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Chapter 12

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Chapter 12

Inference About A Population

- Identify the parameter to be estimated or tested.
- Specify the parameter’s estimator and its sampling distribution.
- Derive the interval estimator and test statistic.

- We will develop techniques to estimate and test three population parameters:
- Population Mean
- Population Variance
- Population Proportion p

- Previously, we looked at estimating and testing the population mean when the population standard deviation ( ) was known or given:
- But how often do we know the actual population variance?
- Instead, we use the Student t-statistic, given by:

- When is unknown, we use its point estimator s
- and the z-statistic is replaced by the the t-statistic, where the number of “degrees of freedom” , is n–1.

- When the population standard deviation is unknown and the population is normal, the test statistic for testing hypotheses about is:
- which is Student t distributed with = n–1 degrees of freedom. The confidence interval estimator of is given by:

- Will new workers achieve 90% of the level of experienced workers within one week of being hired and trained?
- Experienced workers can process 500 packages/hour, thus if our conjecture is correct, we expect new workers to be able to process .90(500) = 450 packages per hour.
- Given the data, is this the case?

IDENTIFY

- Our objective is to describe the population of the numbers of packages processed in 1 hour by new workers, that is we want to know whether the new workers’ productivity is more than 90% of that of experienced workers. Thus we have:
- H1: > 450
- Therefore we set our usual null hypothesis to:
- H0: = 450

COMPUTE

- Our test statistic is:
- With n=50 data points, we have n–1=49 degrees of freedom. Our hypothesis under question is:
- H1: > 450
- Our rejection region becomes:
- Thus we will reject the null hypothesis in favor of the alternative if our calculated test static falls in this region.

COMPUTE

- From the data, we calculate = 460.38, s=38.83 and thus:
- Since
- we reject H0 in favor of H1, that is, there is sufficient evidence to conclude that the new workers are producing at more than 90% of the average of experienced workers.

COMPUTE

- Alternatively, we can use t-test:Mean from
- Tools > Data Analysis Plus in Excel…

rejection

region

:

:

COMPUTE

- In addition to looking at the computed t-statistic and the critical value of t (one tail), we could look at the p-value (0.0323) and see that it is “small” (~3%), so again, we reject the null hypothesis in favor of the alternative…

p-value

IDENTIFY

- Can we estimate the return on investment for companies that won quality awards?
- We have are given a random sample of n = 83 such companies. We want to construct a 95% confidence interval for the mean return, i.e. what is: ??

COMPUTE

- From the data, we calculate:
- For this term
- and so:

INTERPRET

- We are 95% confident that the population mean, , i.e. the mean return of all publicly traded companies that win quality awards, lies between 13.20% and 16.84%
- Tools > Data Analysis Plus > t-Estimate: Mean
- is an alternative to the
- manual calculation…

- The Student t distribution is robust, which means that if the population is nonnormal, the results of the t-test and confidence interval estimate are still valid provided that the population is “not extremely nonnormal”.
- To check this requirement, draw a histogram of the data and see how “bell shaped” the resulting figure is. If a histogram is extremely skewed (say in the case of an exponential distribution), that could be considered “extremely nonnormal” and hence t-statistics would be not be valid in this case.

- Large populations are defined as “populations that are at least 20 times the sample size”
- We can use the confidence interval estimator of a mean to produce a confidence interval estimator of the population total:
- Where N is the size of the finite population.

- For example, a sample of 500 households (in a city of 1 million households) reveals a 95% confidence interval estimate that the household mean spent on Halloween candy lies between $20 & $30.
- We can estimate the total amount spent in the city by multiplying these lower and upper confidence limits by the total population:
- Thus we estimate that the total amount spent on Halloween in the city lies between $20 million and $30 million.

- Factors that identify the t-test and estimator of :

- If we are interested in drawing inferences about a population’s variability, the parameter we need to investigate is the population variance:
- The sample variance (s2)is an unbiased, consistent and efficient point estimator for . Moreover,
- the statistic, , has a chi-squared distribution,
- with n–1 degrees of freedom.

- The test statistic used to test hypotheses about is:
- (which is chi-squared with = n–1 degrees of freedom).

