AP Stats Chapter 10: Estimating with Confidence



LT 8.3: Estimating a Population Mean

How long can you expect an AA battery to last? What proportion of college undergraduates have engaged in binge drinking? Is caffeine dependence real? Rather than collect every battery or ask all college undergraduates, we use a sample to answer our questions. We will then use our sample data to make some conclusions about the population.

[pic]

A. Confidence Intervals

Example #1: The admissions director at Big Bucks University wants to market his school using the IQ score of current students. He chooses a SRS of 50 students from the school’s 5000 freshmen and administers an IQ test. The mean of the sample is 112. What can the director say about the mean score of the population of all 5000 freshmen? Is the mean IQ of all freshmen 112? Somewhere close to 112? How do we know?

How close to 112 is[pic]likely to be? How would the sample mean [pic]vary (suppose the standard deviation is 15 for the population) if we took many samples of 50 freshmen from this same population?

One-Sample z Interval for a Population Mean

Draw an SRS of size n from a population having unknown mean μ and known standard deviation σ. As long as the Normal and Independent conditions are met, a level C confidence interval for μ is:

[pic]

The critical value z* is found from the standard Normal distribution.

This method isn’t very useful in practice, however in most real-world settings. If we don’t know the population mean μ, then we don’t know the population standard deviation σ either. But we can use the one-sample z interval for a population mean to estimate the sample size needed to achieve a specified margin of error. The process mimics what we did for a population proportion.

The Reasoning behind Statistical Estimation in a Nutshell

• To estimate the unknown population mean μ, use the mean [pic] of our random sample.

• Although [pic] is an unbiased estimate of μ, it will rarely be exactly equal to μ, so our estimate has some error.

• In repeated samples, the values of [pic] follow approximately a normal distribution with mean μ and standard deviation of 2.1 ([pic]).

• The 95% part of the 68%-95%-99.7% rule for Normal distributions says that in about 95% of all samples, the mean IQ score for the sample will be within +/- 2 standard deviations or 4.2 points of the population mean.

• Whenever [pic] is within 4.2 points of μ, μ is also within 4.2 points of [pic]. This happens in about 95% of all possible samples. So the unknown μ lies between [pic] - 4.2 and [pic] + 4.2 in about 95% of all samples.

Sampling Distribution for the Mean IQ score of an SRS of Size 50.

In 95% of all samples, the mean lies within +/- 4.2 of the unknown population mean.

[pic]

To say that [pic] +/- 4.2 is a 95% confidence interval for the population mean μ is to say that, in repeated samples, 95% of these intervals capture μ.

Conclusion: Our sample of 50 freshmen gave [pic] = 112. The resulting 95% confidence interval is 112 +/- 4.2 which can be written as (107.8, 116.2). We are 95% confident that the unknown mean IQ score for all university freshmen is between 107.8 and 116.2.

The interval we give is called the Confidence Interval and the 95% is the Confidence Level. The plus or minus 4.2 is the Margin of Error (ME).

[pic]

The calculation of our interval is based on three assumptions:

[pic]

Finding Nemo and Finding z*--Constructing a Confidence Interval

To construct an 80% confidence interval, we must catch the central 80% of the Normal sampling distribution of [pic]. In catching the central 80%, we leave out 20% or 10% in each tail. So z* is the point with 0.1 area to the right and 0.9 area to the left under the standard Normal curve. Looking at Table A the closest entry with 0.9 to the left is z*=1.28. 80% of the area under the normal curve lies between -1.28 and 1.28 standard deviations from the mean.

[pic] [pic]

[pic]

B. Confidence Interval for a Population Mean where σ is Known)

Example #2: Video Terminals

A manufacturer of high resolution video terminals must control the tension on the mesh of fine wires that lies behind the surface of the viewing screen. Some variation is inherent in the production process. The standard deviation of the tension readings is [pic] = 43mV. Here are the readings from an SRS of 20:

269.5 297.0 269.6 283.3 304.8 280.4 233.5 257.4 317.5 327.4

264.7 307.7 310.0 343.3 328.1 342.6 338.8 340.1 374.6 336.1

Construct and interpret a 90% confidence interval for the mean tension μ of all the screens produced on this day using the PANIC procedure.

