252solngr2-01 2/7/01
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Graded Assignment 2
2. A rental firm believes that that the average time that a backhoe is rented is 4.2 days. In order to verify this
statement, a sample of rental records is taken, with the following results.
4, 2, 4, 3, 6, 2, 3, 2, 5, 3, 2, 4, 2, 2
You presumably now know that n  14 , x  3.1429 , s x2  1.6703 or s x  1.2924 . Test the hypothesis that
the mean is 4.2 as follows:
a) Use a test ratio and a 99% confidence level. State your null and alternative hypotheses.
b) Use the test ratio to find an approximate p-value for the null hypothesis. Would you reject the null
hypothesis if   .01 ?   .05 ?   .10 ?
c) Find a critical value or values (for the sample mean) for the null hypothesis and test it at the 99%
confidence level.
3. (Ken Black) It is generally believed that 79% of US companies offer flexible scheduling . A survey of
415 accounting firms finds that 303 have flexible scheduling. Using a test ratio, test the hypothesis that the
proportion of accounting firms that offer flexible hours is below that for other US firms. State your null and
alternative hypotheses and find a p-value for the null hypothesis. If the significance level is 1% would you
reject the null hypothesis?
4. Using a significance level of 1% and a critical value for the observed proportion, repeat 3.
Solution:
2) You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and
stated your conclusion.
a) Test Ratio: n  14 , x  3.1429 , s x2  1.6703 or s x  1.2924 ,   .01 . H 0 :   4.2 , H1 :   4.2 , so
  .005
  4.2 .
t n1  t 13  3.012
0
2

2
.005
s x2
x   0 3.1429  4.2
1.6703

 0.3454 t 

 3.061 . Make a diagram showing an almost
n
14
sx
0.3454
Normal curve with a mean at 0 and shaded 'reject' zones below -3.012 and above 3.012. Since t  3.061 is
below -3.012, we reject H 0 .
sx 
b) To find the p-value, we need (because this is a 2-sided test) pval  2Px  3.1429   2Pt  3.061  .
Make a diagram showing an almost Normal curve with a mean at 0 and the p-value represented by a
shaded area below -3.061 and another shaded area above 3.061. Use the 'df = 13' line of the t table. The
13
13
 3.012 and t .001
 3.852 . In other words
numbers on the t table that are closest to 3.061 are t .005
Pt  3.012   .005 and Pt  3.852   .001 . Since 3.061 is between -3.012 and -3.852, we can say
that .001  Pt  3.061   .005 . If we double this we have .002  pval  .01 . So if   .01,   .05, or
  .10 we would reject the null hypothesis because the p-value is below all of these significance levels.
c) Critical value: We need critical values for x above and below 4.2, so use x cv   0  t n1 s x
2
 4.2  3.012 0.3454   4.2  1.040 or 3.160 to 5.24. Make a diagram showing an almost Normal curve
with a mean at 4.2 and shaded 'reject' zones below 3.16 and above 5.24. Since x  3.1429 is below 3.16, we
reject H 0 .
3) If we interpret the problem as saying that the proportion of firms that offer flexible hours is below 79%,
we have stated an alternative hypothesis because it does not contain an equality. H 0 : p  .79 H 1 : p  .79
The problem also says that x  303 and n  415 , so that p 
x 303

 .7301 and since p 0  .79,
n 415
p0 q0
.79.21

 0.01999 . Since the alternative hypothesis is H 1 : p  .79 , we are worried
n
415

about the proportion being too low, so pval  P p  .7301  . Under the null hypothesis p has the normal
p  p0
distribution with a mean of p 0  .79 and a standard deviation of  p  0.01999 . So z 
, and
p
p 
.7301  .79 

pval  P p  .7301   P z 
  Pz  3.00   .5  .4987  .0013 . The rule on p-value(from
.01999 

the website) says 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.
If   .01 , .0013 is lower and we must reject the null hypothesis.
4) Since the alternative hypothesis is H 1 : p  .79 , we are worried about the proportion being too low, so

we want a critical value for p that is below .79. The critical value formula, pcv  p0  z  p must be
2
modified to give us a number below .79, so pcv  p0  z  p  .79  2.327.01999  .79  .0465  .7435.
Make a diagram with the mean at .79 showing the rejection zone is the area below .7435. Since p  .7301
is below this critical value, reject H 0 .
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