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Math 144. Confidence Interval. In addition to the estimated value of the estimator, some statisticians suggest that we should also consider the variance of the estimator.

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math 144

Math 144

Confidence Interval

slide2

In addition to the estimated value of the estimator,

some statisticians suggest that we should also consider the variance of the estimator.

Use the single value and the variance of the estimator to form an interval that has a high probability to cover the unknown parameter.

This method including the variance of the point estimator is called interval estimation, or "confidence interval".

interval estimation

Assume that and are two functions of a random sample and are determined by a point estimator of an unknown parameter

such that

Interval estimation

where α is a known value between 0 and 1.

interval estimation1

After sampling, if the actual values of and are a and b, respectively, then the interval [a, b] is called

a 100(1-α)% confidence interval (hereafter, C.I.) for θ.

Interval estimation

The quantity 1-αis called the confidence level associated with the confidence interval.

slide5

By the definition, before sampling, we have a random interval estimation

for the unknown parameter θ.

Caution:

After sampling, the confidence interval [a, b] is a fixed (not random) interval. Indeed, it depends on the particular sample observations.

slide6

After sampling, we have observations

a

b

P(a≤ θ≤b) = 0

θ

[

[

Caution:

Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is,

P(a≤ θ≤b) = 0 or 1.

slide7

After sampling, we have observations

a

b

P(a≤ θ≤b) = 1

θ

[

[

Caution:

Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is,

P(a≤ θ≤b) = 0 or 1.

slide8

Recall that before sampling, we have

Caution:

Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is,

P(a≤ θ≤b) = 0 or 1.

slide9

Interpretation of C.I.

The interpretation of a 100(1-α)% C.I. is that when we obtained N (sufficient large) independent sets of random sample and for each set of random sample, we construct one particular interval by using the same point estimator, then there are N(1-α) out of these N intervals will contain the true unknown parameter θ.

However, we do not know which interval will contain θ and which will not contain θ, because θ is unknown.

slide10

Interpretation of C.I.

For instance, if we construct a random interval by drawing different sets of samples repeatedly, say 100 times, then

95% = 100(1-0.05)% C.I. for μmeans that μ is contained in 95 out of the 100 fixed intervals. Again, we do not know what these 95 intervals are, because µ is unknown.

slide11

Steps to construct a confidence interval

Step 1: Find a point estimator of θ

Step 2: Find its EXACT (or approximate) distribution.

Step 3: Based on the exact (or approximate) distribution found in Step 2 to construct the C.I.

Throughout this course, we are only interested in how to construct confidence intervals of parameters µ and σ2 by the sample mean and sample variance S2.

In the following, we will discuss the distributions of

and S2, and then see how to obtain the confidence interval of µ and σ2 case by case.

one sample

One sample

Confidence Interval for µ with NORMAL population

(known variance)

confidence interval for

Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and KNOWN variance σ2. That is,

Then we have a result that the sampling distribution of the sample mean is

Or equivalently,

Confidence interval for µ

Case I: Normal distribution with unknown mean and KNOWN variance:

slide14

How to construct the interval?

Define a quantity such that

α

slide15

How to construct the interval?

By the symmetry of the standard normal distribution, we have

Define a quantity such that

slide16

How to construct the interval?

z1-α/2

= -zα/2

1 - α

α/2

α/2

zα/2

slide17

How to construct the interval?

By the symmetry of the standard normal distribution, we have

θ

Define a quantity such that

slide18

How to construct the interval?

or simply written as

The margin of error

After sampling, we can find an actual value of the sample mean, say . Thus, 100(1-α)% C.I for μ is that

slide19

If all X1,…, Xn are observed, i.e. we have x1,…,xn , then 95% C.I for μ is that

For example, if α = 0.05, then

slide20

Note that μ is an unknown BUT fixed number, and and σ2 are known.

Remark again that it does not mean that μ is inside this interval with a probability 0.95.

So, μ is either inside or outside the fixed interval.

questions
Questions

Page 12

Q1: Given a random sample of 100 observations from a normal distribution for which µ is unknown and σ = 8. Suppose that the sample mean is found to be 42.7 after sampling. Then what is the 95% C.I. for µ?