SOCS, BINS, BITS, DOFS & now PANIC (PHANTOM Comes Next)

Step 1--P: Population & Parameter. Identify the population of interest and the parameter you want to draw conclusions about.

Step 2--A: Assumptions & Conditions--Choose the appropriate inference procedure. Verify conditions for using it.

• SRS:

• Normality:

• Independence:

Normal: Since the sample size is small (n = 20), we must check whether it’s reasonable to believe that the population distribution is Normal. So we examine the sample data. The figure below shows (a) a dotplot, (b) a boxplot, and (c) a Normal probability plot of the tension readings in the sample. Neither the dotplot nor the boxplot shows strong skewness or any outliers. The Normal probability plot looks roughly linear. These graphs give us no reason to doubt the Normality of the population.

[pic]

In Chapter 2, we noted that a data set with an approximately Normal shape will have a Normal probability plot that’s roughly linear. Now we’re trying to get information about the shape of the population distribution from a Normal probability plot of the sample data. This is much harder, because even samples drawn from perfectly Normal populations don’t always look Normal.

Step 3--N: Name & Formula. What is the name and formula for the confidence interval method you are using?

Step 4—I: (Confidence) Interval = Estimate +/- Margin of Error (ME)

Step 5—C: Conclusion in Context. Interpret your results in the context of the problem. Remember: Conclusion, connection and context.

Just a reminder…Larger samples give smaller intervals because the standard deviation is being divided by a larger value.

[pic]

Example #3: How Confidence Intervals Behave

Suppose the manufacturer of the screens in the example wants 99% confidence rather than 90% confidence. The critical value for 99% confidence is z* = __________. The 99% confidence interval for [pic]based on an SRS of 20 with mean = 306.3 is:

Demanding 99% confidence instead of 90% confidence has increased the margin of error from ___________ to ________.

[pic]

Example #4: Sample Size for a Desired Margin of Error

Researchers would like to estimate the mean cholesterol level of μ of a particular variety of monkey that is often used in laboratory experiments. They would like their estimate to be within 1 mg/dl of blood of the true value of[pic]at a 95% confidence level. A previous study involving this variety of monkey suggests that the standard deviation of cholesterol level is about 5 mg/dl. Obtaining monkeys is time-consuming and expensive, so the researchers want to know the minimum number of monkeys they will need to generate a satisfactory estimate.

To determine the sample size n that will yield a confidence interval for a population mean with a specified margin of error (ME), set the expression for the margin of error to be less than or equal to m and solve for n.

Sample Size for a Desired Margin of Error When Estimating μ

To determine the sample size n that will yield a level C confidence interval for a population mean with a specified margin of error ME:

• Get a reasonable value for the population standard deviation σ from an earlier or pilot study.

• Find the critical value z* from a standard Normal curve for confidence level C.

• Set the expression for the margin of error to be less than or equal to ME and solve for n: [pic]

For 95% confidence, the table gives z* = 1.96. We know that [pic] = 5. Set the margin of error to be at most 1 and solve for n.

Always round up to the next whole number when finding n. Notice that the size of the sample determines the margin of error. The size of the population does not influence the sample size as long as the population is much larger than the sample)

C. When σ Is Unknown: The t Distributions

When the sampling distribution of [pic] is close to Normal, we can find probabilities involving [pic] by standardizing:

[pic]

Recall that the sampling distribution of [pic] has mean μ and standard deviation [pic]. What are the shape, center, and spread of the sampling distribution of the new statistic z? z has the standard Normal distribution N(0, 1). Therefore, we can use Table A or a calculator to find the related probability involving z. That’s how we have gotten the critical values for our confidence intervals so far.

When we don’t know σ, we estimate it using the sample standard deviation sx. What happens now when we standardize?

[pic]

This new statistic does not have a standard Normal distribution.

When doing inference about a population mean μ, what happens when we use the sample standard deviation sx to estimate the population standard deviation σ? We’ll start with a Normal population having mean μ = 100 and standard deviation σ = 5.

