MATH 214 (NOTES)math.unm.edu/~alvaro/math214-p3.pdf · MATH 214 (NOTES) – p. 23/62. Z test for a...

MATH 214 (NOTES)Math 214

Al Nosedal

Department of Mathematics

Indiana University of Pennsylvania

MATH 214 (NOTES) – p. 1/62

"Pepsi" problem

A market research consultant hired by the Pepsi-Cola Co. isinterested in determining the proportion of IUP students whofavor Pepsi-Cola over Coke Classic. A random sample of100 students shows that 40 students favor Pepsi over Coke.Use this information to construct a 95% confidence intervalfor the proportion of all students in this market who preferPepsi.

MATH 214 (NOTES) – p. 2/62

Bernoulli Distribution

xi =

{

1 i-th person prefers Pepsi0 i-th person prefers Coke

µ = E(xi) = p

σ2 = V (xi) = p(1 − p)

Let p̂ be our estimate of p. Note that p̂ =∑n

i=1 xi

n = x̄. If n is"large", by the Central Limit Theorem, we know that:x̄ is roughly N(µ, σ√

n), that is,

p̂ is roughly N

(

p,

√

p(1−p)n

)

MATH 214 (NOTES) – p. 3/62

Interval Estimate of p

Draw a simple random sample of size n from a populationwith unknown proportion p of successes. An (approximate)confidence interval for p is:

p̂ ± z∗

(√

p̂(1 − p̂)

n

)

where z∗ is a number coming from the Standard Normal thatdepends on the confidence level required.Use this interval only when:1) n is "large" and2) np̂ ≥ 5 and n(1 − p̂) > 5.

MATH 214 (NOTES) – p. 4/62

Problem 31 (page 322)

A simple random sample of 400 individuals provides 100Yes responses.a. What is the point estimate of the proportion of thepopulation that would provide Yes responses?b. What is the point estimate of the standard error of theproportion, σp̂?c. Compute the 95% confidence interval for the populationproportion.

MATH 214 (NOTES) – p. 5/62

Solution

a. p̂ = 100400 = 0.25

b. Standard error of p̂ =√

(p̂)(1−p̂)n =

√

(0.25)(0.75)400 = 0.0216

c. p̂ ± z∗

(

√

(p̂)(1−p̂)n

)

0.25 ± 1.96(0.0216)

(0.2076, 0.2923)

MATH 214 (NOTES) – p. 6/62


A simple random sample of 800 elements generates asample proportion p̂ = 0.70.a. Provide a 90% confidence interval for the populationproportion.b. Provide a 95% confidence interval for the populationproportion.

MATH 214 (NOTES) – p. 7/62

Solution

a. p̂ ± z∗

(

√

(p̂)(1−p̂)n

)

0.70 ± 1.65

(

√

(0.70)(1−0.70)800

)

0.70 ± 1.65(0.0162)

(0.6732, 0.7267)

b. 0.70 ± 1.96(0.0162)

(0.6682, 0.7317)

MATH 214 (NOTES) – p. 8/62


A survey of 611 office workers investigated telephoneanswering practices, including how often each office workerwas able to answer incoming telephone calls and how oftenincoming telephone calls went directly to voice mail. A totalof 281 office workers indicated that they never need voicemail and are able to take every telephone call.a. What is the point estimate of the proportion of thepopulation of office workers who are able to take everytelephone call?b. At 90% confidence, what is the margin of error?c. What is the 90% confidence interval for the proportion ofthe population of office workers who are able totake every telephone call?

MATH 214 (NOTES) – p. 9/62

Solution

a. p̂ = 281611 = 0.46

b. Margin of error =

z∗

√

(p̂)(1−p̂)n = 1.65

√

(0.46)(0.54)611 = 1.65(0.0201) = 0.0332

c. p̂ ± z∗

(

√

(p̂)(1−p̂)n

)

0.46 ± .0332

(0.4268, 0.4932)

MATH 214 (NOTES) – p. 10/62


In a survey, the planning value for the population proportionis p∗ = 0.35. How large a sample should be taken to providea 95% confidence interval with a margin of error of 0.05?

Solution.n =

(

z∗E

)2p∗(1 − p∗) =

(

1.960.05

)2(0.35)(1 − 0.35) = 350

(Always round up).

