health inspector

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5) A health inspector at a restaurant will enter the kitchen and choose 5 stations to
inspect from a predetermined list of 15 stations present in most restaurant kitchens.
a. How many different sets of 5 stations exist?

5)!
b. If all sets are equally likely, what is the probability of each set?
1
Probability of each set =
3003
c. If the inspector were instead to randomly select 13 stations to inspect, how many
different sets of 13 stations would exist?
15!
C 15,13
(
) =
= 105 different sets
13! 15 − 13
(
)!
d. If all sets were equally likely, what is the probability of each set?
1
Probability of each set =
105
1) When the owner of a small business reviews her list of contracts for 2011, she finds that
35% of the contracts were from clients she met at a large conference at the end of 2010.
Answer the following questions about this situation.
A. Was this measurement obtained by a sample or a census? What words in the
description of the situation make you confident that your answer is correct?
This situation
represents a census of her contracts for 2011. Presumably the
“review” included all
of her 2011 contracts, which would preclude this from
being a sample since all
applicable members of the population were surveyed.
B. Should the owner have taken into account some measure of reliability associated with
the value 35%.
A measure of reliability does not seem warranted here, as nothing is being
“measured”. The percentage of contracts from clients met at the conference is
simply a matter of counting the
number of contracts that meet that criteria, and
then dividing by the total number of 2011 contracts. If she were to repeat the
process, she should get the exact same result each time. This is not the same as
measuring the length or weight of a set of objects, where inaccuracies in the
measuring device, or user error, can lead to variation in the measured values.
8) Since careful records have begun being kept in January, Eric’s small business has
delivered the following quantities of flowers throughout town
January February March April May June July August
Small Bouquets 85
34
26
24 43 29 30 19
Large Bouquets 23
64
27
18 33 23 20 13
Assuming the data is normally distributed construct two separate 90% confidence
intervals one for the number of deliveries of small bouquets in September and one for the
number of large bouquets in September.
Small Bouquets:
xi

Mean:
x =
=
n
290
8
= 36.25
xi − x
(
)
2

Std.Dev
s =
=
3071.5
7
= 20.9472
n − 1
Critical value for a 90% confidence interval, with 8 – 1 = 7 degrees of
freedom is tcrit = 1.895.
The confidence interval limits are then:
x − tα /2
s
n





⎟ < µ < x + tα /2
s
n






36.25 − 1.895
(
)
20.9472
8





⎟ < µ < 36.25 + 1.895
(
)
20.9472
8






22.2157 < µ < 50.2483
In interval notation, and rounded to two decimal places, the confidence
interval is (22.22, 50.25)
Large Bouquets:
xi

Mean:
x =
=
n
221
8
= 27.625
xi − x
(
)
2

Std.Dev
s =
=
1759.875
7
= 15.8559
n − 1
Critical value for a 90% confidence interval, with 8 – 1 = 7 degrees of
freedom is tcrit = 1.895.
The confidence interval limits are then:




x − tα /2
s
n

⎟ < µ < x + tα /2


s
n




27.625 − 1.895
(
)
15.8559
8





⎟ < µ < 27.625 + 1.895
(
)
15.8559
8






17.0018 < µ < 38.2482
In interval notation, and rounded to two decimal places, the confidence
interval is (17.00, 38.25)
7. A quality control analyst measures the number of hours a patient in a low-risk
condition waits for car at the emergency room of a small hospital. The following data are
obtained for 20 patients.
2.26, 2.01, 3.0, 1.22, 1.92, 1.79, 0.78, 1.89, 0.71, 1.58
2.02, 2.77, 2.87, 0.51, 0.74, 1.95, 2.76, 2.61, 3.54, 2.95
a. Compute the sample mean, sample median, and range of data
Sample mean:
xi

x =
=
n
39.88
20
= 1.994
Sample median:
The ordered data set is:
0.51, 0.71, 0.74, 0.78, 1.22, 1.58, 1.79, 1.89, 1.92, 1.95,
2.01, 2.02, 2.26, 2.61, 2.76, 2.77, 2.87, 2.95, 3, 3.54
Since there are an even number of members in the data set, the median is
the average of the two middle values:
Median =
1.95 + 2.01
2
= 1.98
Range:
Range = Maximum value – Minimum value
Range = 3.54 – 0.51
Range = 3.03
b. Compute the sample standard deviation and sample variance
Sample variance:
xi − x
2
(
)
2

