Download chapter 11 review

CHAPTER 11 REVIEW
1. Find the critical value of t required to construct a 99% confidence interval for a population mean
based on a sample of size 15.
ANS:
Looking in the row for df = 15 – 1 = 14, and the column for 99%, we find t* = 2.977.
2. The following data were collected as part of a study. Construct a 90% confidence interval for the
true difference between the means (μ1 – μ2). Does it seem likely that, despite the sample differences,
that there is not real difference between the population means? The samples were SRSs from
independent, approximately normal, populations.
Population
1
2
n
20
18
x-bar
9.87
7.13
s
4.7
4.2
ANS:
The relatively small values of n tell us that we need to use a 2-sample t interval. The conditions
necessary for using this interval are given in the problem: SRSs from independent, approximately
normal, proportions. Using the “conservative” method of choosing the degrees of freedom:
df = min {n1 – 1, n2 – 1} = min {19, 17} = 17  t* = 1.740
4.7 2 4.2 2

 2.74  1.740(1.444)  (0.227, 5.25)
20
18
We are 90% confident that the true difference between the means lies in the interval from 0.227 to
5.25. If the true difference between the means is zero, we would expect to find 0 in the interval.
(9.87  7.13)  1.740
3. The local farmers association in Cass County wants to estimate the mean number of bushels of
corn produced per acre in the county. A random sample of 13 1-acre plots produced the following
results (in number of bushels per acre): 98, 103, 95, 99, 92, 106, 101, 91, 99, 101, 97, 95, 98.
Construct a 95% confidence interval for the mean number of bushels per acre in the entire county.
The local association has been advertising that the mean yield per acre is 100 bushels. Do you think
they are justified in this claim?
ANS:
The population standard deviation is unknown, and the sample size is small (13), so we need to use a
t procedure.
Now, x-bar = 98.1, s = 4.21, df = 13 – 1 = 12  t* = 2.179. The 95% confidence interval is
 4.21 
98.1  2.179
  (95.56, 100.64)
 13 
Because 100 is contained in this interval, we do not have strong evidence against the association’s
claim that the mean number of bushels per acre is 100, even though the sample mean is only 98.1.
4. A study done to determine if male and female 10th graders differ in performance in mathematics
was conducted. Twenty-three randomly selected males and 26 randomly selected females were each
given a 50-question multiple-choice test as part of the study. The scores were approximately
normally distributed. The results of the study were as follows:
Sample size
Mean
St. Deviation
Males
23
40.3
8.3
Females
26
39.2
7.6
Construct a 99% confidence interval for the true difference between the mean score for males and
the mean score for females. Does the interval suggest that there is a difference between the true
means for males and females?
ANS:
The problem states that the samples were randomly selected and that the scores were approximately
normally distributed, so we can construct a two-sample t interval.
df = min {23 – 1, 26 – 1} = 22  t* = 2.819
8.32 7.6 2
(40.3  39.2)  2.819

