Lab 4

.odt

School

Eastern Oregon University *

*We aren’t endorsed by this school

Course

243Z

Subject

Statistics

Date

Jun 4, 2024

Type

odt

Pages

Uploaded by SuperHumanUniverse16204

Tim Kretzschmar STAT-243Z Lab 4 Q1 Statistics Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximu m Sample 100 100 0 97.19 1.51 15.08 54.12 85.68 96.05 107.60 132.49 Sample 30 30 0 104.06 2.49 13.63 77.41 92.51 104.00 114.86 124.28 Descriptive Statistics N Mean StDev SE Mean 95% CI for μ 100 97.19 15.08 1.50 (94.25, 100.13) : population mean of Sample 100 μ Known standard deviation = 15 Descriptive Statistics N Mean StDev SE Mean 95% CI for μ 30 104.06 13.63 2.74 (98.69, 109.43) : population mean of Sample 30 μ Known standard deviation = 15 Compare and contrast the set of confidence intervals for Sample100 with the ones for Sample30. Explain any differences that you see. (Hint: Consider how sample size affects the Standard Error.) The sample100 is narrower (94.25,100.13), due to a larger sample size, and sample30 is wider (98.69,109.43), due to a smaller sample size. Sample100 has a smaller mean, 97.19, compared to sample30, 104.06. Sample100 has a larger StDev, 15.08, and Sample30 has a smaller StDev, 13.63. Q2 Descriptive Statistics Sample N Mean StDev SE Mean Treatment 59 683.41 31.64 4.12 Control 59 647.17 29.75 3.87 Estimation for Paired Difference Mean StDev SE Mean 5% Upper Boun d for μ_difference 36.24 31.48 4.10 29.39 µ_difference: population mean of (Treatment - Control) Test Null hypothesis H : _difference = ₀ μ 0 Alternative hypothesis H : _difference < ₁ μ 0 T-Value P-Value 8.84 1.000 1-Sample T-test: Descriptive Statistics N Mean StDev SE Mean 95% CI for μ 59 36.24 31.48 4.10 (28.03, 44.44) : population mean of Difference μ Test Null hypothesis H : = 0 ₀ μ Alternative hypothesis H : ≠ 0 ₁ μ T-Value P-Value 8.84 0.000

2-Sample T-test: Method : population mean of μ₁ Treatment µ : population mean of Control ₂ Difference: - µ μ₁ ₂ Equal variances are not assumed for this analysis. Descriptive Statistics Sample N Mean StDev SE Mean Treatment 59 683.4 31.6 4.1 Control 59 647.2 29.8 3.9 Estimation for Difference Differenc e 95% Lower Bound for Difference 36.24 26.86 Test Null hypothesis H : - µ = ₀ μ₁ ₂ 0 Alternative hypothesis H : - µ > 0 ₁ μ₁ ₂ T-Value DF P-Value 6.41 115 0.000 What conclusions can you draw from each test? Write at least full sentence for both tests with your analysis. Which test seems better, the Paired-t or the 2-Sample t? Why? Can we always choose either one? (Hint: Consider the possible results for the test and how likely you are to come to an incorrect conclusion.) 1 sample test has a p-value = 0, so we reject the null. There is enough evidence to support the claim that the mean of treatment is different from the mean of control. 2 sample test has a p-value = 0, so we reject the null. There is enough evidence to support the claim that the mean of treatment is different from the mean of control. 1 sample test offers a larger T-value compared to the 2 sample test. 2-sample t-test, we need to assume that the data from both samples are normally distributed and they have the same variances. The paired t-test, we only require that the difference of each pair is normally distributed. 2-sample is used for independent date vs Paired when data is a matched pair. Depending on the data set, each is better in its own regards and if the wrong test is used, your data will be wrong.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version