One-Sample Hypothesis Tests Copyright (c) 2008 by The McGraw-Hill Companies. This material is intended solely for educational purposes by licensed users of LearningStats. It may not be copied or resold for profit.

Copyright Notice Portions of MINITAB Statistical Software input and output contained in this document are printed with permission of Minitab, Inc. MINITABTM is a trademark of Minitab Inc. in the United States and other countries and is used herein with the owner's permission.

Type I and II Error

Definitions Type I error Type II error Power P(Reject H0 | H0 is true) = a Type II error P(Accept H0 | H0 is false) = b Power P(Reject H0 | H0 is false) = 1-b

One-Sample Tests Parameter Left-Tail Test Two-Tail Test Right-Tail Test Mean H0: m  130 H1: m < 130 H0: m = 130 H1: m  130 H0: m  130 H1: m > 130 Proportion H0: p  0.07 H1: p < 0.07 H0: p = 0.07 H1: p  0.07 H0: p  0.07 H1: p > 0.07 Variance H0: s2  5.76 H1: s2 < 5.76 H0: s2 = 5.76 H1: s2  5.76 H0: s2  5.76 H1: s2 > 5.76

Right-Tailed z Values z = 1.282 a = 0.10 z = 1.960 a = 0.025 z = 1.645

Common Two-Tailed z Values

Example: One Mean Given: For a certain heart procedure, last month’s sample of 26 hospital patients shows a mean length of stay of 5.423 days with a standard deviation of 1.206 days. Question: Does the average LOS differ from the U.S. average of 5.05 days? Answer: No. For a two-tailed test at a = 0.05, the p-value (p = 0.127) shows that the sample mean does not differ significantly from 5.05. The 95% CI for m includes 5.05.

Histogram and CI from Minitab Sample Statistics from Excel Statistics: One Mean Histogram and CI from Minitab Sample Statistics from Excel Mean 5.423077 Standard Error 0.236468 Median 5 Mode Standard Deviation 1.205755 Sample Variance 1.453846 Kurtosis 2.318118 Skewness 1.45598 Range Minimum 4 Maximum 9 Sum 141 Length of Stay (n = 26) Count 26

Hypotheses: One Mean H0: m = 5.05 (U.S. benchmark) H1: m  5.05 (suggested by sample) Note Since the population standard deviation is unknown, we use a t-test rather than a z-test. With d.f. = n-1 = 26 – 1 = 25, the two-tailed critical value is tcrit = 2.060 (had we assumed normality, we would have used zcrit = 1.960, which is similar).

One Mean: Minitab

Calculations: One Mean Applying the Formula: Results from Minitab:

Example: One Proportion Given: For the entire nation, the mortality rate for patients over age 65 during a certain type of surgery is 0.014 (i.e., 14 deaths per 1,000 surgeries). At Carver Hospital last year, 473 such surgeries were performed, resulting in 9 deaths (p = 9/473 = 0.019). Question: At a = 0.05, does Carver's death rate exceed the national average? Answer: No. For a right-tailed test, the p-value (p = 0.222 for binomial, or p = 0.176 for normal) shows that the sample proportion does not significantly exceed 0.014. The binomial is exact, but the normal is acceptable since np0 = (473)(0.014) = 6.62 which exceeds 5.

Hypotheses: One Proportion H0: p  0.014 (U.S. benchmark) H1: p > 0.014 (suggested by sample) Note Since np0 = (473)(0.014) = 6.62 exceeds 10, we could justify assuming normality and using a z-test with a critical value of zcrit = 1.645. However, MINITAB will also do the test without assuming normality (the exact binomial test) which would be even better. In this case, the tests agree.

One Proportion: MINITAB

One Proportion: MINITAB Copyright Notice Portions of MINITAB Statistical Software input and output contained in this document are printed with permission of Minitab, Inc. MINITABTM is a trademark of Minitab Inc. in the United States and other countries and is used herein with the owner's permission.

Calculations: One Proportion Applying the Formula: Note Since this is a right-tailed test, the confidence interval is one-sided. Results from Minitab:

Example: One Variance Given: Historical data show that the mean prone resting systolic blood pressure (BP) for male Army aged 18-24 is 126 with a standard deviation of 7. A sample of 67 recent male Army recruits in this age group shows a mean prone resting systolic blood pressure of 128.46 with a standard deviation of 6.59. Question: At a = 0.05, does the recruits’ BP variance differ from the historical variance of 49 (i.e., 72)? Answer: No. The test statistic of 58.49 is within the range of the lower and upper 2.5% critical values for chi-square (45.43 to 90.35) so the sample variance does not differ significantly from 49. We ignore the sample mean since it not relevant to this test.

Statistics: One Variance Minitab’s 95% CI for s Sample Statistics from Excel Mean 128.4627 Standard Error 0.805153 Median 129 Mode Standard Deviation 6.590462 Sample Variance 43.43419 Kurtosis -0.3166 Skewness -0.23836 Range 30 Minimum 113 Maximum 143 Sum 8607 Systolic Pressure (n = 67) Count 67

Hypotheses: One Variance H0: s2 = 49 (historical benchmark) H1: s2  49 (suggested by sample) Note Using the sample data, MINITAB's Graphical Summary shows that the 95% CI for the true standard deviation includes 7. This provides a two-tailed test of the hypothesis that s = 7 at a = 0.05, which is equivalent to testing the preceding hypotheses.

Calculations: One Variance Applying the Formula: Conclusion The test statistic lies between the 95% critical values of chi-square, so we accept the null hypothesis. Critical Values of Chi-Square Lower 2.5% is 45.4314 Upper 2.5% is 90.3488 Excel function CHIINV(0.975,66) Excel function CHIINV(0.025,66)