• Statistical Inference: Use of a subset of a population (the sample) to draw conclusions about the entire population.

  The validity of inference is related to the way the data are obtained, and to the stationarity of the process producing the data.

  For valid inference the data must be obtained using a probability sample. The simplest probability sample is a simple random sample (SRS).

• The Models We will study
  • The C+E model

    $$Y=\mu+\epsilon,$$

    where $\mu$ is a model parameter representing the center of the population, and $\epsilon$ is a random error term (hence the name C+E).
  • The binomial model

• Types of Inference
  • Estimation of model parameters
  • Prediction of a future observation
  • Tolerance interval
• Estimation for the C+E Model: Point Estimation
  • Least absolute errors finds m to minimize

    $$\mbox{SAE}(m)=\sum_{i=1}^{n}\vert y_{i}-m\vert.$$

    For the C+E model, the least absolute errors estimator is the sample median, Q₂.
  • Least squares finds m to minimize

    $$\mbox{SSE}(m)=\sum_{i=1}^{n}(y_{i}-m)^2.$$

    For the C+E model, the least squares estimator is the sample mean, $\overline{Y}$.

• Example:

  Recall the example from Chapter 4:

  One stage of a manufacturing process involves a manually-controlled grinding operation. Management suspects that the grinding machine operators tend to grind parts slightly larger rather than slightly smaller than the target diameter, 0.75 inches while still staying within specification limits, which are 0.75 $\pm$ 0.01 inches. To verify their suspicions, they sample 150 within-spec parts. Summary measures and graphs are displayed on the following output.

  We will assume these data were generated by the C+E model:

  $$Y=\mu+\epsilon.$$

  The data are found in the SAS data set sasdata.grind. Looking at these data, we find the the sample mean is 0.7518 and the sample median is 0.7526. So, using the least absolute errors criterion, the estimate of $\mu$ is 0.7526, and using least squares, the estimate of $\mu$ is 0.7518.
• Estimator or Estimate?
  • The Randomness in a Set of Data From a Designed Study Is in the Production of the Data: Measuring, Sampling, Treatment Assignment, etc.
  • An estimator is a rule for computing a quantity from a sample that is to be used to estimate a model parameter.
  • An estimate is the value that rule gives when the data are taken.

• Estimation for the C+E Model: Sampling Distributions

  The distribution model of an estimator is called its sampling distribution. For example, in the C+E model, the least squares estimator $\overline{Y}$, has a $N(\mu,\sigma^2/n)$ distribution (its sampling distribution):

  • Exactly, if $\epsilon \sim N(0,\sigma^2)$
  • Approximately, if n is large enough. The CLT guarantees it!

• Confidence Intervals

  A level L confidence interval for a parameter $\theta$ is an interval $(\hat{\theta}_1,\; \hat{\theta}_2)$, where $\hat{\theta}_1$ and $\hat{\theta}_2$ are estimators having the property that

  $$P(\hat{\theta}_1<\theta<\hat{\theta}_2)=L.$$

• Estimation for the C+E Model:

  Confidence Interval for $\mu$: Known Variance

  Suppose we know $\sigma^2$. Then if $\overline{Y}$ can be assumed to have a $N(\mu,\sigma^2/n)$ sampling distribution, we know that

  $$Z=\frac{(\overline{Y}-\mu)}{\sigma/\sqrt{n}}=\frac{\sqrt{n} (\overline{Y}-\mu)}{\sigma}$$

  has a N(0,1) distribution, so

  Noting that

  $$z_{\frac{1-L}{2}}=-z_{\frac{1+L}{2}},$$

  we obtain the formula for a level L confidence interval for $\mu$:

  $$\left(\overline{Y}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+L}{2}}, \overline{Y}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+L}{2}}\right).$$

  Denoting the standard error of $\overline{Y}$, $\sigma/\sqrt{n}$, by $\sigma(\overline{Y})$, we have the formula

  $$\left(\overline{Y}-\sigma(\overline{Y})z_{\frac{1+L}{2}}, \overline{Y}+\sigma(\overline{Y})z_{\frac{1+L}{2}}\right).$$

• The Interpretation of Confidence Intervals

  The confidence level, L, of a level L confidence interval for a parameter $\theta$ is interpreted as follows: Consider all possible samples that can be taken from the population described by $\theta$and for each sample imagine constructing a level L confidence interval for $\theta$. Then a proportion L of all the constructed intervals will really contain $\theta$.

