Exam 3: Review of Statistics

A manufacturer claims that a certain brand of VCR player has an average life expectancy of 5 years and 6 months with a standard deviation of 1 year and 6 months. Assume that the life expectancy is normally distributed. (a)Selecting one VCR player from this brand at random, calculate the probability of its life expectancy exceeding 7 years. (b)The Critical Consumer magazine decides to test fifty VCRs of this brand. The average life in this sample is 6 years and the sample standard deviation is 2 years. Calculate a 99% confidence interval for the average life. (c)How many more VCRs would the magazine have to test in order to halve the width of the confidence interval?

You have collected data on the average weekly amount of studying time (T)and grades (G)from the peers at your college. Changing the measurement from minutes into hours has the following effect on the correlation coefficient:

(Advanced)Unbiasedness and small variance are desirable properties of estimators. However, you can imagine situations where a trade-off exists between the two: one estimator may be have a small bias but a much smaller variance than another, unbiased estimator. The concept of "mean square error" estimator combines the two concepts. Let $\hat { \mu }$ be an estimator of μ. Then the mean square error (MSE)is defined as follows: MSE( $\hat { \mu }$ )= E( $\hat { \mu }$ - μ)². Prove that MSE( $\hat { \mu }$ )= bias² + var( $\hat { \mu }$ ). (Hint: subtract and add in E( $\hat { \mu }$ )in E( $\hat { \mu }$ - μ)².)

Degrees of freedom

When you are testing a hypothesis against a two-sided alternative, then the alternative is written as

An estimator is

The t-statistic has the following distribution:

The net weight of a bag of flour is guaranteed to be 5 pounds with a standard deviation of 0.05 pounds. You are concerned that the actual weight is less. To test for this, you sample 25 bags. Carefully state the null and alternative hypothesis in this situation. Determine a critical value such that the size of the test does not exceed 5%. Finding the average weight of the 25 bags to be 4.7 pounds, can you reject the null hypothesis? What is the power of the test here? Why is it so low?

The power of the test is

A type I error is

During the last few days before a presidential election, there is a frenzy of voting intention surveys. On a given day, quite often there are conflicting results from three major polls. (a)Think of each of these polls as reporting the fraction of successes (1s)of a Bernoulli random variable Y, where the probability of success is Pr(Y = 1)= p. Let $\hat { p }$ be the fraction of successes in the sample and assume that this estimator is normally distributed with a mean of p and a variance of $\frac { p ( 1 - p ) } { n }$ Why are the results for all polls different, even though they are taken on the same day? (b)Given the estimator of the variance of $\hat { p }$ , $\frac { \hat { p } ( 1 - \hat { p } ) } { n }$ , construct a 95% confidence interval for $\hat { p }$ For which value of $\hat { p }$ is the standard deviation the largest? What value does it take in the case of a maximum $\hat { p }$ ? (c)When the results from the polls are reported, you are told, typically in the small print, that the "margin of error" is plus or minus two percentage points. Using the approximation of 1.96 ? 2, and assuming, "conservatively," the maximum standard deviation derived in (b), what sample size is required to add and subtract ("margin of error")two percentage points from the point estimate? (d)What sample size would you need to halve the margin of error?

The correlation coefficient

Your textbook mentions that dividing the sample variance by n -1 instead of n is called a degrees of freedom correction. The meaning of the term stems from the fact that one degree of freedom is used up when the mean is estimated. Hence degrees of freedom can be viewed as the number of independent observations remaining after estimating the sample mean. Consider an example where initially you have 20 independent observations on the height of students. After calculating the average height, your instructor claims that you can figure out the height of the 20^th student if she provides you with the height of the other 19 students and the sample mean. Hence you have lost one degree of freedom, or there are only 19 independent bits of information. Explain how you can find the height of the 20^th student.

You have collected weekly earnings and age data from a sub-sample of 1,744 individuals using the Current Population Survey in a given year. (a)Given the overall mean of $434.49 and a standard deviation of $294.67, construct a 99% confidence interval for average earnings in the entire population. State the meaning of this interval in words, rather than just in numbers. If you constructed a 90% confidence interval instead, would it be smaller or larger? What is the intuition? (b)When dividing your sample into people 45 years and older, and younger than 45, the information shown in the table is found. Age Category Average Earnings Standard Deviation N Age \geq45 \ 488.87 \ 328.64 507 Age <45 \ 412.20 \ 276.63 1237 Test whether or not the difference in average earnings is statistically significant. Given your knowledge of age-earning profiles, does this result make sense?

At the Stock and Watson (http://www.pearsonhighered.com/stock_watson)website go to Student Resources and select the option "Datasets for Replicating Empirical Results." Then select the "CPS Data Used in Chapter 8" (ch8_cps.xls)and open it in Excel. This is a rather large data set to work with, so just copy the first 500 observations into a new Worksheet (these are rows 1 to 501). In the newly created Worksheet, mark A1 to A501, then select the Data tab and click on "sort." A dialog box will open. First select "Add level" from one of the options on the left. Then select "sort by" and choose "Northeast" and "Largest to Smallest." Repeat the same for the "South" as a second option. Finally press "ok." This should give you 209 observations for average hourly earnings for the Northeast region, followed by 205 observations for the South. a. For each of the 209 average hourly earnings observations for the Northeast region and separately for the South region, calculate the mean and sample standard deviation. b Use the appropriate test to determine whether or not average hourly earnings in the Northeast region the same as in the South region. c Find the 1%, 5%, and 10% confidence interval for the differences between the two population means. Is your conclusion consistent with the test in part (b)? d In all three cases of using the confidence interval in (c), the power of the test is quite low (5%). What can you do to increase the power of the test without reducing the size of the test?

When testing for differences of means, you can base statistical inference on the

To derive the least squares estimator µ_Y, you find the estimator m which minimizes

The t-statistic is defined as follows:

Adult males are taller, on average, than adult females. Visiting two recent American Youth Soccer Organization (AYSO)under 12 year old (U12)soccer matches on a Saturday, you do not observe an obvious difference in the height of boys and girls of that age. You suggest to your little sister that she collect data on height and gender of children in 4^th to 6^th grade as part of her science project. The accompanying table shows her findings. Height of Young Boys and Girls, Grades 4-6, in inches (a)Let your null hypothesis be that there is no difference in the height of females and males at this age level. Specify the alternative hypothesis. (b)Find the difference in height and the standard error of the difference. (c)Generate a 95% confidence interval for the difference in height. (d)Calculate the t-statistic for comparing the two means. Is the difference statistically significant at the 1% level? Which critical value did you use? Why would this number be smaller if you had assumed a one-sided alternative hypothesis? What is the intuition behind this?

Let Y be a Bernoulli random variable with success probability Pr(Y = 1)= p, and let Y₁,..., Y_n be i.i.d. draws from this distribution. Let $\hat { p }$ be the fraction of successes (1s)in this sample. In large samples, the distribution of $\hat { p }$ will be approximately normal, i.e., $\hat { p }$ is approximately distributed N(p, $\frac { p ( 1 - p ) } { n }$ ). Now let X be the number of successes and n the sample size. In a sample of 10 voters (n=10), if there are six who vote for candidate A, then X = 6. Relate X, the number of success, to $\hat { p }$ , the success proportion, or fraction of successes. Next, using your knowledge of linear transformations, derive the distribution of X.