 (-5.34, 7.54)
23
26
0 is a likely value for the true difference between the means because it is in the interval. Hence, we
do not have evidence that there is a difference between the true means for males and females.
5. A company president believes that there are more absences on Monday than on other days of the
week. The company has 45 workers. The following table gives the number of worker absences on
Mondays and Wednesdays for an 8-week period. Do the data provide evidence that there are more
absences on Mondays?
Monday
Wednesday
Week 1
5
2
Week 2
9
5
Week 3
2
4
Week 4
3
0
Week 5
2
3
Week 6
6
1
Week 7
4
2
Week 8
1
0
ANS:
Because the data are paired on a weekly basis, the data we use for this problem are the difference
between the days of the week for each of the 8 weeks. Adding a row to the table gives the
differences (absences on Monday minus absences on Wednesday), we have:
Monday
Wednesday
Difference
Week 1
5
2
3
Week 2
9
5
4
Week 3
2
4
-2
Week 4
3
0
3
Week 5
2
3
-1
Week 6
6
1
5
Week 7
4
2
2
Week 8
1
0
1
Let μ = the true mean difference between the number of absences on Monday and absences on
Wednesday.
H0: μ = 0 and Ha: μ > 0
We will use a one-sample t test for the difference scores.
1.75  0
x  1.75, s  2.49, t 
 1.99, df  8  1  7  0.025  P  0.05 (from Table C). [Using the
2.49
8
TI-83, P = 0.043)
The P-value is small. This provides us with evidence that there are more absences on Mondays than
on Wednesdays.
6. Twenty-six pairs of identical twins are enrolled in a study to determine the impact of training on
ability to memorize a string of letters. Two programs (A and B) are being studied. One member of
each pair is randomly assigned to one of the two groups and the other twin goes into the other group.
Each group undergoes the appropriate training program, and then the scores for pair of twins is
compared. The means and standard deviations for groups A and B are determined as well as the
mean and standard deviation for the difference between each twins score. Is this study a one-sample
or two-sample situation, and how many degrees of freedom are involved in determining the t value?
ANS:
This is a paired study because the scores for each pair of twins is compared. Hence, it is a onesample situation, and there are 26 pieces of data to be analyzed, which are the 26 differences scores
between the twins. Hence, df = 26 – 1 = 25.
7. An avid reader, Booker Worm, claims that he reads books that average more than 375 pages in
length. A random sample of 13 books on his shelf had the following number of pages: 595, 353,
434, 382, 420, 225, 408, 422, 315, 502, 503, 384, 420. Do the data support Booker’s claim? Test at
the 0.05 level of significance.
ANS:
Let μ = the true average number of pages in the books Booker reads.
H0: μ = 375 and Ha: μ > 375.
We are going to use a one-sample t test at α = 0.05.
n  13, x  412.5, s  91.35, df  13  1  12
412.5  375
 1.48
91.35
13
0.05 < P < 0.10 (from Table C) or P = 0.082 from TI-83.
Because P > 0.05, we cannot reject H0. We do not have strong evidence to back up Booker’s claim
that the books he reads actually average more than 375 pages in length.
t
8. The statistics teacher, Dr. Tukey, gave a 50-point quiz to his class of 10 students and they did not
do very well, at least by Dr. Tukey’s standards. Rather than continuing to the next chapter, he spent
some time reviewing the material and then gave another quiz. The quizzes were comparable in
length and difficulty. The results of the two quizzes were as follows.
Student
Quiz 1
Quiz 2
1
45
42
2
40
38
3
36
34
4
38
37
5
34
36
6
28
26
7
44
44
8
35
32
9
42
38
10
30
31
Do the data indicate that the review was successful, at the 0.05 level, of improving the performance
of the students on this material? Give good statistical evidence for your conclusion.
ANS:
The data are paired by individual students, so we need to test the difference scores for the students
rather than the means for each quiz. The differences are given in the following table.
Student
Quiz 1
Quiz 2
Difference
(Q1 – Q2)
1
45
42
3
2
40
38
2
3
36
34
2
4
38
37
1
5
34
36
-2
6
28
26
2
7
44
44
0
8
35
32
3
9
42
38
4
10
30
31
-1
Let μ = the mean of the differences between the scores of students on Quiz 2 and Quiz 1.
H0: μ = 0 and Ha: μ > 0
This is a matched pairs t test. That is, it is a one-sample t test for a population mean.
n  10, x  1.4, s  1.90, df  10  1  9
1.4  0
 2.33
1.90
10
0.02 < P < 0.025 (from Table C) or P = 0.022 (from TI-83)
Because P < 0.05, we reject the null. The data provide evidence at the 0.05 level that the review was
successful at improving student performance on the material.
t
9. A company uses two different models, call them model A and model B, of a machine to produce
electronic locks for hotels. The company has several hundred of each machine in use in its various
factories. The machines are not perfect, and the company would like to phase out of service the one
that produces the most defects in the locks. A random sample of 13 model A machines and 11
model B machines are tested and the data for the average number of defects per week are given in
the following table:
Model A
Model B
n
13
11
x-bar
11.5
13.1
s
2.3
2.9
Dot plots of the data indicate that there are no outliers or strong skewness in the data and that there
are no strong departures from normal. Do these data provide statistically convincing evidence that
the two machines differ in terms of the number of defects produced?
ANS:
Let μ1 = the true mean number of defects produced by machine A.
Let μ2 = the true mean number of defects produced by machine B.
H0: μ1 – μ2 = 0 and Ha: μ1 – μ2 ≠ 0
We use a two-sample t test for the difference between means. The conditions for this procedure are
given in the problem: both samples are simple random samples from independent, approximately
normal populations.
df = min {13 – 1, 11 – 1} = 10
11.5  13.1
t
 1.48  0.05  P  0.10 (from Table C). However, since it is a two-sided test,
2.3 2 2.9 2