• Example:

  Recall again the example from Chapter 4:

  One stage of a manufacturing process involves a manually-controlled grinding operation. Management suspects that the grinding machine operators tend to grind parts slightly larger rather than slightly smaller than the target diameter, 0.75 inches while still staying within specification limits, which are 0.75 $\pm$ 0.01 inches. To verify their suspicions, they sample 150 within-spec parts. Summary measures and graphs are displayed on the following output.

  We will assume these data were generated by the C+E model:

  $$Y=\mu+\epsilon.$$

  Suppose we know $\sigma=0.0048$. Then

  $$\sigma(\overline{Y})=\frac{\sigma}{\sqrt{n}}= \frac{0.0048}{\sqrt{150}}=0.0004,$$

  and a 95% confidence interval for $\mu$ is

  $$\left(\overline{Y}-\sigma(\overline{Y})z_{0.975}, \overline{Y}+\sigma(\overline{Y})z_{0.975} \right)$$

  =(0.7518-(0.0004)(1.96),0.7518+(0.0004)(1.96))

  =(0.7510,0.7526).

  Based on these data, we estimate that $\mu$ lies in the interval (0.7510,0.7526). As all values in this interval exceed 0.75, we conclude that the true mean diameter, $\mu$, is greater than 0.75. We are 95% confident in our conclusion, meaning that in repeated sampling, 95% of all intervals computed in this way will contain the true value of $\mu$.

• Estimation for the C+E Model:

  Classical Confidence Interval for $\mu$: Unkown Variance If $\sigma$ is unknown, estimate it using the sample standard deviation, S. This means that instead of computing the exact standard error of $\overline{Y}$, we use the estimated standard error,

  $$\hat{\sigma}(\overline{Y})=\frac{S}{\sqrt{n}}.$$

  However, the resulting standardized estimator,

  $$t=\frac{\overline{Y}-\mu}{\hat{\sigma}(\overline{Y})},$$

  now has a t_n-1, rather than a N(0,1), distribution. The result is that a level L confidence interval for $\mu$ is given by

  $$\left(\overline{Y}-\hat{\sigma}(\overline{Y})t_{n-1,\frac{1+... ...line{Y}+\hat{\sigma}(\overline{Y})t_{n-1,\frac{1+L}{2}}\right).$$

• Example:

  Recall the example from Chapter 4:

  For these data, n=150 and s=0.0048, which means that $\hat{\sigma}(\overline{Y})=\frac{0.0048}{\sqrt{149}}=0.0004$.In addition, $t_{n-1,\frac{1+L}{2}}=t_{149,0.975}=1.976$,so a level 0.95 confidence interval for $\mu$ is

  (0.7518-(0.0004)(1.976),0.7518+(0.0004)(1.976))

  =(0.7510,0.7526).

  This interval is identical (to four decimal places) with the interval computed assuming $\sigma$ known because for large n (and 150 is large), the t_n-1 distribution is very close to the N(0,1). This is reflected in the closeness of z_0.975=1.96 to t_149,0.975=1.976.
• Classical Prediction for the C+E Model

  The problem is to predict a new (i.e. not yet available) observation from the C+E model using presently available data. To see what is involved, suppose we know $\mu$. Then it can be shown that we should predict the new observation to be $\mu$. However, even using this knowledge, we will still have prediction error:

  $$Y_{new}-\hat{Y}_{new}=Y_{new}-\mu=(\mu+\epsilon_{new})- \mu=\epsilon_{new},$$

  where Y_new is the new observation and $\hat{Y}_{new}$ is the predictor. The variance of prediction, $\sigma^2(Y_{new}-\hat{Y}_{new})$, is therefore $\sigma^2$, the variance of the model's error distribution.

