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The null and alternate hypothesis are:
H0:
Ha:
Thus, it is a left-tailed test.
Since this a left tailed test, so the p-value is given by:
Hence option D is the correct option.
Consider the data that is summarized in the Minitab output below. Descriptive Statistics: C1, C2 Variable...
Consider the data set that is summarized in the Minitab Output below. -------------------------------------------------------------------------------------------------- Source DF Adj SS Adj MS F-Value P-Value C2 2 67.46 33.73 7.36 0.002 Error 34 155.71 4.58 Total 36 223.17 S = 2.140 R-Sq = 30.23% R-Sq(adj) = 26.12% Level N Mean StDev 1 12 8.386 2.071 2 12 10.638 2.269 3 13 11.608 2.080 Pooled StDev = 2.140 Fisher Individual Tests for Differences of Means Difference of Levels Difference of Means 97.6667% CI T-Value Adjusted...
Problem #4: Consider the data set that is summarized in the Minitab Output below Source DF C2 Error 34 Total 36 Adj SS 86.12 104.98 191.10 Adj MS 43.06 3.09 F-Value 13.95 P-Value 0.000 S 1.157 R-Sq Level NMean 2 45.06% R-Sq ( adj) -41.83% StDev 9 8.4151.458 10 11.4012.106 18 7.807 1.681 Pooled StDev- 1.757 Fisher Individual Tests for Differences of Means Difference Difference of Levels Adjusted P-Value of Means 97。6667% CI T-Value (1.068, 4.903) 0.608(-2.312, 1.095) -3.594 (-5.239,...
- Consider the data set that is summarized in the Minitab Output below. Source DF C2 Error 34 Total 36 Adj SS 122.79 210.34 333.13 Adj MS 61.39 6.19 E-Value 9.92 P-value 0.000 S = 2.487 R-Sq = 36.86% R-Sq(adj) = 33.15% Level N Mean 1 10 12.517 2 9 7.673 18 9.100 St Dev 3.261 0.928 2.518 Pooled St Dev = 2.487 Fisher Individual Tests for Differences of Means Difference of Levels Adjusted p-Value T-Value 2 Difference of Means...
consider the following Minitab 8. (8 points) Descriptive Statistics Consider the following Minitab output shown below. Descriptive Statistics: density Statistics Variable NN" Mean SE Mean StDev Minimum Q1 Median Q3 Maximum density 8 3 51.5 38.2 6.0 18.2 45.7 91.3 96.4 13.5 (a) (2 points) How many missing observations were in the data set. Enter your answer in Akindi line 19. A. 8 B. 3 C. 14 D. 6 E. Correct answer is not provided (b) (2 points) What is...
Consider the data set that is summarized in the MInitab output below. SAMPLE N Mean StDev 95% CI 1 36 109.81 12.12 (?, ?) 2 13 126.57 9.71 (?, ?) 3 14 ? 11.09 (?, ?) Suppose that the following is a Bonferonni confidence interval for μ1 − μ3, (2.04, 8.91). If we were to use ANOVA to analyze the above data, what would be the value of SS(treatment)?
Consider the data set that is summarized in the Minitab Output below. (a) Find a 93% Bonferonni confidence interval for μ2 − μ1. (b) Which pairs of means are significantly different (using the Bonferonni method at the 7% significance level)? Problem #4(a): (A) 2 and 3 only (B) 1 and 3 only (C) 1 and 2, 2 and 3 only (D) all of them (E) none of them (F) 1 and 3, 2 and 3 only (G) 1 and 2,...
Below is a selection of MiniTab output for hypothesis tests about data on homes sold in Stockton, CA in the late 90s. Use this information to evaluate the hypothesis tests A) Two-tailed, c-100000, α 0.05 Descriptive Statistics N Mean StDev SEMean 95% CI for 1500 123694 63251 1633 (120490, 126897) μ mean of spruce B) Evaluate at α 0.01 Test Null hypothesis Alternative hypothesis T-Value P-Value Ho: μ-0.05 H1: μ > 0.05 240 0.008 C) Two-tailed, C-3, α-0.01 Descriptive Statistics...