MATH 214 (NOTES) – p. 11/62

Determining the Sample Size

Sample Size for an Interval Estimate of a PopulationProportion.

n =(z∗

E

)2p∗(1 − p∗)

In practice, the planning value p∗ can be chosen by one ofthe following procedures.1. Use the sample proportion from a previous sample of thesame or similar units.2. Use a planning value of p∗ = 0.5.

MATH 214 (NOTES) – p. 12/62


At 95% confidence, how large a sample should be taken toobtain a margin of error of 0.03 for the estimation of apopulation proportion? Assume that past data are notavailable for developing a planning value for p∗.Solution.n =

(

z∗E

)2p∗(1 − p∗) =

(

1.960.03

)2(0.5)(1 − 0.5) = 1068

(Always round up).

MATH 214 (NOTES) – p. 13/62


The percentage of people not covered by health careinsurance in 2003 was 15.6%. A congressional committeehas been charged with conducting a sample survey toobtain more current information.a. What sample size would you recommend if thecommittee’s goal is to estimate the current proportion ofindividuals without health care insurance with a margin oferror of 0.03? Use a 95% confidence level.b. Repeat part a) using a 99% confidence level.

MATH 214 (NOTES) – p. 14/62

Solution

a. n =(

z∗E

)2p∗(1 − p∗) =

(

1.960.03

)2(0.156)(1 − 0.156) = 563

b. n =(

z∗E

)2p∗(1 − p∗) =

(

2.580.03

)2(0.156)(1 − 0.156) = 974

MATH 214 (NOTES) – p. 15/62

CHAPTER 9HYPOTHESIS TESTS

MATH 214 (NOTES) – p. 16/62

Do you want to become a millionaire?

Let’s say that one of you is invited to this popular show. Asyou probably know, you have to answer a series of multiplechoice questions and there are four possible answers toeach question. Perhaps you also have seen that if you don’tknow the answer to a question you could either "jump thequestion" or you could "ask the audience".Suppose that you run into a question for which you don’tknow the answer with certainty and you decide to "ask theaudience". Let’s say that you initially believe that the rightanswer is A. Then you ask the audience and only 2% of theaudience shares your opinion. What would you do? Changeyour initial belief or reject it?

MATH 214 (NOTES) – p. 17/62

Sweetening colas

Diet colas use artificial sweeteners to avoid sugar. Thesesweeteners gradually lose their sweetness over time.Manufacturers therefore test new colas for loss of sweetnessbefore marketing them. Trained tasters sip the cola alongwith drinks of standard sweetness and score the cola on a"sweetness score " of 1 to 10. The cola is then stored for amonth at high temperature to imitate the effect of fourmonths’ storage at room temperature. Each taster scoresthe cola again after storage. This is a matched pairsexperiment. Our data are the differences (score beforestorage minus score after storage) in the tasters’ scores.The bigger these differences, the biggerthe loss of sweetness.

MATH 214 (NOTES) – p. 18/62

Sweetening colas (cont.)

Suppose we know that for any cola, the sweetness lossscores vary from taster to taster according to a Normaldistribution with standard deviation σ = 1. The mean µ for alltasters measures loss of sweetness, and is different fordifferent colas.The following are the sweetness losses for a new cola, asmeasured by 10 trained tasters: 2.0 0.4 0.7 2.0 -0.4 2.2 -1.31.2 1.1 2.3.Are these data good evidence that the cola lost sweetnessin storage?

MATH 214 (NOTES) – p. 19/62

Solution

µ= mean sweetness loss for the population of all tasters.1. State hypotheses. H0 : µ = 0 vs Ha : µ > 0

2. Test statistic. z∗ = x̄−µ0

σ/√

n= 1.02−0

1/√

10= 3.23

3. P-value. P (Z > z∗) = P (Z > 3.23) = 0.0006

4. Conclusion. We would very rarely observe a samplesweetness loss as large as 1.02 if H0 were true. The smallP-value provides strong evidence against H0 and in favor ofthe alternative Ha : µ > 0, i.e., it gives good evidence that themean sweetness loss is not 0, but positive.

MATH 214 (NOTES) – p. 20/62

Executives’ blood pressures

The National Center for Health Statistics reports that thesystolic blood pressure for males 35 to 44 years of age hasmean 128 and standard deviation 15. The medical directorof a large company looks at the medical records of 72executives in this age group and finds that the mean systolicblood pressure in this sample is x̄ = 126.07. Is this evidencethat the company’s executives have a different mean bloodpressure from the general population?Suppose we know that executives’ blood pressures follow aNormal distribution with standard deviation σ = 15.