  We won't know $\mu$, however, so we estimate it from the present data by computing $\hat{\mu}=\overline{Y}$, and use this as the predictor of the new observation. When $\hat{\mu}$ is used for prediction instead of estimation, we call it $\hat{Y}_{new}$. When using $\hat{Y}_{new}$ to predict a new observation, the prediction error is

  $$Y_{new}-\hat{Y}_{new}=(\mu+\epsilon_{new})-\hat{Y}_{new}= (\mu-\hat{Y}_{new})+\epsilon_{new}.$$

  $\mu-\hat{Y}_{new}$ is the error due to using $\hat{\mu}$ to estimate $\mu$. Its variance, as we have already seen, is $\sigma^2/n$.$\epsilon_{new}$ is the random error inherent in Y_new. Its variance is $\sigma^2$. Since these terms are independent, the variance of their sum is the sum of their variances.

  In most applications $\sigma$ will not be known, so we estimate it with the sample standard deviation S, giving the estimated standard error of prediction

  $$\hat{\sigma}(Y_{new}-\hat{Y}_{new})=S \sqrt{1+\frac{1}{n}}.$$

  A classical level L prediction interval for a new observation is then

  $$\hat{Y}_{new}\pm \hat{\sigma}(Y_{new}-\hat{Y}_{new}) t_{n-1,\frac{1+L}{2}}.$$

• Example:

  We return to the grinding example from Chapter 4. Recall that for these data, $\overline{y}=0.7518$, so that the predicted value is $\hat{y}_{new}=\overline{y}=0.7518$. Also, n=150 and s=0.0048, which means that

  $$\hat{\sigma}(\hat{Y}_{new})=0.0048 \sqrt{1+\frac{1}{150}}=0.00482.$$

  In addition, $t_{n-1,\frac{1+L}{2}}=t_{149,0.975}=1.976$,so a level 0.95 prediction interval for the diameter of a new piece is:

  (0.7518-(0.00482)(1.976),0.7518+(0.00482)(1.976))

  =(0.7422,0.7614).

• Estimation for the Binomial Model:

  Exact Confidence Interval for p

  Suppose we observe Y successes in the n trials. Then a level L confidence interval for p is (p_D,p_U), where

  • if Y>0, p_D is the unique solution of

    $$\sum_{y=Y}^n \frac{n!}{y!(n-y)!}p_D^y(1-p_D)^{n-y}=(1-L)/2,$$

  • if Y=0, p_D=0,

  and

  • if Y<n, p_U is the unique solution of

    $$\sum_{y=0}^Y \frac{n!}{y!(n-y)!} p_U^y(1-p_U)^{n-y}=(1-L)/2,$$

  • if Y=n, p_U=1.
• Estimation for the Binomial Model:

  Classical Estimation for Large Samples

  Suppose $Y\sim b(n,p)$, where n is large (rule of thumb: Y and n-Y exceed 10). Let $\hat{p}=Y/n$ be the sample proportion of successes, and let $\hat{\sigma}(\hat{p})=\sqrt{\hat{p}(1-\hat{p})/n}$be its estimated standard error. Then by the CLT,

  $$\frac{\hat{p}-p}{\hat{\sigma}(\hat{p})} \sim N(0,1),$$

  approximately. This means that an approximate level L confidence interval for p is

  $$(\hat{p}-\hat{\sigma}(\hat{p})z_{\frac{1+L}{2}},\; \hat{p}+ \hat{\sigma}(\hat{p})z_{\frac{1+L}{2}}).$$

• Example: We'll once again consider the grinding example from Chapter 4, but this time in its original form. Recall that 150 parts were sampled at random and that 93 had diameters greater than the specification diameter. We will use these data to obtain level 0.99 confidence intervals for p, the true population proportion of parts with diameters greater than spec.
  • Exact interval The interval is (p_D,p_U), where p_D is the unique solution of

    $$\sum_{y=93}^{150} \frac{150!}{y!(150-y)!}p_D^y(1-p_D)^{150-y}=0.005,$$

    and p_U is the unique solution of

    $$\sum_{y=0}^{93} \frac{150!}{y!(150-y)!} p_U^y(1-p_U)^{150-y}=0.005.$$

  • Large sample classical interval Here the observed value of $\hat{p}$ is $\hat{p}=93/150=0.62$, so that $\hat{\sigma}(\hat{p})=\sqrt{0.62(1-0.62)/150}=0.0396$. Also, z_0.995=2.5758. Therefore a level 0.99 large sample classical interval is

    (0.62-(0.0396)(2.5758),0.62+(0.0396)(2.5758))

    =(0.52,0.72).

