1. One-Sample Hypothesis Tests
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2. 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.

3. Type I and II Error

4. 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

5. 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

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

7. Common Two-Tailed z Values

8. Example: One Mean Given:
For a certain heart procedure, last month’s sample of 26 hospital patients shows a mean length of stay of days with a standard deviation of 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 The 95% CI for m includes 5.05.

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

10. Hypotheses: One Mean H0: m = 5.05 (U.S. benchmark)
H1: m  (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 = (had we assumed normality, we would have used zcrit = 1.960, which is similar).

11. One Mean: Minitab

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

13. Example: One Proportion
Given:
For the entire nation, the mortality rate for patients over age 65 during a certain type of surgery is (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 = for binomial, or
p = for normal) shows that the sample proportion does not significantly exceed The binomial is exact, but the normal is acceptable since np0 = (473)(0.014) = 6.62 which exceeds 5.

14. Hypotheses: One Proportion
H0: p  (U.S. benchmark)
H1: p > (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 = 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.

15. One Proportion: MINITAB

16. 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.

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

18. Example: One Variance Given:
Historical data show that the mean prone resting systolic blood pressure (BP) for male Army aged 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 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 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.

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

20. Hypotheses: One Variance
H0: s2 = (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.

21. 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
Upper 2.5% is
Excel function CHIINV(0.975,66)
Excel function CHIINV(0.025,66)