The figure below shows the results of taking 500 SRSs of size n = 4 and standardizing the value of the sample mean [pic]. The values of z follow a standard Normal distribution, as expected. The standardized values using the sample standard deviation sx in place of the population standard deviation σ, show much greater spread. In fact, in a few samples, the statistic:

[pic]

took values below −6 or above 6. This statistic has a distribution that is new to us, called a t distribution. It has a different shape than the standard Normal curve: still symmetric with a single peak at 0, but with much more area in the tails.

[pic]

The statistic t has the same interpretation as any standardized statistic: it says how far [pic] is from its mean μ in standard deviation units. There is a different t distribution for each sample size. We specify a particular t distribution by giving its degrees of freedom (df). When we perform inference about a population mean μ using a t distribution, the appropriate degrees of freedom are found by subtracting 1 from the sample size n, making df = n − 1. We will write the t distribution with n − 1 degrees of freedom as tn−1 for short.

The t Distributions; Degrees of Freedom

Draw an SRS of size n from a large population that has a Normal distribution with mean μ and standard deviation σ. The statistic:

[pic]

has the t distribution with degrees of freedom df = n − 1. This statistic will have approximately a tn−1 distribution as long as the sampling distribution of [pic] is close to Normal.

Think of degrees of freedom as a way of keeping score. A data set contains a number of observations, say, n. They constitute n individual pieces of information. These pieces of information can be used either to estimate parameters or variability. In general, each item being estimated costs one degree of freedom. The remaining degrees of freedom are used to estimate variability. All we have to do is count properly.

For a single sample: There are n observations. There's one parameter (the mean) that needs to be estimated. That leaves n-1 degrees of freedom for estimating variability. The degree of freedom is the number of values in a calculation that we can vary.

The figure to the right compares the density curves of the standard Normal distribution and the t distributions with 2 and 9 degrees of freedom. The figure illustrates these facts about the t distributions:

• The density curves of the t distributions are similar in shape to the standard Normal curve. They are symmetric about 0, single-peaked, and bell-shaped.

• The spread of the t distributions is a bit greater than that of the standard Normal distribution. The t distributions have more probability in the tails and less in the center than does the standard Normal. This is true because substituting the estimate sx for the fixed parameter σ introduces more variation into the statistic.

• As the degrees of freedom increase, the t density curve approaches the standard Normal curve ever more closely. This happens because sx estimates σ more accurately as the sample size increases. So using sx in place of σ causes little extra variation when the sample is large.

Table B in the back of the book gives critical values t* for the t distributions. Each row in the table contains critical values for the t distribution whose degrees of freedom appear at the left of the row. For convenience, several of the more common confidence levels C (in percents) are given at the bottom of the table. By looking down any column, you can check that the t critical values approach the Normal critical values z* as the degrees of freedom increase.

Example # 5: Finding t*--Using Table B

Suppose you want to construct a 95% confidence interval for the mean μ of a Normal population based on an SRS of size n = 12. What critical value t* should you use?

In Table B, we consult the row corresponding to df = n − 1 = 11. We move across that row to the entry that is directly above 95% confidence level on the bottom of the chart. The desired critical value is t* = 2.201.

Notice that the corresponding standard Normal critical value for 95% confidence is z* = 1.96. We have to go out farther than 1.96 standard deviations to capture the central 95% of the t distribution with 11 degrees of freedom.

D. Constructing a Confidence Interval for μ

When the conditions for inference are satisfied, the sampling distribution of [pic] has roughly a Normal distribution with mean μ and standard deviation[pic]/[pic]. Because we don’t know σ, we estimate it by the sample standard deviation sx.

We then estimate the standard deviation of the sampling distribution by sx/[pic]. This value is called the standard error of the sample mean [pic], or just the standard error of the mean.

DEFINITION: Standard Error of the Sample Mean

The standard error of the sample mean [pic]is sx/[pic]where sx is the sample standard deviation. It describes how far [pic]will be from μ, on average, in repeated SRSs of size n.

