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Statistical Analysis of One-Sample Means and Proportions, Exams of Mathematical Statistics

The results of statistical tests and confidence intervals for the means and proportions of one sample. The tests include one-sample t-tests for means and tests for one proportion. The document also includes calculations for power and sample size. The data presented is related to the variable c1.

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

koofers-user-tod
koofers-user-tod 🇺🇸

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9.30
(a) One-Sample T: C1
Test of mu = 22.5 vs not = 22.5
Variable N Mean StDev SE Mean 95% CI T P
C1 5 22.4960 0.3783 0.1692 (22.0262, 22.9658) -0.02 0.982
The P-value of .982 says we get data favoring HA by this much 98.2% of the time when
H0 is actually true. We do not reject H0 since P-value > 0.05. There is no evidence to
contradict H0, so the results are not significant.
(b) With 5 observations it is pretty hard to tell, but there is no evidence to contradict an
assumption of normality
C1
23.0
22.8
22.6
22.4
22.2
22.0
Boxplot of C1
(c) Don’t worry about calculating power
(d) Power and Sample Size
1-Sample t Test
Testing mean = null (versus not = null)
Calculating power for mean = null + difference
Alpha = 0.05 Assumed standard deviation = 0.3783
Sample Target
Difference Size Power Actual Power
0.25 27 0.9 0.910618
(e) 22.5 is in the 95% CI for µ, so we would not reject H0
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(a) One-Sample T: C

Test of mu = 22.5 vs not = 22. Variable N Mean StDev SE Mean 95% CI T P C1 5 22.4960 0.3783 0.1692 (22.0262, 22.9658) -0.02 0.

The P-value of .982 says we get data favoring HA by this much 98.2% of the time when

H 0 is actually true. We do not reject H 0 since P-value > 0.05. There is no evidence to

contradict H 0 , so the results are not significant.

(b) With 5 observations it is pretty hard to tell, but there is no evidence to contradict an

assumption of normality

C

Boxplot of C

(c) Don’t worry about calculating power

(d) Power and Sample Size

1-Sample t Test Testing mean = null (versus not = null) Calculating power for mean = null + difference Alpha = 0.05 Assumed standard deviation = 0. Sample Target Difference Size Power Actual Power 0.25 27 0.9 0.

(e) 22.5 is in the 95% CI for μ, so we would not reject H 0

(a) One-Sample T: C

Test of mu = 98.6 vs not = 98. Variable N Mean StDev SE Mean 95% CI T P C1 25 98.2640 0.4821 0.0964 (98.0650, 98.4630) -3.48 0.

The P-value of 0.002 says we get data favoring the alternative hypothesis by this much

only 0.2% of the time when μ is actually 98.6. Since this is less than 0.05, we reject H 0

and conclude μ is not 98.6. The results are significant (we have “proven” μ ≠ 98.6).

(b) Don’t worry about calculating power.

(c) Power and Sample Size

1-Sample t Test Testing mean = null (versus not = null) Calculating power for mean = null + difference Alpha = 0.05 Assumed standard deviation = 0. Sample Target Difference Size Power Actual Power 0.4 18 0.9 0.

(d) 98.6 is not in the 95% CI, so we would not find 98.6 to be a plausible value of μ.

(e) There are no outliers or other reason to be very concerned about the normality

assumption.

C

Boxplot of C

The remaining problems in this section (32-25 are done similarly).