- Combining this statistic:
- With the probability statement:
- Yields the confidence interval estimator for :

lower confidence limit

upper confidence limit

IDENTIFY

- Consider a container filling machine. Management wants a machine to fill 1 liter (1,000 cc’s) so that that variance of the fills is less than 1 cc2. A random sample of n=25 1 liter fills were taken. Does the machine perform as it should at the 5% significance level?
- We want to show that:
- H1: < 1
- (so our null hypothesis becomes: H0: = 1). We will use this test statistic:

Variance is less than 1 cc2

COMPUTE

- Since our alternative hypothesis is phrased as:
- H1: < 1
- We will reject H0 in favor of H1 if our test statistic falls into this rejection region:
- We computer the sample variance to be: s2=.8088
- And thus our test statistic takes on this value…

compare

INTERPRET

- Since:
- There is not enough evidence to infer that the claim is true.
- Excel output can also be used for this test…

compare

- As we saw, we cannot reject the null hypothesis in favor of the alternative. That is, there is not enough evidence to infer that the claim is true.
- Note: the result does not say that the variance is greater than 1, rather it merely states that we are unable to show that the variance is less than 1.
- We could estimate (at 99% confidence say) the variance of the fills…

COMPUTE

- In order to create a confidence interval estimate of the variance, we need these formulae:
- we know (n–1)s2 = 19.41 from our previous calculation, and we have from Table 5 in Appendix B:

lower confidence limit

upper confidence limit

COMPUTE

- Thus the 99% confidence interval estimate is:
- That is, the variance of fills lies between .426 and 1.963 cc2.

- Factors that identify the chi-squared test and estimator of :

- When data are nominal, we count the number of occurrences of each value and calculate proportions. Thus, the parameter of interest in describing a population of nominal data is the population proportion p.
- This parameter was based on the binomial experiment.
- Recall the use of this statistic:
- where p-hat ( ) is the sample proportion: x successes in a sample size of n items.

- When np and n(1–p) are both greater than 5, the sampling distribution of is approximately normal with
- mean:
- standard deviation:
- Hence:

- Test statistic for p:
- The confidence interval estimator for p is given by:
- (both of which require that np>5 and n(1–p)>5)

IDENTIFY

- At an exit poll, voters are asked by a certain network if they voted Democrat (code=1) or Republican (code=2). Based on their small sample, can the network conclude that the Republican candidate will win the vote?
- That is:
- H1: p > .50
- And hence our null hypothesis becomes:
- H0: p = .50

COMPUTE

- Since our research hypothesis is:
- H1: p > .50
- our rejection region becomes:
- Looking at the data, we count 407 (of 765) votes for code=2. Hence, we calculate our test statistic as follows…

INTERPRET

- Since:
- …we reject H0 in favor of H1, that is, there is enough evidence to believe that the Republicans win the vote.
- Likewise from Excel:

compare these…

…or look at p-value

- The confidence interval estimator for a population proportion is:
- Thus the (half) width of the interval is:
- Solving for n, we have:

- For example, we want to know how many customers to survey in order to estimate the proportion of customers who prefer our brand to within .03 (with 95% confidence).
- I.e. our confidence interval after surveying will be ± .03, that means W=.03
- Substituting into the equation…

Uh Oh. Since we haven’t taken a sample yet, we don’t have this sample proportion…

- Two methods – in each case we choose a value for then
- solve the equation for n.
- Method 1 : no knowledge of even a rough value of . This is a ‘worst case scenario’ so we substitute = .50
- Method 2 : we have some idea about the value of . This is a better scenario and we substitute in our estimated value.

- Method 1 : no knowledge of value of , use 50%:
- Method 2 : some idea about a possible value, say 20%:
- Thus, we can sample fewer people if we already have a reasonable estimate of the population proportion before starting.

- In much the same way as we saw earlier, when a population is large and finite we can estimate the total number of successes in the population by taking the product of the size of the population (N) and the confidence interval estimator:
- The Nielsen Ratings (used to measure TV audiences) uses this technique. Results from a small sample audience (say 2,000 viewers) is extrapolated to the total number of TV sets (say 100 million)…

COMPUTE

- Problem: describe the population of television shows watched by viewers across the country (population), by examining the results from 2,000 viewers (sample).
- We take these values and multiply them by N=100 million to estimate that between 9.9 million and 12.7 million viewers are watching the “Tonight Show”.

- Factors that identify the z-test and interval estimator of p:

Describe a Population

Data Type?

Interval

Nominal

z test & estimator of p

Type of descriptive measurement?

Central Location

Variability

t test & estimator of u.

X2 test & estimator ofs2