  As can be seen, in this case both intervals agree closely. In particular, as each interval contains only values exceeding 0.5, we can conclude with 99% confidence that more than half the population diameters exceed spec.

• Determination of Sample Size

  One consideration in designing an experiment or sampling study is the precision desired in estimators or predictors. Precision of an estimator is a measure of how variable that estimator is. Another equivalent way of expressing precision is the width of a level L confidence interval. For a given population, precision is a function of the size of the sample: the larger the sample, the greater the precision.

  Suppose it is desired to estimate a population proportion p to within d units with confidence level at least L. If we assume a large enough sample size (so the normal approximation can be used in computing the confidence interval), the requirement is that one half the length of the confidence interval equal d, or

  $$z_{\frac{1+L}{2}}\sqrt{p(1-p)/n}=d$$

  Solving this equation for n gives the required sample size as

  $$n=(p(1-p)\cdot z_{\frac{1+L}{2}}^{2})/d^{2}$$

  If we don't know p, we can get an estimate from a pilot experiment or study. Or, since $p(1-p)\leq .25$, we can use .25 in place of p(1-p) in the formula.

  There is an analogous formula when a simple random sample will be used and it is desired to estimate a population mean $\mu$ to within d units with confidence level at least L. If we assume a large enough sample size (so the normal approximation can be used in computing the confidence interval), the required sample size is

  $$n=(\sigma^2 \cdot z_{\frac{1+L}{2}}^{2})/d^{2}.$$

  Again, this supposes we know $\sigma^2$. If we don't, we can get an estimate from a pilot experiment or study.
• The Two Population C+E Model

  We assume that there are n₁ measurements from population 1 generated by the C+E model

  $$Y_{1,i}=\mu_1+\epsilon_{1,i},\; i=1, \ldots, n_1,$$

  and n₂ measurements from population 2 generated by the C+E model

  $$Y_{2,i}=\mu_2+\epsilon_{2,i},\; i=1, \ldots, n_2.$$

  We want to compare $\mu_1$ and $\mu_2$.

• Estimation for Paired Comparisons

  Sometimes each observation from population 1 is paired with another observation from population 2. For example, each student may take a pre- and post-test. In this case n₁=n₂ and by looking at the pairwise differences, D_i=Y_1,i-Y_2,i, we transform the two population problem to a one population problem for C+E model $D=\mu_D+\epsilon_D$, where $\mu_D=\mu_1-\mu_2$ and $\epsilon_D=\epsilon_1-\epsilon_2$. Therefore, a confidence interval for $\mu_1-\mu_2$ is obtained by constructing a one sample confidence interval for $\mu_D$.

• Example:

  The manufacturer of a new warmup bat wants to test its efficacy. To do so, it selects a random sample of 12 baseball players from among a larger number who volunteer to try the bat. For each player, company researchers compute D, the difference between the player's test year average and his pervious year's average. Assuming that these differences follow a C+E model, they construct a level 0.95 confidence interval for the difference in mean batting average, $\mu_D$.The data (found in SASDATA.BATTING) are:

  PLAYER BEFORE AFTER DIFF

  1 0.254 0.262 0.008

  2 0.274 0.290 0.016

  3 0.300 0.304 0.004

  4 0.246 0.267 0.021

  5 0.278 0.291 0.013

  6 0.252 0.257 0.005

  7 0.235 0.248 0.013

  8 0.313 0.324 0.021

  9 0.305 0.317 0.012

  10 0.255 0.252 -0.003

  11 0.244 0.276 0.032

  12 0.322 0.332 0.010

  An inspection of the differences shows no evidence of nonnormality or outliers. For these data, $\overline{d}=0.0127$, s_d=0.0092 and t_11,0.975=2.201. Then $\hat{\sigma}(\overline{D})=0.0092/\sqrt{12}=0.0027$, so the desired interval is

  $$0.0127\pm (0.0027)(2.201)=(0.0068,0.0185).$$

  Based on this, we estimate that the mean batting average increases over the previous year by somewhere between 0.0068 and 0.0185.
• Classical Estimation for Independent Populations

  Let $\overline{Y}_1$ and $\overline{Y}_2$ denote the sample means from populations 1 and 2, S₁² and S₂² the sample variances. The point estimator of $\mu_1-\mu_2$, is $\overline{Y}_1-\overline{Y}_2$.

