252solngr2 2/12/04 (Open this document in 'Page Layout' view!) Name:
Class days and time:
Please include this on what you hand in!
Solution to Graded Assignment 2
1. Which of the following could be null hypotheses? Which could be an alternate hypothesis? Which could be neither? Why?
(i) p

1 .
2 , (ii) p
 
0 .
35 , (iii)
  
0 .
35 , (iv)
 
0 .
35 , (v) x

0 .
35 , (vi)
  
0 .
35 , (vii) s

0 .
35 , (viii)
 
0 .
35 , (ix)
 
1005 .
37 , (x)
 
1005 .
37 .
2. Make sure that I know what formulas you are using.
Before a new long-distance plan, with lower per minute charges is tested, the population mean for long distance charges in the
Hotzeplatz Metropolitan Area is $23.35. The new plan is made available to a randomly selected group of customers and a sample of
25 of their bills is inspected. From the sample we calculate a mean long distance charge of $24.50 and a standard deviation of $3.92.
The new plan will only be made generally available if customers spend more on long distance calls. Should the new plan be adopted?
Use a 10% significance level. a) State your null and alternative hypotheses. b) Find critical values for the sample mean and test the hypothesis. c) Find a confidence interval for the population mean and test the hypothesis. d) Use a test ratio for a test of the mean e) Find an approximate p-value for the test ratio and use the p-value to test the hypothesis.
3. Assume, in problem 2, that the population standard deviation was $3.10. State the test ratio and find its p-value. Using this pvalue, would we reject the null hypothesis a) If the significance level is 1%, b) If the significance level is 5% and c) If the significance level is 10%?
4. (Keller and Warrack.)
Assume that Bush runs against Tree in the next presidential elections. Five minutes after the polls close, the results of an exit poll of
765 voters show that 407 voted for Bush. The Salamander Network will declare Bush the winner of Pennsylvania’s electoral votes if they can conclude that more than half of the voters voted for Bush. Should they do so? a) State your null and alternative hypotheses. b) Find a test ratio for a test of the proportion c) Find a p-value for the test ratio and use the p-value to test the hypothesis at the 5% significance level..
Solution: 1a) Remember the following:
i) Only numbers like

, p ,
 2
,
 and

(the population mean, proportion, variance, standard deviation and median) that are parameters of the population can be in a hypothesis; x , p , s
2
, s and x
.
50
(the sample mean, proportion, variance, standard deviation and median) are statistics computed from sample data and cannot be in a hypothesis because a hypothesis is a statement about a population; ii) The null hypothesis must contain an equality; iii) p must be between zero and one;
(iv) A variance or standard deviation cannot be negative.
So - Which of the following could be null hypotheses? Which could be an alternate hypothesis? Which could be neither? Why? (i) p

1 .
2 can’t be either since the sample proportion is not a parameter and a proportion can’t be above 1. (ii) p
 
0 .
35 can’t be either since a proportion can’t be negative.
(iii)
  
0 .
35 could be H
0
since it contains a parameter and an equality. It could not be H
1
. (iv)
 
0 .
35 could be H
0
since it contains a parameter and an equality. It could not be since the sample mean is not a parameter. (vi)
  
0 .
35 can’t be H
0
or
H
1
. (v) x

0 .
35 can’t be either
H
1
because the population standard deviation can’t be negative. (vii) a parameter (viii)
 
0 .
35 , can’t be H
0 s

0 .
35 can’t be either since the sample standard deviation is not
because there is no equality. It could be H
1
because

is a parameter. The null hypothesis would be
 
0 .
35 .
(ix)
 
1005 .
37 could be H
0
since it contains a parameter (the median) and an equality. It could not be is no equality. It could be H
1
because
H
1
. (x)
 
1005 .
37 can’t be

is a parameter. The null hypothesis would be
H
0
because there
 
1005 .
37 .