Q2: A wine importer needs to report the average percentage of alcohol in bottles of French wine. From previous experience with different kinds of wine, the importer believes the alcohol concentration is normally distributed with standard deviation 1.2%. The importer randomly samples 60 bottles of the new wine and obtains a sample mean 9.3%. Find a 90% C.I. for the population average percentage.

one sample1

One sample

Confidence Interval for µ with NORMAL population

(unknown variance)

confidence interval for1

Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and UNKNOWN variance σ2. That is,

Then we have a result that the sampling distribution of the sample mean is

Or equivalently,

Confidence interval for µ

Case II: Normal distribution with unknown mean and UNKNOWN variance:

slide24

What is the sampling distribution of

After sampling, we can find an actual value of the sample mean, say . Thus, 100(1-α)% C.I for μ is that

However, σ is UNKNOWN. So, this interval is also unknown.

Replace σ2 by the sample variance S2. However, the next problem is:

Still normal?

NO!

slide25

Then the sampling distribution of

has a Student t distribution (or simply t distribution) with n -1 degrees of freedom. Denote by

where

and

Theorem

Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and UNKNOWN variance σ2.

t k distribution
tk distribution
  • Similar to a standard normal distribution, it is also symmetric about 0, so
  • P(T ≤ -a) = 1 - P(T ≤ a) = P(T ≥ a), if T follows a t distribution.
  • Use a table of a t distribution to find a probability of a t-distributed random variable.
slide27

How to construct the interval?

Define a quantity such that

By the symmetry of the t distribution, we have

slide28

How to construct the interval?

or simply written as

After sampling, we can find the actual values of the sample mean and sample variance, say and s. Thus, 100(1-α)% C.I for μ is

slide34

t

3, 0.05

α

Degree of freedom first

2.353 = ?

questions1

P(-t14, 0.025≤ T14 ≤ t14, 0.005)

= P(T14 ≤ t14, 0.005) – P(T14≤ -t14, 0.025)

By the symmetry of t distribution

Questions

Page 14 Q3

(i) Find P(-t14, 0.025≤ T14 ≤ t14, 0.005)

= [1 - P(T14 > t14, 0.005)] – P(T14> t14, 0.025)

= [1 – 0.005] – 0.025

= 0.97

questions2

0.045 = P( k ≤ T14 ≤ - 1.761)

= P(T14 ≤ - 1.761) – P(T14≤ k)

Questions

Page 14 Q3

(ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045

= P(T14≥ 1.761) – P(T14≥ - k)

By the symmetry of t distribution

questions3
Questions

Page 14 Q3

(ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045

0.045 = P( k ≤ T14 ≤ - 1.761)

= P(T14 ≤ - 1.761) – P(T14≤ k)

= P(T14≥ 1.761) – P(T14≥ - k)

By the symmetry of t distribution

= P(T14≥t14, 0.05) – P(T14≥ - k)

= 0.05 – P(T14≥ - k)

 P(T14≥ - k) = 0.05 – 0.045 = 0.005

questions4
Questions

Page 14 Q3

(ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045

0.045 = P( k ≤ T14 ≤ - 1.761)

= P(T14 ≤ - 1.761) – P(T14≤ k)

= P(T14≥ 1.761) – P(T14≥ - k)

By the symmetry of t distribution

= P(T14≥t14, 0.05) – P(T14≥ - k)

= 0.05 – P(T14≥ - k)

 P(T14≥ - k) = 0.05 – 0.045 = 0.005

= P(T14≥ 2.977)

k = - 2.977

questions5

s2 = 56.424, n = 12, α = 0.05

Questions

Page 14

Frequencies, in hertz (Hz), of 12 elephant calls:

14, 16, 17, 17, 24, 20, 32, 18, 29, 31, 15, 35

Assume that the population of possible elephant call frequencies is a normal distribution, Now a scientist is interested in the average of the frequencies, say µ. Find a 95% confidence interval for µ.

Population variance is UNKNOWN

So, use t distribution to construct the C.I. for µ.

Finally, the 95% C.I. for µ is [17.557, 27.103]

slide42

Remark:

When n > 30, the difference of a t distribution with n -1 degrees of freedom and the standard normal distribution is small. So, we have

Therefore, we can use

to approximate the 100(1-α)% C.I for μwith unknown variance, as n > 30.

two samples

Two samples

Confidence Interval for µX - µY with NORMAL populations

(known variances)

confidence interval for x y

Consider two independent random samples,

and

use

to estimate µX - µY.

Confidence interval for µX - µY

Case I: Normal distributions with unknown means and KNOWN variances:

Want to construct a C.I. for the mean difference µX - µY.

First, choose a point estimator of the mean difference.

slide45

How to construct the interval?