To construct a confidence interval for μ, replace the standard deviation[pic]/[pic] of [pic]by its standard error sx/[pic] in the formula for the one-sample z interval for a population mean. Use critical values from the t distribution with n − 1 degrees of freedom in place of the z critical values. That is,

Statistic +/- (Critical Value) • (Standard Deviation of Statistic) = [pic] +/- t* (sx/[pic]

This one-sample t interval for a population mean is similar in both reasoning and computational detail to the one-sample z interval for a population proportion.

The One-Sample t Interval for a Population Mean

Choose an SRS of size n from a population having unknown mean μ. A level C confidence interval for μ is:

[pic] +/- t* (sx/[pic]

where t* is the critical value for the tn−1 distribution. Use this interval only when (1) the population distribution is Normal or the sample size is large (n ≥ 30), and (2) the population is at least 10 times as large as the sample.

As before, we have to verify three important conditions before we estimate a population mean. When we do inference in practice, verifying the conditions is often a bit more complicated.

Conditions for Inference about a Population Mean

• Random: The data come from a random sample of size n from the population of interest or a randomized experiment. This condition is very important.

• Normal: The population has a Normal distribution or the sample size is large (n ≥ 30).

• Independent: The method for calculating a confidence interval assumes that individual observations are independent. To keep the calculations reasonably accurate when we sample without replacement from a finite population, we should check the 10% condition: verify that the sample size is no more than 1/10 of the population size.

When we use Table B to determine the correct value of t* for a given confidence interval, all we need to know are the confidence level C and the degrees of freedom (df). Unfortunately, Table B does not include every possible sample size. When the actual df does not appear in the table, use the greatest df available that is less than your desired df. This guarantees a wider confidence interval than we need to justify a given confidence level. Better yet, use technology to find an accurate value of t* for any df.

Example #6: Auto Pollution--A one-sample t interval for μ

Environmentalists, government officials, and vehicle manufacturers are all interested in studying the auto exhaust emissions produced by motor vehicles.

The major pollutants in auto exhaust from gasoline engines are hydrocarbons, carbon monoxide, and nitrogen oxides (NOX). Researchers collected data on the NOX levels (in grams/mile) for a random sample of 40 light-duty engines of the same type. The mean NOX reading was 1.2675 and the standard deviation was 0.3332.

a. Construct and interpret a 95% confidence interval for the mean amount of NOX emitted by light-duty engines of this type.

STATE: We want to estimate the true mean amount μ of NOX emitted by all light-duty engines of this type at a 95% confidence level.

PLAN: We should construct a one-sample t interval for μ if the conditions are met:

• Random: The data come from a “random sample” of 40 engines from the population of all light-duty engines of this type.

• Normal: We don’t know whether the population distribution of NOX emissions is Normal. Because the sample size, n = 40, is large (at least 30), we should be safe using t procedures.

• Independent: We are sampling without replacement, so we need to check the 10% condition: we must assume that there are at least 10(40) = 400 light-duty engines of this type.

DO: The formula for the one-sample t interval is: [pic] +/- t* (sx/[pic]

The calculator command invT(.025,39) gives t = –2.023. Using the critical value t* = ± 2.023 for the 95% confidence interval gives:

1.2675 +/- 2.2023(0.332)/[pic][pic] = 1.2675 +/- 0.0166 = (1.1609, 1.3741)

CONCLUDE: We are 95% confident that the interval from 1.1599 to 1.3751 grams/mile contains the true mean level of nitrogen oxides emitted by this type of light-duty engine.

b. The environmental Protection Agency (EPA) sets a limit of 1.0 gram/mile for NOX emissions. Are you convinced that this type of engine has a mean NOX level of 1.0 or less? Use your interval from (a) to support your answer.

The confidence interval from (a) tells us that any value from 1.1609 to 1.3741 g/mi is a plausible value of the mean NOX level μ for this type of engine. Since the entire interval exceeds 1.0, it appears that this type of engine violates EPA limits.

E. Using t Procedures Wisely

The stated confidence level of a one-sample t interval for μ is exactly correct when the population distribution is exactly Normal. No population of real data is exactly Normal. The usefulness of the t procedures in practice therefore depends on how strongly they are affected by lack of Normality.