MATH 214 (NOTES) – p. 21/62

Solution

µ= mean of the executive population.1. State hypotheses. H0 : µ = 128 vs Ha : µ 6= 128

2. Test statistic. z∗ = x̄−µ0

σ/√

n= 126.07−128

15/√

72= −1.09

3. P-value. 2P (Z > |z∗|) = 2P (Z > | − 1.09|) = 2P (Z > 1.09) =

2(1 − 0.8621) = 0.2758

4. Conclusion. More than 27% of the time, a simple randomsample of size 72 from the general male population wouldhave a mean blood pressure at least as far from 128 as thatof the executive sample. The observed x̄ = 126.07 istherefore not good evidence that executives differ from othermen.

MATH 214 (NOTES) – p. 22/62

Tests for a population mean

There are four steps in carrying out a significance test:1. State the hypotheses.2. Calculate the test statistic.3. Find the P-value.4. State your conclusion in the context of your specificsetting.Once you have stated your hypotheses and identified theproper test, you or your calculator can do Steps 2 and 3 byfollowing a recipe. Here is the recipe for the test we haveused in our examples.

MATH 214 (NOTES) – p. 23/62

Z test for a population meanµ

Draw a simple random sample of size n from a Normalpopulation that has unknown mean µ and known standarddeviation σ. To test the null hypothesis that µ has a specifiedvalue, H0 : µ = µ0

calculate the one-sample z statistic

z∗ =x̄ − µ0

σ√n

In terms of a variable Z having the standard Normaldistribution, the P-value for a test of H0 againstHa : µ > µ0 is P (Z > z∗)

Ha : µ < µ0 is P (Z < z∗)

Ha : µ 6= µ0 is 2P (Z > |z∗|)MATH 214 (NOTES) – p. 24/62


Consider the following hypothesis test:H0 : µ = 20

Ha : µ < 20

A sample of 50 provided a sample mean of 19.4. Thepopulation standard deviation is 2.a. Compute the value of the test statistic.b. What is the p-value?c. Using α = 0.05, what is your conclusion?

MATH 214 (NOTES) – p. 25/62

Solution

a. Test statistic.z∗ = x̄−µ0

σ/√

n= 19.4−20

2/√

50= −2.1213

b. P-value.P (Z < z∗) = P (Z < −2.1213) = 0.0169

c. Conclusion.Since P-value = 0.0169 < α = 0.05, we reject H0 : µ = 20. Weconclude that µ < 20.

MATH 214 (NOTES) – p. 26/62



Ha : µ > 25


MATH 214 (NOTES) – p. 27/62

Solution


σ/√

n= 26.4−25

6/√

40= 1.4757

b. P-value.P (Z > z∗) = P (Z > 1.4757) = 0.0700

c. Conclusion.Since P-value = 0.0700 > α = 0.01, we CAN’T rejectH0 : µ = 25. We conclude that we don’t have enoughevidence to claim that µ > 25. (Some of us would say that weaccept that µ = 25).

MATH 214 (NOTES) – p. 28/62



Ha : µ 6= 15


MATH 214 (NOTES) – p. 29/62

Problem 11 (Solution)


σ/√

n= 14.15−15

3/√

50= −2.0034

b. P-value.2P (Z > |z∗|) = 2P (Z > | − 2.0034|) = 2P (Z > 2.0034) = 0.0451

c. Conclusion.Since P-value = 0.0451 < α = 0.05, we reject H0 : µ = 15. Weconclude that µ 6= 15.

MATH 214 (NOTES) – p. 30/62

Tests from confidence intervals

CONFIDENCE INTERVALS AND TWO-SIDED TESTS.

A level α two-sided significance test rejects a hypothesisH0 : µ = µ0 exactly when the value µ0 falls outside a level1 − α confidence interval for µ.

MATH 214 (NOTES) – p. 31/62

Problem 11 (again)

The 95% confidence interval for µ in problem 11 is:

x̄ ± z∗(σ√n

)

14.15 ± 1.96(3√50

)

(13.3184, 14.9815)

The hypothesized value µ0 = 15 in problem 11 falls outsidethis confidence interval, so we reject H0 : µ = 15.