  • Equal Variances

    If the population variances are equal ($\sigma_1^2=\sigma_2^2=\sigma^2$), then we estimate $\sigma^2$ by the pooled variance estimator

    $$S^2_p=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}.$$

    The estimated standard error of $\overline{Y}_1-\overline{Y}_2$ is then given by

    $$\hat{\sigma}_p(\overline{Y}_1-\overline{Y}_2)= \sqrt{S_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}.$$

    $$t^{(p)}=\frac{\overline{Y}_1-\overline{Y}_2-(\mu_1-\mu_2)}{\hat{\sigma}_p(\overline{Y}_1-\overline{Y}_2)}$$

    has a t_n1+n2-2 distribution. This leads to a level L pooled variance confidence interval for $\mu_1-\mu_2$:

    $$\overline{Y}_1-\overline{Y}_2 \pm \hat{\sigma}_p(\overline{Y}_1- \overline{Y}_2)t_{n_1+n_2-2,\frac{1+L}{2}}$$

  • Unequal Variances

    If $\sigma_1^2 \neq \sigma_2^2$, an approximate level L confidence interval for $\mu_1-\mu_2$ is

    $$\overline{Y}_1-\overline{Y}_2 \pm \hat{\sigma}(\overline{Y}_1-\overline{Y}_2)t_{\nu,\frac{1+L}{2}},$$

    where $\nu$ is the largest integer less than or equal to

    $$\frac{\left(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}\right)^2} {\... ...ht)^2}{n_1-1}+\frac{\left(\frac{S_2^2} {n_2}\right)^2}{n_2-1}},$$

    and

    $$\hat{\sigma}(\bar{Y}_1-\bar{Y}_2)=\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2} {n_2}}.$$

• Example:

  A company buys cutting blades used in its manufacturing process from two suppliers. In order to decide if there is a difference in blade life, the lifetimes of 10 blades from manufacturer 1 and 13 blades from manufacturer 2 used in the same application are compared. A summary of the data shows the following (units are hours):

  $$\overline{y}$$ Manufacturer n s

  1 10 118.4 26.9

  2 13 134.9 18.4

  Obtain a level 0.90 confidence interval to compare the mean lifetimes of blades from the two manufacturers.

  The experimenters generated histograms and normal quantile plots of the two data sets and found no evidence of nonnormality or outliers. The estimate of $\mu_1-\mu_2$ is $\overline{y}_1-\overline{y}_2=118.4- 134.9=-16.5$.

  • Pooled variance interval The pooled variance estimate is

    $$s^2_p=\frac{(10-1)(26.9)^2+(13-1)(18.4)^2}{10+13-2}=503.6.$$

    This gives the standard error estimate of $\overline{Y}_1-\overline{Y}_2$ as

    $$\hat{\sigma}_p(\overline{Y}_1-\overline{Y}_2)= \sqrt{503.6\left(\frac{1}{10}+\frac{1}{13}\right)}=9.44.$$

    Finally, t_21,0.95=1.7207. So a level 0.90 confidence interval for $\mu_1-\mu_2$ is

    $$(-16.5-(9.44)(1.7207),\;-16.5+(9.44)(1.7207))$$

    =(-32.7,-0.3).