Procedures that are not strongly affected when a condition for using them is violated are called robust.

DEFINITION: Robust Procedures

An inference procedure is called robust if the probability calculations involved in that procedure remain fairly accurate when a condition for using the procedure is violated.

For confidence intervals, “robust” means that the stated confidence level is still pretty accurate. That is, if we use the procedure to calculate many 95% confidence intervals, about 95% of those intervals would capture the population mean μ. If the procedure isn’t robust, then the actual capture rate might be very different from 95%.

If outliers are present in the sample, then the population may not be Normal. The t procedures are not robust against outliers, because [pic] Correct Answer

We are 95% confident that the interval from 21.53 to 29.81 micrometers per hour captures the true mean healing time for newts.

and sx are not resistant to outliers.

Example #7: More Auto Pollution--t Procedures Not Robust Against Outliers

In the previous example, we constructed a confidence interval for the mean level of NOX emitted by a specific type of light-duty car engine. The original random sample actually included 41 engines, but one of them recorded an unusually high amount (2.94 grams/mile) of NOX. Upon further inspection, this engine had a mechanical defect. So the researchers decided to remove this value from the data set. The Minitab computer output below gives some numerical summaries for NOX emissions in the original sample.

Descriptive Statistics: NOX

[pic]

Did you notice the SE Mean, 0.0656, in the computer output? As you probably guessed, this is the standard error of the mean. The confidence interval based on this sample of 41 engines would be (using df = 40 from Table B)

[pic]

Our new confidence interval is wider and is centered at a higher value than our original interval of 1.1599 to 1.3751.

Fortunately, the t procedures are quite robust against non-Normality of the population except when outliers or strong skewness are present. Larger samples improve the accuracy of critical values from the t distributions when the population is not Normal. This is true for two reasons:

• The sampling distribution of the sample mean [pic] from a large sample is close to Normal (that’s the Central Limit Theorem). Normality of the individual observations is of little concern when the sample size is large.

• As the sample size n grows, the sample standard deviation sx will be an accurate estimate of [pic] whether or not the population has a Normal distribution.

Always make a plot to check for skewness and outliers before you use the t procedures for small samples. For most purposes, you can safely use the one-sample t procedures when n ≥ 15 unless an outlier or strong skewness is present.

Except in the case of small samples, the condition that the data come from a random sample or randomized experiment is more important than the condition that the population distribution is Normal. Here are practical guidelines for the Normal condition when performing inference about a population mean.

Using One-Sample t Procedures: The Normal Condition

• Sample size less than 15: Use t procedures if the data appear close to Normal (roughly symmetric, single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t.

• Sample size at least 15: The t procedures can be used except in the presence of outliers or strong skewness.

• Large samples: The t procedures can be used even for clearly skewed distributions when the sample is large, roughly n ≥ 30.

If your sample data would give a biased estimate for some reason, then you shouldn’t bother computing a t interval.

Example #7: People, Trees, and Flowers--Can we use t?

Determine whether we can safely use a one-sample t interval to estimate the population mean in each of the following settings.

a. Figure 8.15(a) is a histogram of the percent of each state’s residents who are at least 65 years of age.

b. Figure 8.15(b) is a stemplot of the force required to pull apart 20 pieces of Douglas fir.

c. Figure 8.15(c) is a stemplot of the lengths of 23 specimens of the red variety of the tropical flower Heliconia.

a. No. We have data on the entire population of the 50 states, so formal inference makes no sense. We can calculate the exact mean for the population. There is no uncertainty due to having only a sample from the population, and no need for a confidence interval.

b. No. The data are strongly skewed to the left with possible low outliers, so we cannot trust the t procedures for n = 20.

c. Yes. The data are mildly skewed to the right and there are no outliers. We can use the t distributions for such data.

LT 8.3 Problem Set

1. A consumer watchdog organization estimates the mean weight of 1-ounce “Fun-Size” candy bars to see if customers are getting full value for their money. A random sample of 25 bars is selected and weighed, and the organization reports that a 90% confidence interval for the true mean weight of the candy bars is 0.992 to 0.998 ounces.

a) What is the point estimate from this sample?

b) What is the margin of error?

c) Interpret the 90% confidence interval 0.992 to 0.998 in the context of the problem.