13
11
we have to multiply P-value by 2. Therefore, it is 0.10 < P < 0.20.
The P-value is too large to be strong evidence against the null hypothesis that there is no difference
between the machines. We do not have strong evidence that the types of machines actually differ in
the number of defects produced.
10. You must estimate the number of students who will attend the annual Math Meet at your
university this year. Over the past 8 years, the attendance has been rather consistent, with a mean of
122 and a standard deviation of 12. Estimate the mean number of attendees with 95% confidence.
ANS:
x-bar = 122, s = 12, n = 8, and C = 95%
The sample is a simple random sample. The population is normal and σ is unknown. Therefore, we
can use one-sample t interval.
df = 8 – 1 = 7  t* = 2.365
12
Confidence interval = 122  2.365
 (112, 132)
8
We are 95% confident that the number of students who will attend this year’s annual Math Meet will
be between 112 and 132.
11. Two independent random samples were taken to compare the means of the two populations.
The sample statistics are summarized in the table below. Construct a 99% confidence interval for
the estimation of the difference in the means of the two samples.
Sample
A
B
n
35
50
x-bar
63.5
50.4
s
5.2
9.6
ANS:
x1  63.5, x2  50.4, n1  35, n2  50, C  99%
s1  5.2, s 2  9.6
df = min {35 – 1, 50 – 1} = 34  t* = 2.750
(Since our table C does not have df = 34, we err on the conservative side and take the next lowest
value the table contains. In this case, it would be df = 30.)
5.2 2 9.6 2

 (8.8, 17.4)
35
50
We are 99% confident that the difference between means of these two populations is between 11.495
and 14.705.
Confidence interval = (63.5  50.4)  2.750
12. When 10 cars of a new model were tested for gas mileage, the results showed a mean of 27.2
miles per gallon with a standard deviation of 1.8 miles per gallon. What is a 95% confidence
interval estimate for the gas mileage achieved by this model?
ANS:
The population standard deviation is unknown, and so we use the t-distribution. The standard
1.8
deviation of the sample means is s 
 0.569. With 10 – 1 = 9 degrees of freedom  t* =
10
2.262. Thus, we can be 95% certain that the gas mileage of the new model is in the range of
1.8
27.2  2.262
 (25.9, 28.5) miles per gallon.
10
13. A new process for producing synthetic gems yielded six stones weighing 0.43, 0.52, 0.46, 0.49,
0.60, and 0.56 carats, respectively, in its first run. Find a 90% confidence interval estimate for the
mean carat weight from this process.
ANS:
Using 1-Var Stats, we get x-bar = 0.51 and s = 0.0632.
With df = 6 – 1 = 5  t* = 2.015.
Thus, we can be 90% sure that the new process will yield stones weighing
0.0632
0.51  2.015
 (0.458, 0.562) carats.
6
14. A survey is run to determine the difference in the cost of groceries in suburban stores versus
inner city stores. A preselected group of items is purchased in a sample of 45 suburban and 35 inner
city stores, and the following data are obtained.
Suburban stores
Inner city stores
n1 = 45
n2 = 35
x-bar1 = $36.52
x-bar2 = $39.40
s1 = $1.10
s2 = $1.23
Find a 90% confidence interval estimate for the difference in the cost of groceries.
ANS:
df = min {45 – 1, 35 – 1} = 34  t* = 1.697
(1.23) 2 (1.10) 2

 (2.44, 3.32)
Confidence interval = (39.40  36.52)  1.697
35
45
Thus, we are 90% certain that the selected group of items costs between $2.44 and $3.32 more in
suburban stores that in inner city stores.
15. Two varieties of corn are being compared as to difference in maturation time. Ten plots of the
first variety reach maturity in an average of 95 days with a standard deviation of 5.3 days, while
eight plots of the second variety reach maturity in an average of 74 days with a standard deviation of
4.8 days. Determine a 95% confidence interval estimate for the difference in maturation time.
ANS:
n1 = 10
n2 = 8
x-bar1 = 95 x-bar2 = 74
s1 = 5.3
s2 = 4.8
With df = min {10 – 1, 8 – 1} = 7  t* = 2.365
(5.3) 2 (4.8) 2