MATH 214 (NOTES) – p. 32/62


H0 : µ = 12

Ha : µ > 12

A sample of 25 provided a sample mean x̄ = 14 and asample standard deviation s = 4.32.a. Compute the value of the test statistic.b. Use the t distribution table to compute a range for thep-value.c. At α = 0.05, what is your conclusion?

MATH 214 (NOTES) – p. 33/62

Solution

a. t∗ = x̄−µ0

s/√

n= 14−12

4.32/√

25= 2.31

b. Degrees of freedom = n − 1 = 24.P-value = P (T > t∗) = P (T > 2.31)

Using t-table (on page 920), P-value is between 0.01 and0.025.Exact P-value = 0.0147 (using a TI-83 and the functionT-Test).c. Since P-value < α = 0.05, we reject H0.

MATH 214 (NOTES) – p. 34/62


H0 : µ = 18

Ha : µ 6= 18

A sample of 48 provided a sample mean x̄ = 17 and asample standard deviation s = 4.5.a. Compute the value of the test statistic.b. Use the t distribution table to compute a range for thep-value.c. At α = 0.05, what is your conclusion?

MATH 214 (NOTES) – p. 35/62

Solution

a. t∗ = x̄−µ0

s/√

n= 17−18

4.5/√

48= −1.54

b. Degrees of freedom = n − 1 = 47.P-value = 2P (T > |t∗|) = 2P (T > | − 1.54|) = 2P (T > 1.54)

Using t-table (on page 920), P-value is between 0.10 and0.20.Exact P-value = 0.1303 (using a TI-83 and the functionT-Test).c. Since P-value > α = 0.05, we CAN’T reject H0.

MATH 214 (NOTES) – p. 36/62

Another approach

Recall that α = 0.05 and that we have a two-sided alternative(Ha : µ 6= 18) which means that we could solve this problemusing a 95% confidence interval instead of the four-stepprocedure presented above.x̂ ± t∗(

s√n)

17 ± 2.012( 4.5√48

)

(15.6931, 18.3068)

The hypothesized value µ0 = 18 in problem 18 falls INSIDEthis confidence interval, so we CAN’T reject H0 : µ = 18.

MATH 214 (NOTES) – p. 37/62


H0 : µ = 45

Ha : µ < 45

A sample of 36 is used. Identify the p-value and state yourconclusion for each of the following sample results. Useα = 0.01.a. x̄ = 44 and s = 5.2.b. x̄ = 43 and s = 4.6.c. x̄ = 46 and s = 5.0.

MATH 214 (NOTES) – p. 38/62

Solution a

a. x̄ = 44 and s = 5.2.P-value = P (T < t∗) = P (T < −1.1538) = P (T > 1.1538) (Tdistributions are symmetric)Using t-table (on page 921), P-value is between 0.10 and0.20.Exact P-value = 0.1281 (using a TI-83 and the functionT-Test).Since P-value > α = 0.01, we CAN’T reject H0.

MATH 214 (NOTES) – p. 39/62

Solution b

b. x̄ = 43 and s = 4.6.P-value = P (T < t∗) = P (T < −2.6086) = P (T > 2.6086) (Tdistributions are symmetric)Using t-table (on page 921), P-value is between 0.005 and0.01.Exact P-value = 0.0066 (using a TI-83 and the functionT-Test).Since P-value < α = 0.01, we reject H0. We conclude thatµ < 45.

MATH 214 (NOTES) – p. 40/62

Solution c

c. x̄ = 46 and s = 5.0.P-value = P (T < t∗) = P (T < 1.2) = 1 − P (T > 1.2) (Tdistributions are symmetric)Using t-table (on page 921), P-value is between 0.80 and0.90.Exact P-value = 0.8809 (using a TI-83 and the functionT-Test).Since P-value > α = 0.01, we can’t reject H0.

MATH 214 (NOTES) – p. 41/62

Problem 27

The Employment and Training Administration reported theU.S. mean unemployment insurance benefit of $ 238 perweek. A researcher in the state of Virginia anticipated thatsample data would show evidence that the mean weeklyunemployment insurance benefit in Virginia was below thenational level.a. Develop appropriate hypotheses such that rejection of H0

will support the researcher’s contention.b. For a sample of 100 individuals, the sample mean weeklyunemployment insurance benefit was $231 with a samplestandard deviation of $80. What is the p-value?c. At α = 0.05, what is your conclusion?.

