Question

In: Statistics and Probability

A company distributes juice in bottles labeled 12 ounces. A government agency thinks that the company...

A company distributes juice in bottles labeled 12 ounces. A government agency thinks that the company actually puts less than 12 ounces in each bottle. The agency randomly selects 36 of these bottles, measures their contents, and obtains a sample mean of 11.8 ounces and a sample standard deviation of 0.6 ounces.

Which hypothesis test do you use on your calculator?

Question 2 options:

a)

T-Test

b)

One Proportion Z-Test

A manufacturer indicates that its light bulb has a mean lifespan of 600 hours. A consumer advocate claims the mean is less than 600 hours.

A random sample of 36 light bulbs produces a sample mean of 592 hours and a sample standard deviation of 60 hours.

Which hypothesis test do you use on your calculator?

Question 5 options:

a)

T-Test

b)

One Proportion Z-Test

In 2001, 23% of American college students had at least one tattoo. A researcher thinks this proportion has increased. She obtains a simple random sample of 1026 college students and finds that 254 have at least one tattoo.

Which hypothesis test do you use on your calculator?

Question 8 options:

a)

T-Test

b)

One Proportion Z-Test

According to the CDC, 55% of adults have only a cell phone and not a landline phone.

A researcher thinks the percentage of adults under age forty who are cell-phone only is greater than 55%.

In a random sample of 500 adults in that age range, 298 had only a cell phone and not a landline phone.

Which hypothesis test do you use on your calculator?

Question 11 options:

a)

T-Test

b)

One Proportion Z-Test

Question 13 :

A researcher conducts a hypothesis test on the null and alternate hypothesis shown below.

H0:p=0.29,H1:p<0.29

The researcher gets a P-value of 0.0415

At the α=0.05 level of significance, does the researcher reject the null hypothesis?

Question 13 options:

a)

Reject H0{"version":"1.1","math":"\(H_0\)"}

b)

Do Not Reject H0

A researcher conducts a hypothesis test on the null and alternate hypothesis shown below.

H0:p=0.29,H1:p<0.29{"version":"1.1","math":"H_0: p=0.29,H_1:p<0.29"}

The researcher gets a P-value of 0.0415{"version":"1.1","math":"\(0.0415\)"}

At the α=0.05{"version":"1.1","math":"\(\alpha=0.05\)"} level of significance, what is the conclusion?

Question 14 options:

a)

There is sufficient evidence to conclude that p<0.29{"version":"1.1","math":"\(p<0.29\)"}

b)

There is not sufficient evidence to conclude that p<0.29{"version":"1.1","math":"\(p<0.29\)"}

c)

There is sufficient evidence to conclude that p=0.29

Question 15 (1 point)

A researcher conducts a hypothesis test on the null and alternate hypothesis shown below.

H0:μ=300,H1:μ>300{"version":"1.1","math":"H_0: \mu=300,H_1:\mu>300"}

The researcher gets a P-value of 0.1794{"version":"1.1","math":"\(0.1794\)"}

At the α=0.05{"version":"1.1","math":"\(\alpha=0.05\)"} level of significance, does the researcher reject the null hypothesis?

Question 15 options:

a)

Reject H0{"version":"1.1","math":"\(H_0\)"}

b)

Do Not Reject H0

A researcher conducts a hypothesis test on the null and alternate hypothesis shown below.

H0:μ=300,H1:μ>300{"version":"1.1","math":"H_0: \mu=300,H_1:\mu>300"}

The researcher gets a P-value of 0.1794{"version":"1.1","math":"\(0.1794\)"}

At the α=0.05{"version":"1.1","math":"\(\alpha=0.05\)"} level of significance, what is the conclusion?

