p!=0.05
test statistic: 0.9177
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.3588
we do not have enough evidence to support the claim that if only 5% if the processed bottles are not full.
e.
Assume number (n)=35 because not given in the data,
Given that,
population mean(u)=31129
sample mean, x =31279
standard deviation, s =1797
null, Ho: μ=31129
alternate, H1: μ!=31129
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.032
since our test is two-tailed
reject Ho, if to < -2.032 OR if to > 2.032
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =31279-31129/(1797/sqrt(35))
to =0.4938
| to | =0.4938
critical value
the value of |t α| with n-1 = 34 d.f is 2.032
we got |to| =0.4938 & | t α | =2.032
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.4938 ) = 0.6246
hence value of p0.05 < 0.6246,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=31129
alternate, H1: μ!=31129
test statistic: 0.4938
critical value: -2.032 , 2.032
decision: do not reject Ho
p-value: 0.6246
we do not have enough evidence to support the claim that the mean for all Denver first year CMAs is not equal to $31,129.