252solngr2 2/12/04
2. From the formula table:
Interval for
Mean (

known)
Confidence
Interval
  x
 z

2
 x
Mean (

unknown)
 
DF x

 n t


2
1 s x
Hypotheses
H
0
H
1
:
:





0

0
H
0
H
1
:
:





0

0
Test Ratio t z

 x

 x

0 x
 
0 s x
Critical Value x cv x cv
   z

2
 x
   t

2 s x
While we are engaging in generalities, here are three important points from last year’s solution:
1) You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion.
2) The rule on p-value says that if the p-value is less than the significance level (alpha =

) reject the null hypothesis; if the p-value is greater than or equal to the significance level, do not reject the null hypothesis.
3) Make sure that I know what formulas you are using.
Before a new long-distance plan, with lower per minute charges is tested, the population mean for long distance charges in the Hotzeplatz Metropolitan Area is $23.35. The new plan is made available to a randomly selected group of customers and a sample of 25 of their bills is inspected. From the sample we calculate a mean long distance charge of $24.50 and a standard deviation of $3.92. The new plan will only be made generally available if customers spend more on long distance calls. Should the new plan be adopted? Use a 10% significance level. a) State your null and alternative hypotheses: What the problem says: ‘The new plan will only be made generally available if customers spend more on long distance calls. Should the new plan be adopted?’
-----This definitely implies that "
 
23 .
35 " goes somewhere, but it does not contain an equality, so it must be an alternate hypothesis. The other clue is that the alternate hypothesis is often an ‘action’ hypothesis, while the null hypothesis generally says ‘keep things as they are.’ So the null hypothesis is H
0
:
 
23 .
35 and the alternate hypothesis is H
1
:
 
23 .
35 .
What else the problem says: A random sample of n

25 bills gives a sample mean of 24.50 and a sample standard deviation of 3.92. Use a 10% significance level. So

0

23 .
35 , n

25 , x

24 .
50 , s x

3 .
92 and
 
.
10 .
This implies that s x
24.50 in the hypotheses when you know x
 s
2 x 
3 .
92
2 n 25

24 .
50 is true.

3 .
92
5

0 .
784 .
It makes no sense to put t n




1
 

23

.
25 b) Find critical values for the sample mean and test the hypothesis: Degrees of freedom are
35 t

 
.
10
1


1
24
.
.
We must use
318 .
t because we do not know
, we need a critical value for x above 23.35. Since this means a one-sided test, we use
The two-sided formula , x cv
 
0
 t


 
2

1
. Because the alternate hypothesis is s x
, becomes x cv
 
0
 t
 
 n

1 s x

23 .
35

1 .
318

0 .
784


23 .
35

1 .
033

24 .
383 . Make a diagram showing an almost Normal curve with a mean at 23.35 and a shaded 'reject' zone above 24.383. Since x

24 .
50 is above 24.383, we reject H
0
. c) Find a confidence interval for the population mean and test the hypothesis: Because the alternate hypothesis is
 
23 .
35 we need a '

' confidence interval.
  x
 t

2 s x
becomes
  x
 t  s x

24 .
50

1 .
318

0 .
784

or
 
23 .
467 . Make a diagram showing an almost Normal curve with a mean at x

24 .
50 , and the confidence interval above 23.467 shaded. Since

0

23 .
35 is below 23.467 and thus not in the confidence interval, we reject H
0
.
2

252solngr2 2/12/04 d) Use a test ratio for a test of the mean: The test ratio is t calc
 x
 
0 
24 .
50

23 .
35
0 .
784

1 .
467 . s x
Since this is a one-sided, right-tail test, pick t
 

 t
 
.
10

1
Since 1.467 is in the ‘reject’ zone, reject the null hypothesis.
.
318 .
The ‘reject’ zone is the area above 1.318. e) Find an approximate p-value for the test ratio and use the p-value to test the hypothesis: . On the 24 df line of the t-table, note that 1.467 is between 1.318 and 1.711. Because 1.711 is in the .05 column and 1.318 is in the .10 column, the table is telling us that t
.
10
 t calc
 t
.
05
, so .
05

P
 t

1 .
467


.
10 ,
P
which means
 t

.
05
1 .
711


.
05
 p
 value
and

.
10 .
P
 t

1 .
318


.
10 or
Since these p-values are below the 10% significance level, reject H
0
.
Since we rejected the null hypothesis the new plan should be adopted.
3. Assume, in problem 2, that the population standard deviation was $3.10. State the test ratio and find its pvalue. Using this p-value, would we reject the null hypothesis a) If the significance level is 1%, b) If the significance level is 5% and c) If the significance level is 10%?
Note!!! The only thing that has changed from problem 2 is that the sample standard deviation has been replaced by the population standard deviation of 3.10 – not the sample mean, not the hypotheses. The only reason for this section was because some find e) above difficult, and this is easier since you can use the Normal table.
We still have H
0
:
 
23 .
35 and the alternate hypothesis is H
1
:
 
23 .
35 .
What else the problem says: Assume, in problem 2, that 3.10 was a population standard deviation.
Everything else is unchanged, except that there is no significance level at first. So

