Conduct the following test at the a= 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume that the samples were obtainedindependently using simple random sampling.Test whether p₁ #P2. Sample data are x₁ = 28, n₁ = 255, x₂ = 38, and n₂ = 301.(a) Determine the null and alternative hypotheses. Choose the correct answer below.O A. Ho: P₁ = 0 versus H₁: P₁ = 0OC. Ho: P₁ P2 versus H₁: P₁ P2(b) The test statistic zo is(Round to two decimal places as needed.)(c) The P-value is. (Round to three decimal places as needed.)Test the null hypothesis. Choose the correct conclusion below.OB. Ho: P₁ P2 versus H₁: P₁ P2O D. Ho: P₁P2 versus H₁: P₁ P2conclude that P₁ P2-O A. Reject the null hypothesis because there is not sufficient evidenceO B. Do not reject the null hypothesis because there is not sufficient evidence to conclude that p₁ #P2.OC. Reject the null hypothesis because there is sufficient evidence to conclude that p₁ #P₂.OD. Do not reject the null hypothesis because there is sufficient evidence to conclude that p₁ > P2-

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Answered: (a) Determine the null and alternative…

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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Solve the problem. A researcher conducts a hypothesis test on a population proportion. Her null and alternative hypothesis are Ho: p = 0.6 and Ha: P < 0.6. The test statistic and p-value for the test are z = -1.51 and p-value = 0.0655. For a significance level of a = 0.05, choose the correct conclusion regarding the null hypothesis. O None of these. O There is sufficient evidence to warrant rejection of the null hypothesis. There is insufficient evidence to warrant rejection of the null hypothesis. STEP 1/3 Routin We're findi answer you In the mea DIVE DEEP
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Consider the following data for a dependent variable y and two independent variables, x, and X2. X1 | X2 y 30 12 93 47 10 108 17 112 51 16 178 40 94 51 19 175 74 7 170 36 12 117 59 13 142 76 16 213 The estimated regression equation for these data is ý = -20.24 + 2.04x, + 4.78x2. Here, SST = 15,563.6, R = 14,417.6, = 0.2488, and = 0.9548. zas 25
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A null and alternative hypothesis are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Ho: H₂: O O ab What type of test is being conducted in this problem? OA. Two-tailed test OB. Right-tailed test OC. Left-tailed test O O os 6.5 06.5 esc Statcrunch ! 1 Q 2 W #3 E $ ► 4 R % 45 F T MacBook Pro A ^ 6 Y & 7 U * 0
A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Full data set O Carpeted Uncarpeted 9.4 9.1 16 10.8 6.4 10.8 14.8 8.8 11.5 11.6 13.7 9.4 10.7 12.4 13.4 7.8 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the a = 0.05 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms. OA Ho P1 H2 B. Ho H1 F2 OC Ho H= H2
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is mu greater than 98.6 degrees F. The sample size is n equals 5 and the test statistic is t equals 1.486. P-value equals (Round to three decimal places as needed.)
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Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is μ>98.6°F. The sample size is n=8 and the test statistic is t=1.289. P-value= nothing (Round to three decimal places as needed.)
Consider a drug testing company that provides a test for marijuana usage. Among 335 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than 10 percent of the test results are wrong. Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.1 H1:p0.1 OC. Ho: p<0.1 H1:p=0.1 O D. Ho: p=0.1 H1:p+0.1
HYPOTHESIS TESTING 1. Formulate the null and the alternative Ho: hypotheses In symbol: Ho: μ = 50 Ha: The average weight of bag of chemical is not equal to 50 kilograms. In symbol: Ha: 2. Specify the following: A. Level of Significance A. B. Types of Hypothesis Testing [one-tailed test B. two tailed test (left-tailed or right-tailed) or a two-tailed test] 3. Specify the following: A. The test statistic to be used. (t-test: one sample mean or t-test: two sample means) A. B. Find the Degree of Freedom Formula: df = n-1 (for one-sample mean test) B. df = n -1 df = (n + n₂) - 2 (for two-sample mean test) df = 16-1 C. The critical or tabular value (t critical). (Can be found on the given table of Tabular Value of T) df = 15 C. 2.947 4. Compute for the value of the test statistic Formula: (t computed) using the data from the population (x-μ)√n t= and the sample/s. Given: хэ 48.1 μl - n-29 s-0.9 Substitute the given to the formula and solve:
Describe type I and type II errors for a hypothesis test of the indicated claim. A furniture store claims that at least 40% of its new customers will return to buy their next piece of furniture. Describe the type I error. Choose the correct answer below. OA. A type I error will occur when the actual proportion of new customers who return to buy their next piece of furniture is at least 0.40, but you fail to reject Ho: p20.40. OB. A type I error will occur when the actual proportion of new customers who return to buy their next piece furniture is at least 0.40, but you reject Ho: p20.40. OC. A type I error will occur when the actual proportion of new customers who return to buy their next piece of furniture is no more than 0.40, but you reject Ho: p ≤ 0.40. OD. A type I error will occur when the actual proportion of new customers who return to buy their next piece f furniture is no more than 0.40, but you fail to reject Ho: p ≤0.40. Describe the type II error. Choose the correct answer…

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