Problems: Q&A chapter 6, problems 1-10. Chapter 6: 6.1 6.5 6.11 6.27 6.35 6.41 6.45

Note: Examples in chapter 6 are based on data that come from a normal distribution: N(, ) and most usually we are estimating  based on the unrealistic assumption that  is known. Formal methods for estimating  when  is unknown follow in later chapters.

Example: Suppose we wish to compute the true mean score on the US college entrance exam, the population mean SAT score. These scores are known to be distributed normally with mean  and standard deviation . That is, if x1, x2, .., xn, represents an SRS from the population of students taking the SAT, we say that the x’s are normally distributed with mean  and standard deviation , or, more formally: x1, x2, .., xn ~ N(, ). We don’t know the mean, . Also, in a real situation, we will not know the standard deviation, . But we will proceed by assuming that we know  and that  = 100. Also, assume that the sample size is n=500.

Our objective is to estimate  from the sample data x1, x2, .., xn. We know (from chapter 5) that an unbiased estimator of  is the sample mean We also know (from chapter 5) that the sample mean also has a normal distribution with mean, the parameter we are estimating (because the sample mean is an unbiased estimator) and variance equal to  divided by the square root of the number of observations in the SRS. In other words ~ N(, /n).

The task in chapter 6 is to use this information to make formal probability statements about the unknown value . All of these statements pertain to the fact that the sample mean has a normal distribution with known variance 2.

Three important points that follow from the normal distribution: The 68-95-99.7 rule states that 95% of the time, the sample mean will lie within  2 standard deviations of the population mean . This implies that 95% of the time the population mean  lies within  2 standard deviations of the sample mean. Hence, 95% of all sample means calculated from SRS’s will contain the true value in the interval

Suppose we take an SRS and calculate the sample mean at 495. We construct the interval 4959 = (495-9, 495+9) = (486, 504). Either (486, 504) or (486, 504). The first situation arises 95% of the time and the second situation arises 5% of the time if we repeated the experiment (take an SRS, calculate the mean, calculate the interval) a very large number of times.

Confidence interval A level C confidence interval for a parameter is an interval computed from the sample data by a method that has probability C of producing an interval containing the true value of the parameter. Convention is to use C = 0.95 or C = 0.90.

Formal construction of confidence intervals This relies on two facts. First, all normal distributions have the same standardized forms. That is, Second, the central limit theorem (chapter 5) tells us that the normal distribution is approximately correct, even if the population is not normal.

Now, we find the central C area under the standard normal curve and use Table A to construct the “critical values”, -z* and +z*. For C = 0.95, z* = 1.645. For C = 0.95, z* = 1.960. For C = 0.99, z* = 2.576.

Formally:

We call the component the “margin of error.” Opportunities for reducing the margin of error are three, namely, reducing z*, reducing , or increasing n.

Controlling the margin of error Usually, we will choose conventional levels of C, for example, C = 0.95 and C = 0.90. The population standard deviation is often out of our control. Only the sample size, n, is within our control. We can select n, to set the margin of error,

Formally:

6.2 Tests of significance

Example: (example 6.6 page 453) Someone states their weight as “187 pounds.” We take four measures over a one month period, x1 = 190.5, x2 = 189.0, x3 = 195.5, x1 = 187.0. Sample mean = 190.5, and we know that the weights are normally distributed with (unknown) mean  and (known) standard deviation  = 3. Are these measures compatible with a true population mean of 187.0.

Example: (continued) First, restate: What is the probability of observing a sample mean of 190.5 or larger from a population with a true mean 187.0? Second, formalize: Null hypothesis, Ho:  = 187.0. Alternative hypothesis, Ha:   187.0. (This is a one-sided hypothesis.)

Example: (continued) Alternatively: First, restate: What is the probability of observing a sample mean of 190.5 or larger from a population with a true mean 187.0? Second, formalize: Null hypothesis, Ho:  = 187.0. Alternative hypothesis, Ha:   187.0. (This is a two-sided hypothesis.)

In this first example, we will compute the probability of observing a sample mean of 190.5 or larger when the true population mean is 1.87 to be prob = 0.01, which is 1% and, thus small. Hence, we conclude that the data are not compatible with the hypothesis, or, more precisely, we conclude that it is only 1% of the time likely that the computed sample mean could be consistent with the null hypothesis.

Suppose we take another sample and find that the computed sample mean is 188.5. This number is much closer to the hypothesized value 187.0 and, hence, it seems more likely that this result could have been generated by the hypothesized population mean 187.0. We will compute the probability of the event “obtaining a sample mean greater than or equal to 188.5 when the true population mean is 187.0 to be 0.16. So, about 16% of the time we expect such a result. This proportion is not particularly small, so we say that the data are not inconsistent with the hull hypothesis.

Calculations: All based on the z distribution, the standard-normal N(0, 1). There are two ways to proceed. First is based on a test statistic. Second is based on P-values. Both approaches exploit the z distribution.

Formally:

Once a critical value is chosen, z Once a critical value is chosen, z*, we reject the null hypothesis in favour of the alternative hypothesis if the computed value, z, lies further in the tail than the critical value z*. Important: Note the rules for one sided tests in the greater-than direction. Note the rules for one sided tests in the less-than direction. Note the rules for two-sided tests.

For one sided tests of the greater-than variety: Ho:  = o Ha:   o We reject Ho in favor of Ha if z  z*.

For one sided tests of the less-than variety: Ho:  = o Ha:   o We reject Ho in favor of Ha if z  z*.

For one sided tests of the not-equal-to variety: Ho:  = o Ha:   o We reject Ho in favor of Ha if z  zL* or z  zU*.

Example: Two-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 You select C = 0.95 and compute 1-C/2 = 0.025. From Table D (last line) you find zL* = -1.96, and zU* = +1.96. You note z > zU* and you reject the null in favour of the alternative.

Example: One-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 You select C = 0.95 and compute 1-C = 0.05. From Table D (last line) you find z* = +1.645. You note z > z* and you reject the null in favour of the alternative.

Example: One-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 You select C = 0.95 and compute 1-C = 0.05. From Table D (last line) you find z* = -1.645. You note z > z* and you do not reject the null in favour of the alternative.

Example: Two-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 You select C = 0.95 and compute 1-C/2 = 0.025. From Table D (last line) you find zL* = -1.96, and zU* = +1.96. You note zL* < z < zU* and you do not reject the null hypothesis.

Example of P-value calculations. Remember, the probability, computed assuming Ho is true, that the test statistic would take a value as extreme or more extreme than the observed value is the P-value. The smaller the P-value, the stronger the evidence against Ho.

Example: One-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 Now, if Ho is true, then z = 1.00 is a single observation from the z, N(0,1), distribution. The probability of observing a z value greater than 1.00 is (from table A) p = 1.00 - 0.8413 = 0.1587.

Example: Two-sided significance test. Ho:  = o = 187.0 Ha:   o = 187.0 Now, if Ho is true, then z = 2.33 is a single observation from the z, N(0,1), distribution. The probability of observing a z value greater than 2.33 is (from table A) p = 1-0.9901 = 0.0099, but, because this is a two-sided test, we say that the P-value is p = 2  0.0099 = 0.0198.

We may cover this section next week, or not. Please look at chapter 7 in advance, especially sections 7.1 and 7.2.