Second, find the sampling distribution of . Indeed, we have a result that

Or equivalently,

slide46

How to construct the interval?

or

Similar to Case 1 in the one-sample case. After sampling, the 100(1-α)% C.I for μX - μY is given by

confidence interval for x y1

then the 100(1-α)% C.I for μX - μY becomes

Confidence interval for µX - µY

Case I: Normal distributions with unknown means and KNOWN variances:

In particular, if two variances are EQUAL, say σX2 = σY2 = σ2,

example
Example

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB.

example1

n = m = 50

α = 0.05

σX = 5.6

σY = 6.3

Example

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB.

Two samples

Known variances

example2
Example

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB.

= [-11.24, -6.56]

two samples1

Two samples

Confidence Interval for µX - µY with NORMAL populations

(unknown variances)

confidence interval for x y2

and

Confidence interval for µX - µY

Case II: Normal distributions with unknown means and UNKNOWN variances:

Consider two independent random samples,

(i) In a case that BOTH UNKNOWN variances are EQUAL:

(ii) In a case that BOTH UNKNOWN variances are DIFFERENT:

slide53

has a t distribution with n-1 degrees of freedom.

Recall that, in the one-sample case with UNKNOWN variance, we replace the population variance σ2 by the sample variance S2. Then we have a result that

So, in two-sample cases, we will also replace the unknown variances by their estimators.

Then what estimators should we use to estimate the variances?

confidence interval for x y3
Confidence interval for µX - µY

Case II: Normal distributions with unknown means and UNKNOWN variances:

(i) In a case that BOTH UNKNOWN variances are EQUAL:

Use a statistic

which is called a pooled estimator of σ2 or pooled sample variance.

confidence interval for x y4

Based on

Confidence interval for µX - µY

Case II: Normal distributions with unknown means and UNKNOWN variances:

(i) In a case that BOTH UNKNOWN variances are EQUAL:

slide56

So, after sampling, the 100(1-α)% C.I for μX - μY is given by

If n+m-2 > 30, then the confidence interval can be approximated by

example3
Example

Page 17

Two tomato fertilizers are compared to see if one is better than the other.

The weight measurements of two independent random samples of tomatoes grown using each of the two fertilizers (in ounces) are as follows:

Fertilizer A(X): 12, 11, 7, 13, 8, 9, 10, 13

Fertilizer B(Y): 13, 11, 10, 6, 7, 4, 10

Assume that two populations are normal and their population variances are equal. Consider a confidence level 1-α = 0.95.

slide58

Since n = 8, m = 7,

and

Thus, the 95% C.I. for µX - µY is given by

= [-1.366, 4.688].

Fertilizer A(X): 12, 11, 7, 13, 8, 9, 10, 13

Fertilizer B(Y): 13, 11, 10, 6, 7, 4, 10

Assume that two populations are normal and their population variances are equal. Consider a confidence level 1-α = 0.95.

question
Question

Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 24 students in the section with labs made an average examination grade of 84 with a standard deviation of 4, and 36 students in the section without labs made an average grade of 77 with a standard deviation of 6. Then find a 99% confidence interval for the difference between the average grades for the two courses.

Assume that the population variances are equal.

confidence interval for x y5
Confidence interval for µX- µY

Case II: Normal distributions with unknown means and UNKNOWN variances:

(ii) In a case that BOTH UNKNOWN variances are DIFFERENT:

We do not have a statistic such that its exact distribution can be found to construct a C.I. for µX - µY in this case. However, it is still possible for us to construct an APPROXIMATE confidence interval.

Now, both variances are different, so we cannot use the pooled sample variance. In this case, we use the sample variance SX2 for σX2 and SY2 for σY2.

slide61

That is, we consider

It can be shown that the sampling distribution of the above statistic is an approximatet distribution with v degrees of freedom, where

slide62

Before sampling, v is random and unknown.

After sampling, the actual value of v is fixed and can be found.

Remark that after sampling, the actual value of the degree of freedom v is not always an integer. So, in practice, we must round down to the nearest integer to achieve the desired confidence interval.