 (15.36, 26.64)
10
8
Thus, we can be 95% certain that the first variety of corn will take between 15.36 and 26.64 more
days to read maturity than the second variety.
Confidence interval = (95  74)  2.365
16. In a test for acid rain, an SRS of 49 water samples showed a mean pH level of 4.4 with a
standard deviation of 0.35. Find a 90% confidence interval estimate for the mean pH level.
(a) 4.4 ± 0.01
(b) 4.4 ± 0.08
(c) 4.4 ± 0.32
(d) 4.4 ± 0.35
(e) 4.4 ± 0.58
ANS:
B – Using t-scores: df = 49 – 1 = 48  t* = 1.676
0.35
Confidence interval = 4.4  1.676
 4.4  0.084
49
17. One gallon of gasoline is put in each of 30 test autos, and the resulting mileage figures are
tabulated with x-bar = 28.5 and s = 1.2. Determine a 95% confidence interval estimate of the mean
mileage.
(a) (28.46, 28.54)
(b) (28.42, 28.58)
(c) (28.1, 28.9)
(d) (27.36, 29.64)
(e) (27.3, 29.7)
ANS:
C – Using t-scores: df = 30 – 1 = 29  t* = 2.045
1.2
Confidence interval = 28.5  2.045
 (28.1, 28.9)
30
18. A company owns 335 trucks. For an SRS of 30 of these trucks, the average yearly road tax paid
is $9540 with a standard deviation of $1205. What is a 99% confidence interval estimate for the
total yearly road taxes paid for the 335 trucks?
(a) $9540 ± $103
(b) $9540 ± 567
(c) $3,196,000 ± $567
(d) $3,196,000 ± $35,000
(e) $3,196,000 ± $203,000
ANS:
E – Using t-scores: df = 30 – 1 = 29  t* = 2.576
1205
Confidence interval = 9540  2.756
 9540  606.3
30
335 (9540 ± 606.3) = $3,196,000 ± $203,000
19. In a study aimed at reducing developmental problems in low-birth-weight (under 2500 grams)
babies, 347 infants were exposed to a special educational curriculum while 561 did not receive any
special help. After 3 years the children exposed to the special curriculum showed a mean IQ of 93.5
with a standard deviation of 19.1; the other children had a mean IQ of 84.5 with a standard deviation
of 19.9. Find a 95% confidence interval estimate for the difference in mean IQs of low-birth-weight
babies who receive special intervention and those who do not.
(a) 9.0 ± 2.60
(b) 9.0 ± 4.42
(c) 9.0 ± 6.24
(d) 89.0 ± 19.5
(e) 89.0 ± 39.0
ANS:
A – With df = min {347 – 1, 561 – 1} = 346  t* = 1.984
n1 = 347, x-bar1 = 93.5, s1 = 19.1
n2 = 561, x-bar2 = 84.5, s2 = 19.9
Confidence interval = (93.5  84.5)  1.984
19.12 19.9 2

 9.0  2.60
347
561
20. Does socioeconomic status relate to age at time of HIV infection? For 274 high-income HIVpositive individuals the average of infection was 33.0 years with a standard deviation of 6.3, while
for 90 low-income individuals the average was 28.6 years with a standard deviation of 6.3. Find a
90% confidence interval estimate for the difference in ages of high- and low-income people at the
time of HIV infection.
(a) 4.4 ± 0.963
(b) 4.4 ± 1.26
(c) 4.4 ± 2.51
(d) 30.8 ± 2.51
(e) 30.8 ± 6.3
ANS:
B – With df = min {274 – 1, 90 – 1} = 89  t* = 1.664
n1 = 274, x-bar1 = 33, s1 = 6.3
n2 = 90, x-bar2 = 28.6, s2 = 6.3
Confidence interval = (33  28.6)  1.664
6.3 2 6.3 2