MATH 214 (NOTES) – p. 42/62

Solution

a. H0 : µ = 238 vs Ha : µ < 238.b. t∗ = x̄−µ0

s/√

n= 231−238

80/√

100= −0.88

Degrees of freedom = n − 1 = 99.Using t table (page 922), P-value is between 0.10 and 0.20

c. P-value > 0.05, we CAN’T reject H0. Cannot concludemean weekly benefit in Virginia is less than the nationalmean.

MATH 214 (NOTES) – p. 43/62


The National Association of Professional Baseball Leagues,Inc., reported that attendance for 176 minor league baseballteams reached an all-time high during the 2001 season. Ona per-game basis, the mean attendance for minor leaguebaseball was 3530 people per game. Midway through the2002 season, the president of the association asked for anattendance report that would hopefully show that the meanattendance for 2002 was exceeding the 2001 level.

MATH 214 (NOTES) – p. 44/62

Problem 28 (cont.)

a. Formulate hypotheses that could be used to determinewhether the mean attendance per game in 2002 was greaterthan the previous year’s level.b. Assume that a sample of 92 minor league baseballgames played during the first half of the 2002 seasonshowed a mean attendance of 3740 people per game with asample standard deviation of 810. What is the p-value?c. At α = 0.01, what is your conclusion?

MATH 214 (NOTES) – p. 45/62

Solution

a.H0 : µ = 3530 vs Ha : µ > 3530

b. t∗ = x̄−µ0

s/√

n= 3740−3530

810/√

92= 2.49

Degrees of freedom = n − 1 = 91. Using t-table (on page922), P-value is between 0.005 and 0.01.Exact P-value = 0.007 (using a TI-83 and the function T-Test).c. Since P-value < α = 0.01, we reject H0. We conclude thatthe mean attendance per game has increased.

MATH 214 (NOTES) – p. 46/62

Hypotheses Tests for a Proportion

To test the hypothesis H0 : p = p0, compute the z∗ statistic,z∗ = p̂−p0

√

p0(1−p0)

n

In terms of a variable Z having the standard Normaldistribution, the approximate P-value for a test of H0 againstHa : p > p0 : is : P (Z > z∗)

Ha : p < p0 : is : P (Z < z∗)

Ha : p 6= p0 : is : 2P (Z > |z∗|)Use this test when the sample size n is so large that bothnp0 and n(1 − p0) are 10 or more.

MATH 214 (NOTES) – p. 47/62


Consider the following hypothesis test:H0 : p = 0.75

Ha : p < 0.75

A sample of 300 items was selected. Compute the p-valueand state your conclusion for each of the following sampleresults. Use α = 0.05.a. p̂ = 0.68

b. p̂ = 0.72

c. p̂ = 0.70

d. p̂ = 0.77

MATH 214 (NOTES) – p. 48/62

Solution a

z∗ = p̂−p0√p0(1−p0)/n

= 0.68−0.75√0.75(1−0.75)/300

= −2.80

Using Normal table, P-value =P (Z < z∗) = P (Z < −2.80) = 0.0026

P-value<α = 0.05, reject H0.

MATH 214 (NOTES) – p. 49/62

Solution b

z∗ = p̂−p0√p0(1−p0)/n

= 0.72−0.75√0.75(1−0.75)/300

= −1.20


P-value>α = 0.05, do not reject H0.

MATH 214 (NOTES) – p. 50/62

Solution c

z∗ = p̂−p0√p0(1−p0)/n

= 0.70−0.75√0.75(1−0.75)/300

= −2.00


P-value<α = 0.05, reject H0.

MATH 214 (NOTES) – p. 51/62

Solution d

z∗ = p̂−p0√p0(1−p0)/n

= 0.77−0.75√0.75(1−0.75)/300

= 0.80

Using Normal table, P-value =P (Z < z∗) = P (Z < 0.80) = 0.7881

P-value>α = 0.05, do not reject H0.

MATH 214 (NOTES) – p. 52/62


Consider the following hypothesis test:H0 : p = 0.20

Ha : p 6= 0.20

A sample of 400 provided a sample proportion p̂ = 0.175.a. Compute the value of the test statistic.b. What is the p-value?c. At the α = 0.05, what is your conclusion?d. What is the rejection rule using the critical value? What isyour conclusion?

