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 wasrecorded. Tables 1 and 2 present the Shapiro-Wilk normality test for both types of forest plots and the IndependentSamples t-test results. Use the information presented in these tables to answer the questions.Table 1: Tests of NormalityTests of NormalityKolmogorov-SmirnovShapiro-WilkForest PlotStatisticdfSig.StatisticdfSig.Tree_Species Logged.18112.200.93612.444Unlogged.11014.200.94514.480*. This is a lower bound of the true significance.a. Lilliefors Significance CorrectionTable 2: Independent Samples TestIndependent Samples TestLevene's Test for Equality ofVariancest-test for Equality of Means95% Confidence Interval of theDifferenceMeanStd. ErrorFSig.tdfSig. (2-tailed)DifferenceDifferenceLowerUpperEqual variancesassumedTree_Species2.751.110-2.62724.015-3.011901.14651-5.37818-.64563Equal variances notassumed-2.69723.271.013-3.011901.11661-5.32030-.70351We want to test, at the 1% level of significance, whether the mean number of species in a logged forest plot is different tothat found in an unlogged forest plot. To perform this test, we assume that the population standard deviations of loggedand unlogged forest plots are equal. 5. An hypothesis test is of the form Ho : µ1 – Hz = 0 vs Ha : µ1 – H2 < 0. Both populations are assumed normalwith equal variance and samples are assumed independent. We draw 21 observations for the first sample and 37observations for the second sample. State the distribution of the test statistic for this test.t-distribution with 60 degrees of freedomt-distribution with 58 degrees of freedom)Normal distribution with 57 degrees of freedomO t-distribution with 56 degrees of freedom

Answered: Image /qna-images/answer/c82f9302-743a-493e-a76d-65818a0f322d.jpg

Answered: 5. An hypothesis test is of the form Ho…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A sample of 33 lights in Austin gives an average of 46 cars/min while the city SD=3.4 A sample of 34 Dallas lights give a mean of 44 cars/mim with the city SD=1.4 At the 91% level, do the towns differ? 1. Is the test (A-Left Tailed, B-Right Tailed, C-2-Tailed) (separate with comma if there are 2) 2. Critical Value 3. Test statistic 4. 3-decimal p-Value
2. A pet association claims that the mean annual costs of food for dogs and cats are the same. The results for samples of the two types of pets are shown in the following table. On wished to test the claim. Assume both populations follow normal distributions Dogs Cats 21 - 16 T1= 239 n2 = 18 Iz= 203 S1 = 32 82 =28 with equal variances. Which of the following formula should be used to calculate the standardized test statistic? (a) Z = (21-32)- Do lu A n2 = (E1-2)-Do (b) Z V n1+2 (c) T (1-2)-Do (21-1)a7+(n2-1)s I1 11+11タ-3 V 1'n2 (d) T = d- Do sp/Vn (e) Z
Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Diet Regular μ μ1 μ2 n 20 20 x 0.78646 lb 0.80233 lb s 0.00449 lb 0.00755 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1≠μ2 B. H0: μ1=μ2 H1: μ1>μ2 C. H0: μ1≠μ2 H1: μ1<μ2 D. H0: μ1=μ2 H1: μ1<μ2 The test statistic, t, is nothing. (Round to two decimal places as needed.) The P-value is…
My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?
You have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.
Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Diet Regular μ μ1 μ2 n 26 26 x 0.78073 lb 0.80038 lb s 0.00447 lb 0.00745 lb Question content area bottom Part 1 a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1>μ2 B. H0: μ1=μ2 H1: μ1<μ2 Your answer is correct. C. H0: μ1≠μ2 H1: μ1<μ2 D. H0: μ1=μ2 H1: μ1≠μ2 Part 2 The test statistic, t, is…
A researcher wants to measure average cardiovascular health of university students and compare those scores to the average scores in the general population. Assuming that population variance is known, what statistical test is most appropriate for this study? independent-samples t-test single-sample t-test z-test for sample mean related-samples t-test
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.10 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm Left arm 143 179 OA. Ho: H0 H₁: Hd=0 OC. Ho: P = 0 H₁: Hd 0 sufficient evidence to support the claim of a difference in measurements between the two arms.

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