 4.4 ± 1.26
274
90
21. Acute renal graft rejection can occur years after the graft. In one study, 21 patients showed such
late acute rejection when the ages of their grafts (in years) were 9, 2, 7, 1, 4, 7, 9, 6, 2, 3, 7, 6, 2, 3, 1,
2, 3, 1, 1, 2, and 7, respectively. Establish a 90% confidence interval estimate for the ages of renal
grafts that undergo late acute rejection.
(a) 2.024 ± 0.799
(b) 2.024 ± 1.725
(c) 4.048 ± 0.799
(d) 4.048 + 1.041
(e) 4.048 ± 1.725
ANS:
D – x-bar = 4.048, s = 2.765, df = 21 – 1 = 20  t* = 1.725
2.765
Confidence interval = 4.048  1.725
 4.048 ± 1.041
21
22. In a sample of ten basketball players the mean income was $196,000 with a standard deviation
of $315,000.
a) Assuming all necessary assumptions are met, find a 95% confidence interval estimate of the
mean salary of basketball players.
b) What assumptions are necessary for the above estimate? Do they seem reasonable here?
ANS:
a) With df = 10 – 1 = 9  t* = 2.262
315000
Confidence interval = 196000  2.262
 196000 ± 225000
10
b) We muse assume the sample is an SRS and that basketball salaries are normally distributed. This
does not seem reasonable—the salaries are probably strongly skewed to the right by a few high ones.
23. A cigarette industry spokesperson remarks that current levels of tar are no more than 5
milligrams per cigarette. A reported does a quick check on 15 cigarettes representing a cross section
of the market.
a) What conclusion is reached if the sample mean is 5.63 milligrams of tar with a standard deviation
of 1.61? Assume a 10% significance level.
b) What is the conclusion at the 5% significance level?
c) What is the P-value?
ANS:
a) We have H0: μ < 5 and Ha: μ > 5, α = 0.10, and with df = 15 – 1 = 14  t* = 1.345.
5.63  5
 1.52
T-statistic =
1.61
15
Since 1.52 > 1.345, the industry spokesperson’s remarks should be rejected at the 10% significance
level.
b) α = 0.05, df = 14, t* = 1.761
Since 1.52 < 1.761, the remarks cannot be rejected at the 5% significance level.
c) Since 1.52 is between 1.345 and 1.761, we can conclude that P-value is between 0.05 and 0.10.
[On the TI-83, P = 0.0759]
24. A local chamber of commerce claims that the mean sale price for homes in the city is $90,000.
A real estate salesperson notes eight recent sales of $75,000, $102,000, $80,000, $85,000, $79,000,
$95,000, $98,000, and $62,000. How strong is the evidence to reject the chamber of commerce
claim?
ANS:
H0: μ = 90000 and Ha: μ ≠ 90000
x-bar = 84,500 and s = 13,341.70.
84500  90000
 1.17
t-statistic =
13341.7
8
With df = 8 – 1 = 7, we note that 1.17 is between 1.119 and 1.415, corresponding to 0.15 and 0.10,
respectively. Doubling these values because this is a two-sided test, we see that the P-value is
between 0.20 and 0.30. With such a high P, we conclude that there is no evidence to reject the
chamber of commerce claim. [The TI-83 gives P = 0.2818]
25. A store manager wishes to determine whether there is a significant difference between two
trucking firms with regard to the handling of egg cartons. In a simple random sample of 200 cartons
on one firm’s truck there was an average of 0.7 broken eggs per carton with a standard deviation of
0.31, while a sample of 300 cartons on the second firm’s truck showed an average of 0.775 broken
eggs per carton with a standard deviation of 0.42. Is the difference between the averages significant
at a significance level of 5%? At a significance level of 1%?
ANS:
H0: μ1 = μ2 and Ha: μ1 ≠ μ2
(where μ1 = mean number of broken eggs per carton for the first firm and μ2 = mean number of
broken eggs per carton for the second firm)
0.7  0.775
 2.29
t-statistic =
0.312 0.42 2