MATH 214 (NOTES) – p. 53/62

Solution

a. z∗ = p̂−p0√

p0(1−p0)

n

= 0.175−0.20√

(0.20)(0.80)400

= −1.25

b. Using Normal table, P-value = 2P (Z > |z∗|) = 2P (Z >

| − 1.25|) = 2P (Z > 1.25) = 2(0.1056) = 0.2112

c. P-value > α = 0.05, we CAN’T reject H0.

MATH 214 (NOTES) – p. 54/62


A study found that, in 2005, 12.5% of U.S. workers belongedto unions. Suppose a sample of 400 U.S. workers iscollected in 2006 to determine whether union efforts toorganize have increased union membership.a. Formulate the hypotheses that can be used to determinewhether union membership increased in 2006.b. If the sample results show that 52 of the workersbelonged to unions, what is the p-value for your hypothesistest?c. At α = 0.05, what is your conclusion?

MATH 214 (NOTES) – p. 55/62

Solution

a. H0 : p = 0.125 vs Ha : p > 0.125

b. p̂ = 52400 = 0.13

z∗ = p̂−p0√

p0(1−p0)

n

= 0.13−0.125√

(0.125)(0.875)400

= 0.30

Using Normal table, P-value =P (Z > z∗) = P (Z > 0.30) = 1 − 0.6179 = 0.3821

c. P-value = > 0.05, do not reject H0. We cannot concludethat there has been an increase in union membership.

MATH 214 (NOTES) – p. 56/62


A study by Consumer Reports showed that 64% ofsupermarket shoppers believe supermarket brands to be asgood as national name brands. To investigate whether thisresult applies to its own product, the manufacturer of anational name-brand ketchup asked a sample of shopperswhether they believed that supermarket ketchup was asgood as the national brand ketchup.

MATH 214 (NOTES) – p. 57/62

Problem 38 (cont.)

a. Formulate the hypotheses that could be used todetermine whether the percentage of supermarket shopperswho believe that the supermarket ketchup was as good asthe national brand ketchup differed from 64%.b. If a sample of 100 shoppers showed 52 stating that thesupermarket brand was as good as the national brand, whatis the p-value?c. At α = 0.05, what is your conclusion?

MATH 214 (NOTES) – p. 58/62

Solution

a. H0 : p = 0.64 vs Ha : p 6= 0.64

b. p̂ = 52100 = 0.52

z∗ = p̂−p0√

p0(1−p0)

n

= 0.52−0.64√

(0.64)(0.36)100

= −2.50

Using Normal table, P-value = 2P (Z > |z∗|) = 2P (Z >

| − 2.50|) = 2P (Z > 2.50) = 2(0.0062) = 0.0124

c. P-value = < 0.05, reject H0. Proportion differs from thereported 0.64.

MATH 214 (NOTES) – p. 59/62


The National Center for Health Statistics released a reportthat stated 70% of adults do not exercise regularly. Aresearcher decided to conduct a study to see whether theclaim made by the National Center for Health Statisticsdiffered on a state-by-state basis.a. State the null and alternative hypotheses assuming theintent of the researcher is to identify states that differ from70% reported by the National Center for Health Statistics.b. At α = 0.05, what is the research conclusion for thefollowing states:Wisconsin: 252 of 350 adults did not exercise regularly.California: 189 of 300 adults did not exerciseregularly.

MATH 214 (NOTES) – p. 60/62

Solution

a. H0 : p = 0.70 vs Ha : p 6= 0.70

b. Wisconsin p̂ = 252350 = 0.72

z∗ = p̂−p0√

p0(1−p0)

n

= 0.72−0.70√

(0.70)(0.30)350

= 0.82

Using Normal table, P-value = 2P (Z > |z∗|) = 2P (Z > |0.82|) =

2P (Z > 0.82) = 2(0.2061) = 0.4122

c. P-value = > 0.05, do not reject H0.

MATH 214 (NOTES) – p. 61/62

TO BE CONTINUED...

MATH 214 (NOTES) – p. 62/62

MATH 214 (NOTES)math.unm.edu/~alvaro/math214-p3.pdf · MATH 214 (NOTES) – p. 23/62. Z test for a...

Documents

Transcript of MATH 214 (NOTES)math.unm.edu/~alvaro/math214-p3.pdf · MATH 214 (NOTES) – p. 23/62. Z test for a...