s
=
=
14.4511
19
= 0.7606
n − 1
Sample standard deviation:
s = s
2
= 0.7606 = 0.8721
10) A quality control experiment is to be done on a machine that fills tubes with
toothpaste. Its specifications require that it fill tubes with 4.7 oz. A random sample of 40
tubes filled by machine is taken and each tube is weighed. The resulting data are below,
with the weight of the tube already having been subtracted from each. Perform a
hypothesis test at the 90% confidence level to determine if the machine is performing
according to specifications.
4.66, 4.61, 4.71, 4.63, 4.70, 4.62, 4.63, 4.61, 4.70, 4.56
4.60, 4.66, 4.68, 4.57, 4.67, 4.72, 4.67, 4.64, 4.66, 4.75
4.69, 4.64, 4.67, 4.65, 4.69, 4.65, 4.75, 4.53, 4.57, 4.74
4.68, 4.67, 4.66, 4.68, 4.64, 4.65, 4.64, 4.80, 4.71, 4.69
Data:
xi

186.45
Mean:
x =
=
= 4.6613
n
40
xi − x
(
)
2

Stdev:
s =
=
0.1152
39
= 0.0543
n − 1
Hypotheses:
Null:
H 0
: µ = 4.7
Alternative: Ha : µ ≠ 4.7
Critical value:
With a sample size of 40, the z-statistic can be used.
The critical z-values for a two-tailed test with α = 0.10 are -1.282 and 1.282.
The decision rule is reject the null hypothesis if the test statistic is less than –
1.282, or greater than 1.282.
Alternatively, the null hypothesis will be rejected if the p-value is less than 0.10.
Test value:
ztest =
x − µ
s / n
=
4.6613− 4.7
0.0543 / 40
= −4.5076
The corresponding p-value is 0.00000656.
Decision:
Since the test statistic is less than the negative critical value, the decision is to
reject the null hypothesis.
Alternatively, since the p-value is less than the value of α , the decision is to
reject the null hypothesis.
Summary:
There is sufficient evidence at the 0.10 level of significance to support the claim
that the machine is not operating according to the specifications, as the mean
weight of each tube is significantly different from the specified mean of 4.7
ounces.
6) The company policy for customer service representatives gives time off for positive reviews. If, in the
first 20 calls a customer serive agent handles in a day, 13 or more elect to take a subsequent survery and
the rate of service as “excellent”, then the company goves the agent his or her final hour of work that day
off, paid. Ellie receives excellent reviews from about 30% of the calls she handles. Assuming she always
receives at least 20 calls in the first 7 hours of a workday, on what percentage of her 8 hour workdays
does Ellie ger the final hour off?
Mean of the distribution of Ellie’s review scores is 0.3
We estimate the standard error of the distribution of Ellie’s review scores:
sqrt[p * (1-p) / n ]
sqrt[0.3 * 0.7 / 20 ]
sqrt [0.21 / 20]
sqrt 0.0105
= 0.1025
Hypothesized proportion (criteria for the time off) is 13/20 = 0.65
z = (x – mu) / s
z = (0.65 – 0.3) / 0.1025
z = 0.35 / 0.1025
z = 3.4
p = 0.9997
Interpretation: approximately 99.7% of the distribution of ratings Ellie will get is lower than the
hypothesized/target proportion.
The probability of her getting more than 65% excellent ratings out of 20 — and therefore her hour off — is
about 0.03%
4) Researchers are studying a new chemical process for producing dyes. Although they don’t know it, the yield
of the process is normally distributed with u=550kg and O=75kg. The researchers plan to estimate the products
mean yield by running the process 35 times, recording the yield each time. Have they chosen a large enough
sample to be 90% confident that their computed mean will be within 20kg of the actual mean?
z = the critical value at the desired confidence level i.e 90%