200
300
Using Table C, with df = min {200 – 1, 300 – 1} = 199, we find that -2.29 is between 2.081 and
2.364, corresponding to 0.01 and 0.02. The test is two-sided, and so the P-value is between 0.02 and
0.04. So the observed difference is statistically significant at the 5% level but not at the 1% level.
[On the TI-83, P = 0.0222.]
26. An automotive company executive claims that a mean of 48.3 cars per dealership are being sold
each month. A major stockholder believes this claim is high and runs a test by sampling 30
dealerships. What conclusion is reached if the sample mean is 45.4 cars with a standard deviation of
15.4?
(a) There is sufficient evidence to prove the executive’s claim is true.
(b) There is sufficient evidence to prove the executive’s claim is false.
(c) The stockholder has sufficient evidence to reject the executive’s claim.
(d) The stockholder does not have sufficient evidence to reject the executive’s claim.
(e) There is not sufficient data to reach any conclusion.
ANS:
D
We have H0: μ = 48.3 and Ha: μ < 48.3.
45.4  48.3
 1.03
t-statistic =
15.4
30
Using Table C, with df = 30 – 1 = 29, we find that 1.03 < 1.055, corresponding to 0.15. [On the T83, P = 0.1554.] With such a large P-value, there is little evidence against H0 and the stockholder
should not reject the executive’s claim.
27. A pharmaceutical company claims that a medication will produce a desired effect for a mean
time of 58.4 minutes. A government researcher runs a hypothesis test of 250 patients and calculates
a mean of x-bar = 59.5 with a standard deviation of s = 8.3. In which of the following intervals is
the P-value located?
(a) P < 0.01
(b) 0.01 < P < 0.02
(c) 0.02 < P < 0.05
(d) 0.05 < P < 0.10
(e) P > 0.10
ANS:
C
We have H0: μ = 58.4 and Ha: μ ≠ 58.4
59.5  58.4
 2.10
t-statistic =
8.3
250
Using Table C, with df = 250 – 1 = 249, we find the corresponding probability to be between 0.01
and 0.02. Doubling this value because the test is two-sided results in a P-value between 0.02 and
0.04. [On the TI-83, P = 0.0371.]
28. In a one-sided hypothesis test for the mean (>), in a random sample of size 10 the t-score of the
sample mean is 2.79. Is this significant at the 5% level? At the 1% level?
(a) Significant at the 1% level but not at the 5% level.
(b) Significant at the 5% level but not at the 1% level.
(c) Significant at both the 1% and 5% levels.
(d) Significant at neither the 1% nor 5% level.
(e) Cannot be determined from the given information.
ANS:
B
With df = 10 – 1 = 9, the critical t-scores for the 0.05 and 0.01 tail probabilities are 1.833 and 2.821,
respectively. We have 2.79 > 1.833, but 2.79 < 2.821, and so 2.79 is significant at the 5% level but
not at the 1% level.
29. A recent study of health service costs for coronary angioplasty versus coronary artery bypass
surgery at a London hospital showed an average cost of £6176 with a standard deviation of £329 for
231 angioplasties and an average cost of £8164 with a standard deviation of £264 for 221 bypass
surgeries. Is this sufficient evidence to say that the average cost of angioplasty is less than the
average cost of bypass surgery?
(a) P < 0.001, so this is a very strong evidence that angioplasty costs less.
(b) P is between 0.001 and 0.01, so this is strong evidence that angioplasty costs less.
(c) P is between 0.01 and 0.05, so this is moderate evidence that angioplasty costs less.
(d) P is between 0.05 and 0.10, so there is some evidence that angioplasty costs less.
(e) P > 0.10, so there is little evidence that angioplasty costs less.
ANS:
A
6176  8164
 71, and the P-value is 0.0000. This is very strong evidence that
t-statistic =
2
2
329
264

231
221
angioplasty costs less than bypass surgery.
30. The 26 contestants in the pentathlon event of the 1992 Olympics ran the 200-meter dash in the
following times (in seconds): 25.44, 24.39, 25.66, 23.93, 23.34, 25.01, 24.27, 24.54, 24.86, 23.95,
23.31, 24.60, 23.12, 25.29, 24.40, 25.24, 24.43, 23.70, 25.20, 25.09, 24.48, 26.13, 25.28, 24.18, and
23.83. In testing the null hypothesis that the mean time required for pentathlon contestants to run
this race is 25 seconds against the alternative hypothesis that the mean time required is less than 25
seconds, in which of the following intervals is the P-value located?
(a) P < 0.005
(b) 0.005 < P < 0.01
(c) 0.01 < P < 0.05
(d) 0.05 < P < 0.10
(e) 0.10 < P
ANS:
B
We have H0: μ = 25 and Ha: μ < 25
x-bar = 24.581 and s = 0.7793
24.581  25
 2.74
t-statistic =
0.7793
26
In looking at the df = 26 – 1 = 25 row in Table C, note that 2.74 is between 2.485 and 2.787, so P is
between 0.005 and 0.01. [On the TI-83, P = 0.0056.]