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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%