That is, if v = 1.4, then take 1; if v = 2.9, then take 2.

confidence interval for x y6
Confidence interval for µX- µY

Case II: Normal distributions with unknown means and UNKNOWN variances:

(ii) In a case that BOTH UNKNOWN variances are DIFFERENT:

Thus, the approximate 100(1-α)% C.I for μX - μY is

If v > 30, then the confidence interval becomes

question1
Question

A study was conducted by the Department of Zoology at the Virginia Polytechnic Institute and State University to estimate the difference in the amount of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus is measured in milligrams per liter. Fifteen samples were collected from station 1 and 12 samples were obtained from station 2. The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 samples from station 2 had an average content of 1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter. Find a 95% confidence interval for the difference in the true average orthophosphorus contents at these two stations, assuming that the observations came from normal populations with different variances.

question2
Question

A study was conducted by the Department of Zoology at the Virginia Polytechnic Institute and State University to estimate the difference in the amount of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus is measured in milligrams per liter.Fifteen samples were collected from station 1 and 12 samples were obtained from station 2. The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 samples from station 2 had an average content of 1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter.Find a 95% confidence interval for the difference in the true average orthophosphorus contents at these two stations, assuming that the observations came from normal populations with different variances.

Two sample problem with α=0.05!!

question3
Question

A study was conducted by the Department of Zoology at the Virginia Polytechnic Institute and State University to estimate the difference in the amount of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus is measured in milligrams per liter.Fifteen samples were collected from station 1 and 12 samples were obtained from station 2.The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 samples from station 2 had an average content of 1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter.Find a 95% confidence interval for the difference in the true average orthophosphorus contents at these two stations, assuming that the observations came from normal populations with different variances.

Two sample problem with α=0.05!!

Normal!! Different Variances

question4

and

Question

A study was conducted by the Department of Zoology at the Virginia Polytechnic Institute and State University to estimate the difference in the amount of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus is measured in milligrams per liter.Fifteen samples were collected from station 1 and 12 samples were obtained from station 2.The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 samples from station 2 had an average content of 1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter.Find a 95% confidence interval for the difference in the true average orthophosphorus contents at these two stations, assuming that the observations came from normal populations with different variances.

Two sample problem with α=0.05!!

Normal!! Different Variances

question5

and

Question

Two sample problem with α=0.05!!

Normal!! Different Variances

Consider µ1 - µ2, where µi is the true average orthophosphorus contents at station i, i = 1 and 2.

Since the population variances are assumed to be unequal, we can only find an approximate 95% C.I. based on the t distribution with v degrees of freedom, where

question6

So, for α = 0.05, we have

Thus, the 95% C.I. for µ1 - µ2 is

Question

Two sample problem with α=0.05!!

Normal!! Different Variances

question7

Thus, the 95% C.I. for µ1 - µ2 is

Hence, we can say that we are 95% confident that the interval from 0.60 to 4.10 milligrams per liter contains the difference of the true average orthophosphorus contents for stations 1 and 2.

Question
one or two sample s

One- (or Two-) sample(s)

Confidence Interval for µX (or µX - µY) with NON-NORMAL population(s)

approximate c i in one sample case
Approximate C.I. in One-sample case

Note that, so far, all results are based on the normal population(s). Then a natural question is:

how to construct a C.I. with NON-Normal distribution.

Unfortunately, in general, it is not easy to find a statistic such that its exact distribution is easily found in this case.

However, if the sample size is large enough, then we can use a normal approximation to approximate the distribution of the statistic used to construct the C.I.

central limit theorem clt

If is the sample mean of a random sample X1,…, Xn of size nfrom any distribution witha finite mean µ and a finite positive variance σ2, then the distribution of

Central Limit Theorem (CLT)

is the standard normal distribution N(0,1) in the limit as n goes to infinity.

approximate c i for

After sampling, we can find an actual value of the sample mean, say . Thus, the APPROXIMATE 100(1-α)% C.I for μ is

Approximate C.I. for µ

Case I: Any distribution with unknown mean and KNOWN variance:

Consider a random sample of size n, {X1, X2, …, Xn}, from a distribution with unknown mean µ and KNOWN variance σ2. That is,

slide75

If n is large enough, then the approximate 100(1-α)% C.I for μ becomes

After sampling, we can find the actual values of the sample mean and sample variance, say and s. Thus, the APPROXIMATE 100(1-α)% C.I for μ is

Case II: Any distribution with unknown mean and UNKNOWN variance:

approximate c i in two sample case

if n+m-2 is large enough.

or

Approximate C.I. in Two-sample case

Consider two independent random samples from distributions with means µX and µY and variance σX2 and σY2, respectively.