(d) Interpret the confidence level of 90% in the context of the problem.

2. A manufacturer of flashlights wants to know how well one of their newer styles is selling in a chain of large home-improvement stores. They select a simple random sample of 20 stores, record how many of the flashlights were sold in a 30-day period, and construct a 95% confidence interval for the mean number of flashlights sold.

a) Discuss whether this study meets the necessary conditions for constructing a confidence interval. If you think one of the conditions has not been met, what additional information would be required or what change in the study would you recommend?

b) If, instead of constructing a 95% confidence interval, the flashlight manufacturer constructed a 98% confidence interval, would the 98% interval be wider, narrower, or the same width as the 95% interval? Explain.

(c) How would the width of confidence interval change if the flashlight manufacturer took a larger sample? Explain.

3. Suppose you know that the distribution of finishing times for a certain crossword puzzle has a mean of 25 minutes, a standard deviation of 8 minutes, and is moderately skewed left. You take an SRS of 45 finish times from this distribution and calculate the mean finish time [pic] .

a) Describe the shape, center, and spread of the sampling distribution of [pic].

b) Find a number, k, such that 95% of the values in the sampling distribution will lie within k minutes of the mean of the distribution.

(c) If you take repeated samples of size 45 from this population, what proportion of the time will the interval [pic] +/- k contain the number 25? Explain.

4. To the right are the graphs of a standard Normal distribution and a t-distribution with 3 degrees of freedom.

a) Indicate which graph is which and explain how you know.

b) On the same figure sketch a graph of a t-distribution with 1 degree of freedom.

4. Find the critical t* value for each of the following confidence intervals:

a) 95% confidence interval with 8 degrees of freedom.

(b) 80% confidence interval when n = 20.

6. You want to estimate the mean fuel efficiency of Ford Focus automobiles with 99% confidence and a margin of error of no more than 1 mile per gallon. Preliminary data suggests that σ = 2.4 miles per gallon is a reasonable estimate of the standard deviation for all cars of this make and model. How large a sample do you need?

7. National Fuelsaver Corporation manufactures the Platinum Gasaver, a device they claim “may increase gas mileage by 30%.” Here are the percent changes in gas mileage for 15 identical, randomly-selected vehicles, as presented in one of the company’s advertisements:

|–2.4 |6.9 |10.4 |10.8 |24.8 |

|28.7 |28.7 |33.7 |34.6 |38.5 |

|40.2 |44.6 |46.8 |46.9 |48.3 |

a) The sample mean is [pic] = 29.43 and the sample standard deviation is sx = 16.23. Calculate and interpret the standard error of the mean for these data.

(b) Construct and interpret a 90% confidence interval to estimate the mean change (in percent) in gas mileage. Does the data support the company’s claim? Use the four-step process.

8. Below are graphical representations of three different samples from three different populations. In each case, discuss whether the Normality condition for constructing a t-confidence interval has been satisfied.

[pic]

9. About 130,000 high school students took the AP Statistics exam in 2010. The free-response section of the exam consisted of five open-ended problems and an investigative task. Each free-response question is scored on a 0 to 4 scale (with 4 being the best). For one of the problems, a random sample of 30 student papers yielded the scores that are graphed in the dot plot of part (a) in the previous problem. The mean score for this sample is [pic] = 1.267 and the standard deviation is sx = 1.230

a) Find and interpret the standard error of the mean.

(b) Construct and interpret a 99% confidence interval to estimate the mean score on this question. Use the four-step process.

10. A cereal maker’s container machine is designed to fill boxes so that the mean weight of cereal in the boxes is 18 ounces. A simple random sample of 30 boxes produced by the machine yields a mean weight of 17.92 ounces and a standard deviation of 0.2 ounces. The distribution of box weights is summarized in the Minitab output below:

|MEAN |STDEV |SEMEAN |MIN |Q1 |MEDIAN |Q3 |MAX |

|WEIGHT 17.920 |0.200 |0.0365 |17.55 |17.76 |17.90 |18.00 |18.25 |

a) Construct and interpret a 90% confidence interval to estimate the true mean weight of cereal in the boxes.