Question 16 options:

a)

There is sufficient evidence to conclude that μ>300{"version":"1.1","math":"\(\mu>300\)"}

b)

There is not sufficient evidence to conclude that μ>300{"version":"1.1","math":"\(\mu>300\)"}

c)

There is sufficient evidence to conclude that μ=300{"version":"1.1","math":"\(\mu=300\)"}

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Solutions

Expert Solution

a.
Given that,
population mean(u)=12
sample mean, x =11.8
standard deviation, s =0.6
number (n)=36
null, Ho: μ=12
alternate, H1: μ<12
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.69
since our test is left-tailed
reject Ho, if to < -1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =11.8-12/(0.6/sqrt(36))
to =-2
| to | =2
critical value
the value of |t α| with n-1 = 35 d.f is 1.69
we got |to| =2 & | t α | =1.69
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -2 ) = 0.02665
hence value of p0.05 > 0.02665,here we reject Ho
ANSWERS
---------------
null, Ho: μ=12
alternate, H1: μ<12
test statistic: -2
critical value: -1.69
decision: reject Ho
p-value: 0.02665
we have enough evidence to support the claim that the company actually puts less than 12 ounces in each bottle.
b.
Given that,
population mean(u)=600
sample mean, x =592
standard deviation, s =60
number (n)=36
null, Ho: μ=600
alternate, H1: μ<600
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.69
since our test is left-tailed
reject Ho, if to < -1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =592-600/(60/sqrt(36))
to =-0.8
| to | =0.8
critical value
the value of |t α| with n-1 = 35 d.f is 1.69
we got |to| =0.8 & | t α | =1.69
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.8 ) = 0.21455
hence value of p0.05 < 0.21455,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=600
alternate, H1: μ<600
test statistic: -0.8
critical value: -1.69
decision: do not reject Ho
p-value: 0.21455
we do not have enough evidence to support the claim that the mean is less than 600 hours.
c.
Given that,
possibile chances (x)=254
sample size(n)=1026
success rate ( p )= x/n = 0.248
success probability,( po )=0.23
failure probability,( qo) = 0.77
null, Ho:p=0.23
alternate, H1: p>=0.23
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.24756-0.23/(sqrt(0.1771)/1026)
zo =1.337
| zo | =1.337
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.337 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.33682 ) = 0.09064
hence value of p0.05 < 0.09064,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.23
alternate, H1: p>=0.23
test statistic: 1.337
critical value: 1.645
decision: do not reject Ho
p-value: 0.09064
we do not have enough evidence to support the claim that , 23% of American college students had at least one tattoo.
d.
Given that,
possibile chances (x)=298
sample size(n)=500
success rate ( p )= x/n = 0.596
success probability,( po )=0.55
failure probability,( qo) = 0.45
null, Ho:p=0.55
alternate, H1: p>0.55
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.596-0.55/(sqrt(0.2475)/500)
zo =2.068
| zo | =2.068
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.068 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.06755 ) = 0.01934
hence value of p0.05 > 0.01934,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.55
alternate, H1: p>0.55
test statistic: 2.068
critical value: 1.645
decision: reject Ho
p-value: 0.01934
we have enough evidence to support the claim that the percentage of adults under age forty who are cell-phone only is greater than 55%.
14.
Given that,
possibile chances (x)=21.14
sample size(n)=100
success rate ( p )= x/n = 0.211
success probability,( po )=0.29
failure probability,( qo) = 0.71
null, Ho:p=0.29
alternate, H1: p<0.29
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.2114-0.29/(sqrt(0.2059)/100)
zo =-1.732
| zo | =1.732
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.732 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -1.73219 ) = 0.04162
hence value of p0.05 > 0.04162,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.29
alternate, H1: p<0.29
test statistic: -1.732
critical value: -1.645
decision: reject Ho
p-value: 0.04162 =0.0415
we have enough evidence to support the claim that population proportion is less than 29%
15.
Given that,
population mean(u)=300
sample mean, x =350
standard deviation, s =30
number (n)=36
null, Ho: μ=300
alternate, H1: μ>300
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.69
since our test is right-tailed
reject Ho, if to > 1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =350-300/(30/sqrt(36))
to =10
| to | =10
critical value
the value of |t α| with n-1 = 35 d.f is 1.69
we got |to| =10 & | t α | =1.69
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 10 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=300
alternate, H1: μ>300
test statistic: 10
critical value: 1.69
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that population mean is greater than 300.


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