Question

Explain your process with these problems. Details should include the hypothesis used, Type of test to be used and why, decision obtained. If the null hypothesis is rejected then build a confidence interval using the same level of significance to estimate the interval for the true population mean.

1. A sample of 20 ASUN freshmen had a mean GPA of 2.8 over all their courses taken in their first semester at ASUN. This had a variance of .25. Perform a hypothesis test at the 95 percent level to determine if the first semester GPA of all OU freshmen is less than a B (3.0).

  
2. An environmentalist collects a liter of water from 45 different locations along the banks of a stream. He measures the amount of dissolved oxygen in each specimen. The mean oxygen level is 4.62 mg, with the overall standard deviation of 0.92. A water purifying company claims that the mean level of oxygen in the water is 5 mg. Conduct a hypothesis test with α=0.01 to determine whether the mean oxygen level is less than 5 mg.

  
3. A bus company advertised a mean time of 150 minutes for a trip between two cities. A consumer group had reason to believe that the mean time was more than 150 minutes. A sample of 40 trips showed x̄= 153 minutes and s=7.5 minutes. At the .05 level of significance, test the consumer group’s belief.

  
4. A manufacturing process produces ball bearings with diameters that have a normal distribution with known standard deviation of .04 centimeters. Ball bearings with diameters that are too small or too large are undesirable. To test the claim that μ = 0.50 centimeters, perform a twotailed hypothesis test at the 5% level of significance. Assume that a random sample of 25 gave a mean diameter of 0.51 centimeters. Perform a hypothesis test and state your decision.

Answer 1

1.
Given that,
population mean(u)=3
sample mean, x =2.8
standard deviation, s =0.5
number (n)=20
null, Ho: μ=3
alternate, H1: μ<3
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 =2.8-3/(0.5/sqrt(20))
to =-1.789
| to | =1.789
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =1.789 & | t α | =1.729
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -1.7889 ) = 0.0448
hence value of p0.05 > 0.0448,here we reject Ho
ANSWERS
---------------
null, Ho: μ=3
alternate, H1: μ<3
test statistic: -1.789
critical value: -1.729
decision: reject Ho
p-value: 0.0448
we have enough evidence to support the claim that the first semester GPA of all OU freshmen is
less than 3.

2.
Given that,
population mean(u)=5
sample mean, x =4.62
standard deviation, s =0.92
number (n)=45
null, Ho: μ=5
alternate, H1: μ<5
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.414
since our test is left-tailed
reject Ho, if to < -2.414
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.62-5/(0.92/sqrt(45))
to =-2.771
| to | =2.771
critical value
the value of |t α| with n-1 = 44 d.f is 2.414
we got |to| =2.771 & | t α | =2.414
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -2.7708 ) = 0.00408
hence value of p0.01 > 0.00408,here we reject Ho
ANSWERS
---------------
null, Ho: μ=5
alternate, H1: μ<5
test statistic: -2.771
critical value: -2.414
decision: reject Ho
p-value: 0.00408
we have enough evidence to support the claim that mean oxygen level is less than 5 mg.

3.
Given that,
population mean(u)=150
sample mean, x =153
standard deviation, s =7.5
number (n)=40
null, Ho: μ=150
alternate, H1: μ>150
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.685
since our test is right-tailed
reject Ho, if to > 1.685
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =153-150/(7.5/sqrt(40))
to =2.53
| to | =2.53
critical value
the value of |t α| with n-1 = 39 d.f is 1.685
we got |to| =2.53 & | t α | =1.685
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 2.5298 ) = 0.00778
hence value of p0.05 > 0.00778,here we reject Ho
ANSWERS
---------------
null, Ho: μ=150
alternate, H1: μ>150
test statistic: 2.53
critical value: 1.685
decision: reject Ho
p-value: 0.00778
we have enough evidence to support the claim that mean time was more than 150 minutes.

4.
Given that,
population mean(u)=0.5
sample mean, x =0.51
standard deviation, s =0.04
number (n)=25
null, Ho: μ=0.5
alternate, H1: μ!=0.5
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =0.51-0.5/(0.04/sqrt(25))
to =1.25
| to | =1.25
critical value
the value of |t α| with n-1 = 24 d.f is 2.064
we got |to| =1.25 & | t α | =2.064
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 != 1.25 ) = 0.2234
hence value of p0.05 < 0.2234,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=0.5
alternate, H1: μ!=0.5
test statistic: 1.25
critical value: -2.064 , 2.064
decision: do not reject Ho
p-value: 0.2234
we do not have enough evidence to support the claim that μ = 0.50 centimeters