(b) Does the interval in (a) give you reason to suspect that the machine is not filling boxes with the correct amount of cereal? Explain your reasoning.

Multiple Choice. Identify the choice that best completes the statement or answers the question.

1. You want to calculate a 98% confidence interval for a population mean from a sample of n = 18. What is the appropriate critical t*?

|A. |2.110 |

|B. |2.326 |

|C. |2.539 |

|D. |2.552 |

|E. |2.567 |

2. In checking conditions for constructing confidence intervals for a population mean, it’s important to plot the distribution of sample data. Below are dot plots describing samples from three different populations. For which of the three samples would it be safe to construct a t-interval?

[pic]

| | |

|A. |Sample X only |

|B. |Sample Y only |

|C. |Sample Z only |

|D. |Samples Y and Z |

|E. |None of the samples |

3. A random sample of 900 individuals has been selected from a large population. It was found that 180 are regular users of vitamins. Thus, the proportion of the regular users of vitamins in the population is estimated to be 0.20. The standard error of this estimate is approximately:

|A. |0.1600 |

|B. |0.0002 |

|C. |0.4000 |

|D. |0.0133 |

|E. |0.0267 |

4. To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed a total of n times and the mean [pic]of the weighings is computed. Suppose the scale readings are Normally distributed with unknown mean[pic]and standard deviation [pic] = 0.01 g. How large should n be so that a 95% confidence interval for [pic] has a margin of error of ± 0.0001?

|A. |100 |

|B. |196 |

|C. |27,061 |

|D. |10,000 |

|E. |38,416 |

5. A 95% confidence interval for the mean reading achievement score for a population of third-grade students is (44.2, 54.2). Suppose you compute a 99% confidence interval using the same data. Which of the following statements is correct?

|A. |The intervals have the same width. |

|B. |The 99% interval is narrower. |

|C. |The 99% interval is wider. |

|D. |The 99% interval could be wider or narrower—it depends on the sample. |

|E. |The answer can’t be determined from the information given. |

6. What is the critical value t* that satisfies the condition that the t distribution with 8 degrees of freedom has probability 0.10 to the right of t*?

|A. |0.90 |

|B. |1.282 |

|C. |1.397 |

|D. |1.415 |

|E. |1.860 |

7. The weights of 9 men have mean [pic] = 175 pounds and standard deviation s = 15 pounds. What is the standard error of the mean?

|A. |58.3 |

|B. |15 |

|C. |5 |

|D. |1.67 |

|E. |1.29 |

8. The government claims that students earn an average of $4500 during their summer break from studies. A random sample of students gave a sample average of $3975, and a 95% confidence interval was found to be $3525 < µ < $4425. Which of the following is a correct interpretation of 95% confidence?

|A. |If the study were to be repeated many times, there is a 95% probability that the true average summer earnings is not $4500 as the |

| |government claims. |

|B. |Because our specific confidence interval does not contain the value $4500 there is a 95% probability that the true average summer |

| |earnings is not $4500. |

|C. |If we were to repeat our survey many times, then about 95% of all the confidence intervals will contain the value $4500. |

|D. |If we repeat our survey many times, then about 95% of our confidence intervals will contain the true value of the average earnings|

| |of students. |

|E. |There is a 95% probability that the true average earnings are between $3525 and $4425 for all students. |

9. You want to estimate the mean SAT score for a population of students with a 90% confidence interval. Assume that the population standard deviation is [pic] = 100. If you want the margin of error to be no more than 10, you will need a minimum sample size of approximately:

|A. |17 |

|B. |38 |

|C. |271 |

|D. |385 |

|E. |1646 |

10. The t-confidence interval for a population mean is robust with respect to which of the following concerns?

|A. |Undercoverage. |

|B. |Lack of randomness in sampling. |

|C. |Skew in the population distribution. |

|D. |The presence of extreme outliers. |

|E. |None of the above. |

[pic]

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download