Can weude, at the 0.05 level of significance, that the proportion p₁ of all potential voters in Texas who plan to vote differs from the proportion p, of allpotential voters in Oklahoma who plan to vote?Perform a two-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places and round your answers as specified in the parts below. (If necessary, consult a list offormulas.)(a) State the null hypothesis H, and the alternative hypothesis H₁.H:0(b) Determine the type of test statistic to use.(Choose one) ▼(c) Find the value of the test statistic. (Round to three or more decimal places.)0(d) Find the p-value. (Round to three or more decimal places.)0(e) Can we conclude that the proportion of voters in Texas who plan to votediffers from the proportion of voters in Oklahoma who plan to vote?Yes NoExplanationCheck3I|x5*XaSロ=ロ OSOA200<РSpolo200>0Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibi OIn a random sample of 222 potential voters registered in the state of Texas, 68 indicated that they planned to vote in the next general election. In anindependently chosen, random sample of 228 potential voters registered in Oklahoma, 98 indicated that they planned to vote in the next general election.Can we conclude, at the 0.05 level of significance, that the proportion p, of all potential voters in Texas who plan to vote differs from the proportion p, of allpotential voters in Oklahoma who plan to vote?Perform a two-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places and round your answers as specified in the parts below. (If necessary, consult a list offormulas.)(a) State the null hypothesis Ho and the alternative hypothesis H₁.H:0H₁:0(b) Determine the type of test statistic to use.(Choose one) ▼(c) Find the value of the test statistic. (Round to three or more decimal places.)0(d) Find the p-value. (Round to three or more decimal places.)0ExplanationCheckF|xX96.XOS20=0 OSO<5Р60020>O2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center AccessibilityAug 16

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Answered: (a) State the null hypothesis H. and…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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3132 You'd like to test the null hypothesis that the means of the two samples (column A and column B) are the same. The alternative hypothesis is that they are not the same. You have no reason to believe that the standard deviations of the two samples are equal. Test at the alpha = 0.10 level. After using Excel, what do you conclude? Are the means the same? Group of answer choices You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. X1 X2 98.69 99.96 100.31 98.95 104.50 91.88 103.23 98.48 96.91 103.32 102.29 100.90 100.26 96.75 96.70 101.13 99.82 102.01 96.26…
c. What is the P-value? P-value = 0.0154 (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. A. Ho: p=0.12 O B. Ho: p 0.12 C. Ho: p>0.12 O D. Ho: p<0.12 Decide whether to reject the null hypothesis. Choose the correct answer bylow. O A. Reject the null hypothesis because the P-value is greater than the significance level, a. O B. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. e. What is the final conclusion? O A. There is not sufficient evidence to warrant rejection of the claim that less than 12% of treated subjects experienced headaches. B. There is sufficient evidence to support the claim that less than 12% of treated subjects…
3042 You'd like to test the null hypothesis that the means of the two samples (column A and column B) are the same. The alternative hypothesis is that they are not the same. You have no reason to believe that the standard deviations of the two samples are equal. Test at the alpha = 0.10 level. After using Excel, what do you conclude? Are the means the same? Group of answer choices You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. X1 X2 97.88 98.66 105.98 102.29 96.01 111.40 99.70 106.15 96.96 100.10 99.80 104.89 92.73 99.24 106.30 104.79 97.90 101.51 95.04…
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
p 15 Is sunscreen with an SPF 30 rating more effective at preventing sunburn than sunscreen with an SPF 15 rating? At the alpha = 0.05 level of significance, test the claim that SPF 30 sunscreen is better at preventing sunburn than 15 sunscreen. Let represent the proportion of SPF 15 sunscreen wearers who get a sunburn and represent the proportion of SPF 30 sunscreen wearers who get a sunburn. (Round your results to three decimal places) 1. Which would be correct hypotheses for this test? O O H 0 :p 15 =p 30 , H 1 :p 15 ne p 30 H 0 :p 15 =p 30 , H H 1 :p 15 <p 30 O H 0 :p 15 <p 30 ,H 1 :p 15 >p 30 O H 0 :p 15 =p 30 H 1 :p 15 >p 30 A random sample of 250 beach-goers were given SPF 15 sunscreen and asked to lay in the sun for 1 hour10 of them later reported a sunburn. Another random sample of 250 beach- goers were given SPF 30 sunscreen and asked to lay in the sun for 1 hour. 7 of them later reported a sunburn. 2.Find the test statistic 3. Give the P-value: answer only…
In testing the hypotheses H0:p = 0.62 vs Ha: p≠ 0.62, the test statistic is calculated to be z = 2.11. Which of the following is the correct p-value? A. 0.0348 B. 0.0174 C. 0.9862 D. 1.9652
Time left 1:41. Consider the following hypothesis test: The above test is Select one: O a. an upper tail test O b. a lower tail test O c. a two-tailed test The variable age has the Select one: tof Oa. interval scale of measurement O b. ratio scale of measurement bp f7 f8 fg f10 f11 12 10
K A sports reporter suggests that professional baseball players must be, on average, older than professional football players, since football is a contact sport. Researchers selected 35 professional football players and 33 professional baseball players at random. The data collected are summarized by the accompanying table. Suppose the researchers decide to test the hypothesis. The degrees of freedom formula gives 56.25 df. Test the null hypothesis at x = 0.05. A. Ho: H₁-H₂ = 0 HA: H1 H₂0 O C. Ho: Hi-H2 = 0 HA: H1 H₂ <0 Compute the test statistic. ... Let μ₁ be the mean age of professional baseball players and let μ₂ be the mean age of professional football players. Identify the null and alternative hypotheses. Choose the correct answer below. t= (Round to two decimal places as needed.) n y Baseball 33 27.47 4.07 B. Hoi H1-H2=0 HA: H1 H₂ #0 Football 35 26.47 2.79 D. Ho: H1-H2#0 HA: H1 H₂=0
You are conducting a Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Report all answers correct to three decimal places. Category ObservedFrequency ExpectedFrequency (obs-exp)^2/exp A 25 B 8 C 19 D 11 E 19 What is the chi-square test-statistic for this data?χ2=
2448 You'd like to test the null hypothesis that the means of the two samples (column A and column B) are the same. The alternative hypothesis is that they are not the same. You have no reason to believe that the standard deviations of the two samples are equal. Test at the alpha = 0.10 level. After using Excel, what do you conclude? Are the means the same? Group of answer choices You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. X1 X2 102.04 96.13 100.47 97.63 103.17 95.42 95.09 104.41 103.81 102.08 95.36 103.33 103.38 112.25 100.26 96.93 97.24 100.19 99.61…
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. Ho: p=0.6 versus H,: p>0.6 n= 125; x= 90; a 0.05 Click here to view page 1 of the table. Click here to view page 2 of the table. Calculate the test statistic, zo: Zo !D (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer.
The null and alternate hypotheses are: Ho Hy H2 H₁ H₁ H₂ A random sample of 11 observations from one population revealed a sample mean of 23 and a sample standard deviation of 1.1. A random sample of 4 observations from another population revealed a sample mean of 24 and a sample standard deviation of 1.3. The population standard deviations are unknown but assumed to be equal. At the 0.05 significance level, is there a difference between the population means? Required: a. State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.) Answer is complete but not entirely correct. -1.761 or 1.761 The decision rule is to reject Ho b. Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.) Pooled estimate of the population variance

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(a) State the null hypothesis H. and the alternative hypothesis H₁. H:O (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 μ a P X SÔ Do 8 00 D