Math 333 Homework: Note: in some of these problems, the distribution of x-bar ma
ID: 3314108 • Letter: M
Question
Math 333 Homework: Note: in some of these problems, the distribution of x-bar may be a t-distribution, based on the same criteria as we used in the confidence interval section 1. The mean water temperature downstream from a power plant cooling tower discharge pipe should be no more than 100°F. An environmental company wishes to show that the plant is operating above this boundary with 95% certainty. The water temperature is measured on 40 randomly chosen days, and the average temperature is found to be 101.5°F with a sample standard deviation of 4°F. Test the environmental company's claim with an appropriate hypothesis test. 2. A dealer advertises that a certain brand of tire averages 50,000 miles of use before needing to be replaced. Testing this claim, an investigator finds that a sample of 19 randomly selected tires from this dealer had an average lifespan of 48,700 miles with a standard deviation of 4,500 miles. Assume tire life spans are approximately normally distributed. At the = 0.05 level, find if the investigator can prove the dealer is exaggerating the average lifespan of their tires by performing the hypothesis test 3. A manufacturing company claims the mean life of its AAA batteries is 1000 hours. A potential customer, who is a wholesaler, will go ahead with a long term purchase agreement with the manufacturer, unless there is evidence that the mean lifetime of the batteries is shorter than the manufacturer's claim. A random sample of 41 of these batteries was found to have an average lifetime of 990 hours and a sample variance of 256 hours2. The wholesaler designs a test to see if the batteries do not meet specifications. Carry out the test at 5% level of significance, and state what is the wholesaler's decision regarding the proposed purchase. Compute the P-value of the test 4. The correct thickness for silicon wafers used in a certain circuit is 245 m. To see if the production line is correctly calibrated, a random sample of 45 wafers is obtained, and it shows a mean thickness of 246.18 m with a sample variance of 31.36 m. Test the claim that the machine is incorrectly calibrated at the 0.01 level. Carry out the test at 1% level of significance, and state what the evidence shows about the calibration of the production line. Compute the P-value of the test.Explanation / Answer
1.
Given that,
population mean(u)=100 degrees of F
sample mean, x =101.5 degrees of F
standard deviation, s =4 degrees of F
number (n)=40
null, Ho: =100
alternate, H1: >100
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.68
since our test is right-tailed
reject Ho, if to > 1.68
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =101.5-100/(4/sqrt(40))
to =2.372
| to | =2.372
critical value
the value of |t | with n-1 = 39 d.f is 1.68
we got |to| =2.372 & | t | =1.68
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.3717 ) = 0.01137
hence value of p0.05 > 0.01137,here we reject Ho
ANSWERS
---------------
null, Ho: =100
alternate, H1: >100
test statistic: 2.372
critical value: 1.68
decision: reject Ho
p-value: 0.01137
2.
Given that,
population mean(u)=50000
sample mean, x =48700
standard deviation, s =4500
number (n)=19
null, Ho: =50000
alternate, H1: >50000
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.73
since our test is right-tailed
reject Ho, if to > 1.73
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =48700-50000/(4500/sqrt(19))
to =-1.259
| to | =1.259
critical value
the value of |t | with n-1 = 18 d.f is 1.73
we got |to| =1.259 & | t | =1.73
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > -1.2592 ) = 0.88799
hence value of p0.05 < 0.88799,here we do not reject Ho
ANSWERS
---------------
null, Ho: =50000
alternate, H1: >50000
test statistic: -1.259
critical value: 1.73
decision: do not reject Ho
p-value: 0.88799
we do not have enough evidence to support the claim
3.
Given that,
population mean(u)=1000
sample mean, x =990
standard deviation, s =16
number (n)=41
null, Ho: =1000
alternate, H1: <1000
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.68
since our test is left-tailed
reject Ho, if to < -1.68
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =990-1000/(16/sqrt(41))
to =-4.002
| to | =4.002
critical value
the value of |t | with n-1 = 40 d.f is 1.68
we got |to| =4.002 & | t | =1.68
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -4.002 ) = 0.00013
hence value of p0.05 > 0.00013,here we reject Ho
ANSWERS
---------------
null, Ho: =1000
alternate, H1: <1000
test statistic: -4.002
critical value: -1.68
decision: reject Ho
p-value: 0.00013
we have enough evidence to support the claim that mean life time of batteries is short than manufacture film
4.
Given that,
population mean(u)=245
sample mean, x =246.18
standard deviation, s =5.6
number (n)=45
null, Ho: =245
alternate, H1: !=245
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.69
since our test is two-tailed
reject Ho, if to < -2.69 OR if to > 2.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =246.18-245/(5.6/sqrt(45))
to =1.414
| to | =1.414
critical value
the value of |t | with n-1 = 44 d.f is 2.69
we got |to| =1.414 & | t | =2.69
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.4135 ) = 0.1645
hence value of p0.01 < 0.1645,here we do not reject Ho
ANSWERS
---------------
null, Ho: =245
alternate, H1: !=245
test statistic: 1.414
critical value: -2.69 , 2.69
decision: do not reject Ho
p-value: 0.1645we do not have enough evidence to support the claim that calibration of the production line