0

23 .
35 , n

25 and find x

24 .
50 .
But
 x

3 .
10 and
 x
The table says that the test ratio is z p
 value

P
 x

24 .
50
  z

 2 x 
3 .
10
2

3 .
10

0 .
62 .
We still have a right-tailed test. n 25 5
 x



1 .
855 x

0



.
5
24 .
50

0 .
62

.
4678
23 .
35

.
0322

1 .
855 .
So If we use the Normal table, we
or .5 - .4686 =.0314. Actually, the average of these two, .0318, seems best. To do the last part of this problem, make a diagram of the Normal distribution with a mean of zero and shade the area above 1.855.
P
 z

1 .
855
  z

0
 
0
 z

1 .
855


.
5


.
4678 or .
4682 or .
4686 a) If
 
.
0322 or .
0318 or .
0314
.
01 , since the p-value of about .03 is not below the significance level, we do not reject

.
H
0
. b) If
 
.
05 , since the p-value of about .03 is below the significance level, we reject H
0
. c) If
 
.
10 , since the p-value of about .03 is below the significance level, we reject H
0
.
You did not answer this question if you did the problem 3 different ways and never found the pvalue!
3

252solngr2 2/12/04
4. (Keller and Warrack.) Assume that Bush runs against Tree in the next presidential elections. Five minutes after the polls close, the results of an exit poll of 765 voters show that 407 voted for Bush. The Salamander
Network will declare Bush the winner of Pennsylvania’s electoral votes if they can conclude that more than half of the voters voted for Bush. Should they do so?
From the formula table:
Interval for Confidence
Proportion
Interval p
 p
 z

2 s s p
 p q n q

1
 p p
Hypotheses
H
0
H
1
:
: p
 p
 p
0 p
0
Test Ratio z
 p

 p p
0
Critical Value p cv
 p
0
 z

2
 p
 q
0 p
 p
0

1
 p
0 n q
0 a) State your null and alternative hypotheses: What the problem says: ‘The Salamander Network will declare Bush the winner of Pennsylvania’s electoral votes if they can conclude that more than half of the voters voted for Bush. Should they do so?’ A half is a proportion. There is nothing here about the mean.
There is no reason to multiply n

765 by .5, since 765 is not the size of the population, which is unknown.
The problem asks if the proportion is more than half. Since there is no equality in this question, it must be an alternate hypothesis. So the null hypothesis is H
0
: p

.
5 and the alternate hypothesis is H
1
: p

.
5 .
What else the problem says: A sample of 765 voters found that 407 voted for Bush. So p
0

.
5 , n

765 and x

407 . This implies that p
 x n

407
765

.
532 and q
0

1
 p
0

1

.
5

.
5 .
b) Find a test ratio for a test of the proportion: The table says z
 p

 p p
0
and
 p
 p
0 q
0 n

.
5
 
765

0 .
000326797

0 .
018078 .
This means that z

.
532

.
5

1 .
77 . (It makes no
.
018078 sense to use the test ratio, z
 p

 p p
0
, if you have already said the null hypothesis is about the mean.)

P
 z
 c) Find a p-value for the test ratio and use the p-value to test the hypothesis at the 5% significance level: We use the Normal table to find that the
1 .
77


(.
5

.
4616 )

.
0384 . So, if
 
.
05 , p
 value

P
 p

.
532

since the p-value is below the significance level, we reject H
0
and declare Bush the winner.
Note: You should be able to do this problem using a critical value for p or a confidence interval.
Critical Value: The table says value above p
0

.
5 , so we use p cv p cv

 p
0 p
0

 z

2 z


 p p
, but this is a one-sided test and we want one critical

.
5

1 .
645

.
018078


.
530 . Make a diagram showing an Normal curve centered at p
0

.
5 and a shaded 'reject' zone above .530. Since p

.
532 is above .530, we reject H
0
.
Confidence Interval: The table says p
 p
 z

2 s p
, where, because q

1
 p

1

.
532

.
468 , s p
 p q n

.
532

.
468

765

.
000325459

.
01804 . Since H
1
: p

.
5 implies a 1-sided test we want a 1sided confidence interval in the same direction as the alternate hypothesis. This would be p
 p
 z  s p

.
532

1 .
645

.
018078


.
5011 . Make a diagram showing an Normal curve centered at .532 with the confidence interval above .5011 shaded. Since p
0

.
5 is below .5011 and thus not in the confidence interval, we reject H
0
.
4