  • Separate variance interval The standard error estimate of $\overline{Y}_1-\overline{Y}_2$ is

    $$\hat{\sigma}(\overline{Y}_1-\overline{Y}_2)=\sqrt{\frac{(26.9)^2}{10}+\frac{(18.4)^2}{13}}=9.92.$$

    The degrees of freedom $\nu$ is computed as the greatest integer less than or equal to

    $$\frac{\left(\frac{(26.9)^2}{10}+\frac{(18.4)^2}{13}\right)^2... ...}{10-1}+\frac{\left(\frac{(18.4)^2}{13}\right)^2}{13-1}}=15.17,$$

    so $\nu=15$. Finally, t_15,0.95=1.7530. So a level 0.90 confidence interval for $\mu_1-\mu_2$ is

    $$(-16.5-(9.92)(1.753),\; -16.5+(9.92)(1.753))$$

    =(-33.9,0.89).

• Comparing Two Population Proportions: Classical Estimation for Large Samples

  $Y_1 \sim b(n_1,p_1)$ and $Y_2 \sim b(n_2,p_2)$ are observations from two independent populations. Estimator of p₁-p₂ is

  $$\hat{p}_1-\hat{p}_2=\frac{Y_1}{n_1}-\frac{Y_2}{n_2}.$$

  Its estimated standard error is

  $$\hat{\sigma}(\hat{p}_1-\hat{p}_2)=\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+ \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$$

  If Y₁, Y₂, n₁-Y₁ and n₂-Y₂ >10 we may use the following approximate level L confidence interval for p₁-p₂:

  $$(\hat{p}_1-\hat{p}_2-\hat{\sigma}(\hat{p}_1-\hat{p}_2)z_{\fr... ...1-\hat{p}_2+\hat{\sigma}(\hat{p}_1-\hat{p}_2)z_{\frac{1+L}{2}})$$

• Example:

  In a recent survey on academic dishonesty 26 of the 200 female college students surveyed and 26 of the 100 male college students surveyed agreed or strongly agreed with the statement ``Under some circumstances academic dishonesty is justified.'' With 95% confidence estimate the difference in the proportions p_f of all female and p_m of all male college students who agree or strongly agree with this statement.

  The point estimate of p_f-p_m is

  $$\hat{p}_f-\hat{p}_m=26/200-26/100=-0.13.$$

  It's estimated standard error is

  $$\hat{\sigma}(\hat{p}_1-\hat{p}_2)=\sqrt{\frac{0.13(1-0.13)}{200}+ \frac{0.26(1-0.26)}{100}}$$

  =0.05.

  Since Y_f=26, 200-Y_f=174, Y_m=26, and 100-Y_m=74 all exceed 10, we may use the normal approximation, which gives the interval

  (-0.13-(0.05)(1.96),-0.13+(0.05)(1.96))

  =(-0.228,-0.032).

• Tolerance Intervals

  Tolerance intervals are used to give a range of values which, with a pre-specified confidence, will contain at least a pre-specified proportion of the measurements in the population. Suppose T₁ and T₂ are estimators with $T_1\leq T_2$, and that $\gamma$ is a real number between 0 and 1. Let $A(T_1,T_2,\gamma)$ denote the event

  {The proportion of measurements in the population between T₁ and T₂ is at least $\gamma$}.

  Then a level L tolerance interval for a proportion $\gamma$ of a population is an interval $(T_1,\; T_2)$, where T₁ and T₂ are estimators, having the property that

  $$P(A(T_1,T_2,\gamma))=L.$$

• Normal Theory Tolerance Intervals

  If we can assume the data are from a normal population, a level L tolerance interval for a proportion $\gamma$ of the population is given by

  $$\overline{Y} \pm KS,$$

  where $\overline{Y}$ and S are the sample mean and standard deviation, and K is a mathematically derived constant depending on n, L and $\gamma$(Found in Table A.8, p. 359 in the book).
• Example:

  Refer again to the grinding data. The mean diameter of the n=150 parts is 0.7518 and the standard deviation is 0.0048. For level 0.90 normal theory tolerance interval for a proportion 0.95 of the data, the constant K is obtained by simple interpolation to be 2.137. The interval is then

  $$(0.7518-(2.137)(0.0048),\;0.7518+(2.137)(0.0048))$$

  $$=(0.7415,\;0.7621).$$

So with 90% confidence, we estimate that at least 95% of the population diameters lie between 0.7415 and 0.7621.