Chapter 14. Introduction to Inference

Transcription

1 Chapter 14. Introduction to Inference Topics covered in this chapter: Estimating µ with a 95% Confidence Interval Estimating µ with a 99% Confidence Interval Significance Test with a One-sided P-value Significance Test with a Two-sided P-value Estimating µ with a 95% Confidence Interval (s is known) Example 14.3: Healing of skin wounds The Problem: Biologists studying the healing of skin wounds measured the rate at which new cells closed a razor cut made in the skin of an anesthetized newt. Estimate the mean healing rate using a 95% confidence interval. 1. Open dataset eg14-03.por. 2. Find the sample statistics. a. Click on Analyze. b. Click on Descriptive Statistics. c. Click on Descriptives. d. Move Rate to the Variable(s) box. e. Click OK. 103

2 Introduction to Inference 104 Here we are given the sample mean and sample standard deviation. However, we are given σ = 8 in this example. We will simply use the sample mean from this output. 3. Find the confidence interval. a. Click on Variable View. b. Add the variables xbar, sigma, and n. Change the number of decimals of xbar to 4. c. Click on Data View. d. Under xbar, type for the sample mean given in the table above. e. Under sigma, type 8 for the population standard deviation. f. Under n, type 18 for the sample size.

3 105 Chapter Compute the confidence interval. For 95% confidence intervals, remember to use z * = a. Click on Transform. b. Click on Compute Variable. c. Under Target Variable, type lowerbound. d. Under Numeric Expression, type xbar 1.96 * sigma / SQRT(n). e. Click OK. f. Repeat steps c through e, using upperbound and the expression: xbar * sigma / SQRT(n).

4 Introduction to Inference 106 The 95% confidence interval for the mean healing rate for all newts of this particular species in micrometers per hour is (21.97, 29.36). Estimating µ with a 99% Confidence Interval (s is known) Example (Apply Your Knowledge 14.5): IQ test scores The Problem: Make a stemplot and construct a 99% confidence interval for the IQ scores of 31 seventh-grade girls in a Midwest school district. 1. Make a stemplot. a. Open the data set ex14-05.por. b. Click on Analyze. c. Click on Descriptive Statistics. d. Click on Explore. e. Choose the variable IQ to be in the Dependent List. f. Under Display, choose Plots. g. Click on the Plots button at the right side of the window. h. For Descriptive, choose Stem-and-leaf. i. Click Continue. j. Click OK. The stemplot will show as below in the output:

5 107 Chapter Get the necessary descriptive statistics. a. Click on Analyze. b. Click on Descriptive Statistics. c. Click on Descriptives. d. Choose IQ as the Variable(s). e. Click OK. 3. Add the variables to compute the 99% confidence interval. a. Click on Variable View. b. Add the variables xbar, sigma and n. c. Click on Data View. d. Under xbar, type in the first row. e. Under sigma, type in 15 since the population standard deviation is 15. f. Under n, type in 31 for the sample size.

6 Introduction to Inference Compute the 99% confidence interval. Recall that z * = when computing 99% confidence intervals. a. Click on Transform. b. Click on Compute Variable. c. Under Target Variable, type lowerbound. d. Under Numeric Expression, type: xbar * sigma / SQRT(n). e. Click OK. f. Repeat steps c through e, instead using upperbound and the expression: xbar * sigma / SQRT(n).

7 109 Chapter 14 The 99% confidence interval for the mean IQ scores for all seventh-grade girls in the school district is (98.90, ). Significance Test with a One-sided P-value Example 14.7: Sweetening colas: one-sided P-value The Problem: Diet colas use artificial sweeteners to avoid sugar. These sweeteners gradually lose their sweetness over time. Manufacturers therefore test new colas for loss of sweetness before marketing them. Suppose that we know that for any cola, the sweetness loss scores vary from taster to taster according to a Normal distribution with standard deviation σ = 1. The mean μ for all tasters measures loss of sweetness and is different for different colas. The study of sweetness low tests the hypotheses: H 0 : μ = 0 H a : μ > 0 Given the sample statistics, a one-sided P-value is calculated to determine the evidence against the null hypothesis. 1. Compute a one-sided p-value. a. Click on Variable View. b. Enter the variables z and probability. c. Increase the number of decimal places of probability to four. d. Click on Data View. e. Type 3.23 under z. 2. Calculate a one-sided p-value. a. Click on Transform. b. Click on Compute Variable. c. For target variable, type in probability.

8 Introduction to Inference 110 d. In numeric expression, type 1. e. Under Function group, click on CDF and Noncentral CDF. f. Double-click on Cdfnorm. g. Click on z, then click on the right-facing arrow. h. Click OK. i. Change existing variable? Click OK. The probability of obtaining a sample mean of greater than or equal to 1.02 is The small p-value leads us to believe that the null hypothesis is false.

9 111 Chapter 14 Significance test with a two-sided P-value Example 14.8: Job satisfaction: two-sided p-value The Problem: Does the job satisfaction of assembly workers differ when their work is machine-paced rather than self-paced? Assign workers either to an assembly line moving at a fixed pace or to a self-paced setting. All subjects work in both settings in random order. This is a matched pairs design. After two weeks in each setting, the workers take a test of job satisfaction. The parameter of interest is the mean μ of the differences in scores in the population of all assembly workers. Similar to example 14.7, the sample statistics are given, but in this case the twosided p-value is computed. This example will be solved using nearly identical methods in example However, in this example, z = 1.20 instead of The one method that is different is in step 2d. The value computed needs to be multiplied by two, for a two-sided p-value. Hence, the numeric expression will read: In this example, the p-value =.2301.

