The correct hypothesis statement for this test is: H: μι - -0vs H: μι - <0 (α-0.01) Hz < 0 (a= 0.01) H: μ- 0vs H, : μι- >0 (α-0.01) : µ1 – µ2 O vs Ha : µ1 μ2 >0 (α 0.01) | H: μ- 0vs H, : μι- μ<0 (α0.01) : µ1 – µ2 0 vs Ha : µ1 O Ho : µ1 – µ2 = 0 0 vs Ha : µ1 – Mz #0 (a = 0.01)

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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1. A study compared the number of tree species in unlogged forest plots to similar plots logged 8 years earlier.
Independent random samples of both logged and unlogged plots were analyzed and the number of tree species was
recorded. Tables 1 and 2 present the Shapiro-Wilk normality test for both types of forest plots and the Independent
Samples t-test results. Use the information presented in these tables to answer the questions.
Table 1: Tests of Normality
Tests of Normality
Kolmogorov-Smirnov
Shapiro-Wilk
Forest Plot
Statistic
df
Sig.
Statistic
df
Sig.
Tree_Species Logged
.181
12
.200
.936
12
.444
Unlogged
.110
14
.200
.945
14
.480
*. This is a lower bound of the true significance.
a. Lilliefors Significance Correction
Table 2: Independent Samples Test
Independent Samples Test
Levene's Test for Equality of
Variances
t-test for Equality of Means
95% Confidence Interval of the
Difference
Mean
Std. Error
F
Sig.
t
df
Sig. (2-tailed)
Difference
Difference
Lower
Upper
Equal variances
assumed
Tree_Species
2.751
.110
-2.627
24
.015
-3.01190
1.14651
-5.37818
-.64563
Equal variances not
assumed
-2.697
23.271
.013
-3.01190
1.11661
-5.32030
-.70351
We want to test, at the 1% level of significance, whether the mean number of species in a logged forest plot is different to
that found in an unlogged forest plot. To perform this test, we assume that the population standard deviations of logged
and unlogged forest plots are equal.
Transcribed Image Text:1. A study compared the number of tree species in unlogged forest plots to similar plots logged 8 years earlier. Independent random samples of both logged and unlogged plots were analyzed and the number of tree species was recorded. Tables 1 and 2 present the Shapiro-Wilk normality test for both types of forest plots and the Independent Samples t-test results. Use the information presented in these tables to answer the questions. Table 1: Tests of Normality Tests of Normality Kolmogorov-Smirnov Shapiro-Wilk Forest Plot Statistic df Sig. Statistic df Sig. Tree_Species Logged .181 12 .200 .936 12 .444 Unlogged .110 14 .200 .945 14 .480 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction Table 2: Independent Samples Test Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Difference Mean Std. Error F Sig. t df Sig. (2-tailed) Difference Difference Lower Upper Equal variances assumed Tree_Species 2.751 .110 -2.627 24 .015 -3.01190 1.14651 -5.37818 -.64563 Equal variances not assumed -2.697 23.271 .013 -3.01190 1.11661 -5.32030 -.70351 We want to test, at the 1% level of significance, whether the mean number of species in a logged forest plot is different to that found in an unlogged forest plot. To perform this test, we assume that the population standard deviations of logged and unlogged forest plots are equal.
b. The correct hypothesis statement for this test is:
O Họ : µ1 – Mz
0 vs Ha : µ1 – Hz < 0 (a= 0.01)
O Họ : µ1 – Mz
0 vs Ha : µ1 – Hz > 0 (a = 0.01)
O Ho : µ1 – H2 = 0 vs Ha : µ1 – H2 <0 (a = 0.01)
O Ho : µ1 – µz = 0 vs Ha : µi – H2 #0 (a = 0.01)
Transcribed Image Text:b. The correct hypothesis statement for this test is: O Họ : µ1 – Mz 0 vs Ha : µ1 – Hz < 0 (a= 0.01) O Họ : µ1 – Mz 0 vs Ha : µ1 – Hz > 0 (a = 0.01) O Ho : µ1 – H2 = 0 vs Ha : µ1 – H2 <0 (a = 0.01) O Ho : µ1 – µz = 0 vs Ha : µi – H2 #0 (a = 0.01)
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