Download MATH 2000 TEST 4

MATH 2280 TEST 4 SOLUTIONS
1. A grocery store advertises that the mean wait at its checkout lines is 5 minutes or less.
You survey 64 customers and find that they had a mean wait 6 minutes with a population
standard deviation of 2.5 minutes. At   .05 should you reject the store’s claim?
You will reject the store’s claim if the sample mean is significantly greater than 5. This
makes the test right-tailed. The picture for the rejection region
is:
Shaded area =
 = 0.05
z
You will reject H 0 if z  z , where z = invnorm(.95) = 1.645. You may also get z
from a list of critical values.
H0 :   5
H1 :   5
Reject H 0 if z  1.645
z
x 
65

 3.2
 n 2.5
64
Reject H 0 , the wait time is more than five minutes.
[Or, if you use the p-value method
p = normalcdf(3.2,10000) = .00069. Reject H 0 because p   ]
2. A grocery store advertises that the mean weight of its onions is 5 ounces. You buy 36
onions and find that their mean weight is 6 ounces with a population standard deviation
of 3.5 ounces. At   .05 can you conclude that the mean weight is different from 5
ounces?
The question asks if the mean is different from 5, which makes the test two-tailed.
You find z from the following picture. Or you may look it up in a table of values.
2
Shaded area =

= 0.025
2
Shaded area =

= 0.025
2
z = invnorm(.975) =1.96
2
H0 :   5
H1 :   5
Reject H 0 if z  1.96 or z  1.96
x 
65
z

 1.714
 n 3.5
36
Do not reject H 0
[Or, if you use the p-value method
p = 2*normalcdf(1.714 ,10000) = .0865. Do not eject H 0 because p   ]
3. A professor believes that the mean time it should take students to complete a
questionnaire is at least 30 minutes. The mean time for a sample of 18 students to
complete the questionnaire was 27 minutes with a sample standard deviation of 5
minutes. At   .1 can you conclude that the mean time is less than 30 minutes?
Since the sample size is less than 30 you use a t-test with d.f. = 17.
H 0 :   30
H1 :   30
Reject H 0 if t  1.333
x   27  30
t

 2.546
5
s n
18
Reject H 0
[Or, if you use the p-value method
p = tcdf(10000,2,546, 17) = .0104. Reject H 0 because p   ]
4. The King of a small European nation is willing to provide funding for state parks if it
is determined that at least 30% of the population would make use of the facilities. In a
sample of 124 citizens it was found that 34 said they would use the parks. Can the king
claim at   .01 that fewer than 30% of citizens would use the parks?
34
pˆ 
 .2742 ; p  .3 ; q  1  .3  .7
124
H 0 : p  .3
H1 : p  .3
Reject H 0 if z  2.326
pˆ  p
.2742  .3

 .6271
pq / n
(.3)(.7) /124
Do not reject H 0
The King cannot claim (but probably will) that fewer than 30% of citizens will use the
parks.
z
[Or, if you use the p-value method
p = normalcdf(10000,.6271) = .2653 Do not reject H 0 because p   ]
5. Chef Heet la Greece is concerned that the variance in the weights of his baked
potatoes is too high. His supplier claims that the variance is 10. The good chef takes a
sample of 23 potatoes and finds the variance to be 15. At   .1 should Monsieur la
Greece reject the supplier’s claim?
Note that this is a right tailed test because the chef will only reject the supplier’s claim if
the variance is too high.
H 0 :  2  10
H1 :  2  10
Reject H 0 if  2  30.813
 
2
(n  1) s 2

2

(22)(15)
 33
10
Reject H 0
[Or, if you use the p-value method
p =  2cdf (33,10000,22) = .0619. Reject H 0 because p   ]
6. A manufacturer of batteries wants to advertise that his batteries last longer than a
competitor’s batteries. A sample of 100 batteries has a mean life of 7.1 hours with a
standard deviation of 1.3 hours. A sample of 80 of the competitor’s batteries has a mean
life of 6.9 hours with a standard deviation of 1.5 hours. At   .05 can the manufacturer
claim that his batteries last longer? Use the p-value method.
Recall d.f. is the smaller of n1 - 1 and n2 - 1.
n1  100
n2  80
x1  7.1
x2  6.9
s1  1.3
s2  1.5
H 0 : 1  2
H1 : 1  2
Reject H 0 if p  .05
x x
7.1  6.9
t 1 2 
 .943
s12 s22
1.32 1.52


100 80
n1 n2
p  tcdf (.943,10000, 79) [I do not have a
calculator handy!]
Do not reject H 0
7. A researcher gathered the following data:
n1  27; s1  2.3; x1  28
n2  40; s2  1.8; x2  27.5
Find the 99% confidence interval for 1  2 . At   .01 can you claim that the
population means are different?
For 99% confidence from Table F, t = 2.779
2
x1  x2  t
28  27.5  2.779
2
s12 s22
s12 s22
  1  2  x1  x2  t

2
n1 n2
n1 n2
2.32 1.82
2.32 1.82

 1  2  28  27.5  2.779

27
40
27
40
0.5  1.462  1  2  0.5  1.462
0.962  1  2  1.962
You cannot claim the means are different because 0 is in the confidence interval.
8. A yogurt producer is concerned that the sugar content is too variable in cups of yogurt.
She tests a new stirring machine to see if it reduces the variance in the sugar content in
cups of yogurt. Below is a summary of the data collected:
Her Machine New Machine
n1  41
n2  28
s 7
s22  5
2
1
At   .05 can she claim that  12   22 ? Use the p-value method.
H 0 :  12   22
H1 :  12   22
s12 7
  1.4
s22 5
On the TI83 p = Fcdf(F, 10000, dfN, dfD) for a right-tailed test and p =2* Fcdf(F, 10000,
dfN, dfD) for a two-tailed test.
p = Fcdf(1.4, 10000, 40, 27) = .18064
Do not reject H 0 because p   .
F