10 Introduction to Inference 112 Chapter 14 Exercises 14.3 Find a critical value IQ test scores Sweetening Colas: find the P-value Measuring conductivity Measuring conductivity Significance from a table Testing a random number generator I want more muscle I want more muscle Bone loss by nursing mothers Bone loss by nursing mothers Eye grease.

11 385 Chapter 14 SPSS Solutions **NOTE: SPSS does not do inference based on Z distributions, nor does it perform inference on variables that are already summarized. If you really want to use SPSS for these problems or chapters, follow the instructions below (you ll be basically using Transform, Compute Variable as a calculator) or use another technology (such as a graphing calculator or another statistics program like Minitab or Crunchit.) 14.3 We ll use Transform, Compute Variable and IDF.Normal from the Inverse DF function group. With 97.5% in the center of a standard Normal distribution, there is 0.025/2 = on each end. Due to the symmetry of the distribution, we can find either z* (and remove the negative sign) or z*. z* = As always, if you do not see all the decimal places you want, go to the Variable View and change them Open data file ex To create the stemplot, use Analyze, Descriptive Statistics, Explore. IQ Stem-and-Leaf Plot Frequency Stem & Leaf 2.00 Extremes (=<74) Stem width: 10 Each leaf: 1 case(s) This stemplot indicates there are two low outliers (=<74, namely 72 and 74). In the Descriptives block of the output, a confidence interval is given (set the confidence level using the Statistics button in the Explore dialog box). Descriptives Statistic Std. Error IQ Mean % Confidence Interval for Mean Lower Bound Upper Bound

12 386 This confidence interval is based on a distribution we won t meet until Chapter 17 ( the t distributions). To find the confidence interval, we ll calculate it by hand. We will, however, make use of the mean given above. Based on this sample, with 99% confidence, the average IQ score of all seventh-grade girls in this school district is between 98.9 and When μ = 0, the distribution of x will be N (0, 1/ 10 = 0.316). The P-value is the area to the right of x = 0.3 on this distribution. We use Transform, Compute Variable and CDF.Normal to find the P-value is = For n = 6 measurements, the standard deviation of the sampling distribution is σ =.2 / 6 = For this two-sided alternative, we ll double the area to the left of our observed sample mean (since these are less than the claimed mean). We ll find this area using Transform, Compute Variable and CDF.Normal.

13 387 A sample mean of 4.98 (very close to 5) has P-value , while the sample mean of 4.7 (much farther away from 5) has P-value ; this is much better evidence against the null as it is farther away We ll enter the data and use Analyze, Basic Statistics, Descriptives to find the mean of these data, then compute the test by hand. Descriptive Statistics N Minimum Maximum Mean Std. Deviation Conduct Valid N (listwise) 6 With the not equal (two-tailed) alternate hypothesis, the P-value of the test is twice the area below z = These data are not good evidence that the mean conductivity is not The P-value of this test is the area to the right of z = We find this using CDF.Normal, The P-value is , so this result is significant at the α = 0.05 level (P < α ), but not at the 0.01 level We compute the test statistic and P-value below.

14 388 The test statistic is z = 2.20 with P-value This result is significant at the α = 0.05 level (P < α ), but not at the 0.01 level Again, we compute the confidence interval by hand. If you don t know the value of z*, use IDF.Normal to find it. Based on this sample, the mean muscle gap for American young men (this is where the sample was from) is between 2.06 and 2.64 kg/m 2, with 90% confidence This is a continuation of Exercise (above). If we assume that μ is the difference women s preference minus what they have, we have hypotheses H0 : μ = 0, H a : μ > 0. Since the alternative is greater than the P-value will be the area above the computed test statistic. The test statistic is z = with P-value essentially 0. We know this P-value should be very small because the Rule states that being more than 3 standard deviations above or below the mean is extremely unusual Open data file ex14-51 and use Analyze, Descriptive Statistics, Explore to make the stemplot.

15 389 Change Stem-and-Leaf Plot Frequency Stem & Leaf Stem width: 1.0 Each leaf: 1 case(s) Based on the graph above, there are no strong departures from Normality, so proceeding with inference is reasonable. We use the mean computed by SPSS in calculating the confidence interval. Descriptives Change Statistic Std. Error Mean % Confidence Interval for Lower Bound Mean Upper Bound We are 99% confident the mean bone loss of all breast-feeding mothers is between 4.53% and 2.65%, based on this sample of 47 mothers If you did Exercise 14.51, you should have noted that the interval does not contain 0, indicating that breast-feeding mothers do lose bone mineral, on average, and that this result is statistically significant. We ll compute the z test statistic first for this test.

16 390 We have a test statistic of z = with P-value 0 (being almost 10 standard deviations below the mean ahs essentially no chance of happening). This confirms that breastfeeding mothers lose bone mineral, on average If μ is the mean difference in sensitivity, (with without grease), we have hypotheses H0 : μ = 0, H a : μ > 0, since the question is if grease increases sensitivity increases sensitivity. We use Analyze, Descriptive Statistics, Descriptives to find the mean of the data, then compute the test and P-value. Descriptive Statistics N Minimum Maximum Mean Std. Deviation Diff Valid N (listwise) 16 Our test statistic is z = 1.84 with P-value At the 5% level, these results indicate that eye grease does increase sensitivity, on average.