  • In a case of SAME variance (say, σX2 = σY2 = σ2), the APPROXIMATE 100(1-α)% C.I for µX - µY is

(if variance σ2is known)

(if variance σ2is unknown )

approximate c i in two sample case1

if v is large enough OR n and m are large enough.

or

Approximate C.I. in Two-sample case

Consider two independent random samples from distributions with means µX and µY and variance σX2 and σY2, respectively.

  • In a case of Different variances, the APPROXIMATE 100(1-α)% C.I for µX - µY is

(if variances are known )

(if variances are unknown )

confidence interval for 2 with normal population
Confidence Interval for σ2

with NORMAL population

confidence interval for 2

has a chi-squared (or ) distribution with n – 1 degrees of freedom. We denote it by

Confidence interval for σ2

Case : Normal distribution with UNKNOWN variance:

Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with UNKNOWN mean and UNKNOWN variance σ2. Then, a statistic

slide81

How to construct the interval?

Found from the table of chi squared distribution with k degrees of freedom

Define a quantity such that

So, we have

slide83

How to construct the interval?

Found from the table of chi squared distribution with k degrees of freedom

Define a quantity such that

So, we have

After sampling, we can find an actual value of the sample variance, say s2. Thus, 100(1-α)% C.I for σ2 is

slide89

20.483 = ?

With 10 degrees of freedom

questions6
Questions

Page 21

For a chi-squared distribution with v degrees of freedom,

a) If v = 5, then

slide91

16.750 =

With 5 degrees of freedom

questions7
Questions

Page 21

For a chi-squared distribution with v degrees of freedom,

a) If v = 5, then

b) If v = 19, then

slide95

37.652 =

With 25 degrees of freedom

questions12

For a chi-squared distribution with v degrees of freedom, find

such that

b) when v = 10;

Questions

Page 21

a) when v = 6;

how about the confidence interval for not 2

Recall that

How about the confidence interval for σ, not σ2?

A 100(1 - α)% confidence interval for σ can be obtained by taking the square root of each endpoint of the interval for σ2. That is,

example4
Example

The following are the weights, in decagrams, of 10 packages of grass seed distributed by a certain company:

46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 45.8, 46.9, 45.2 and 46.0.

Find a 95% C.I. for the variance of all such packages of grass seed distributed by this company, assuming that a normal population is used.

example5

n = 10

Example

The following are the weights, in decagrams, of 10 packages of grass seed distributed by a certain company:

46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 45.8, 46.9, 45.2 and 46.0.

Find a 95% C.I. for the variance of all such packages of grass seed distributed by this company, assuming that a normal population is used.

example6

n = 10

Thus, the 95% C.I. for the variance is

Example

The following are the weights, in decagrams, of 10 packages of grass seed distributed by a certain company:

46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 45.8, 46.9, 45.2 and 46.0.

sample size determination
Sample size determination

Before we end the topic of estimation, let’s consider the problem of how to determine the sample size.

Often, we wish to know how large a sample is necessary to ensure that the error in estimating an unknown parameter, say µ, will be less than a specified amount e.

Consider a 100(1-α)% C.I. for µwith known variance. The (marginal) error is

slide104

implies that the required sample size is

Thus, solving for the sample size n in the equation

question8
Question

Page 23

A marketing research firm wants to conduct a survey to estimate the average amount spent on entertainment by each person visiting a popular resort. The people who plan the survey would like to have an estimate close to the true value such that we will have 95% confidence that the difference between them is within $120. If the population standard deviation is $400, then how large should the sample be?

question9

e

Question

Page 23

A marketing research firm wants to conduct a survey to estimate the average amount spent on entertainment by each person visiting a popular resort.The people who plan the survey would like to havean estimateclose tothe true valuesuch that we will have95% confidence that the difference between them is within $120. If the population standard deviation is $400, then how large should the sample be?

question10

e

Question

Page 23

A marketing research firm wants to conduct a survey to estimate the average amount spent on entertainment by each person visiting a popular resort.The people who plan the survey would like to havean estimateclose tothe true valuesuch that we will have95% confidence that the difference between them is within $120. If the population standard deviation is $400, then how large should the sample be?

question11

Then, the required sample size is

Question

Page 23

A marketing research firm wants to conduct a survey to estimate the average amount spent on entertainment by each person visiting a popular resort.The people who plan the survey would like to havean estimateclose tothe true valuesuch that we will have95% confidence that the difference between them is within $120. If the population standard deviation is $400, then how large should the sample be?