Show work when necessary. Each part is worth the same amount. Assume that all necessary conditions are metfor the hypothesis tests we have studied.1) Following are observed frequencies. Perform a goodness of fit test for the null hypothesisHo: P1 = 0.35, P2 = 0.25, P3 = 0.1, p4 = 0.1, p5 = 0.2.Category 1 2 3 4 5Observed 20 3020 3 40i). Compute the expected frequencies.ii). Compute the value of using the formula.iii). Give the p-value using a calculator.iv). State a conclusion, at the a= 0.05 level of significance.

(1) That is, there is no evidence to conclude that the proportions have changed. That is, there is…

Answered: Show work when necessary. Each part is…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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In each problem show all steps of the hypothesis test using PHANTOMS. If some of the assumptions are not met, note that the results of the test may not be correct and then continue the process of the hypothesis test. The FDA regulates that fish that is consumed is allowed to contain at most 1.0 mg/kg of mercury. In Florida, bass fish were collected in 53 different lakes to measure the mean amount of mercury in the sample of fish from each lake. The data for the average amount of mercury in each lake is in table #7.3.5 ("Multi-disciplinary niser activity," 2013). Do the data provide enough evidence to show that the fish in all Florida lakes have more mercury than the allowable amount? Test at the 10% level. Table #7.3.5: Average Mercury Levels (mg/kg) in Fish 1.23 1.33 0.04 0.44 1.20 0.27 0.48 0.19 0.83 0.81 0.71 0.5 0.49 1.16 0.05 0.15 0.19 0.77 1.08 0.98 0.63 0.56 0.41 0.73 0.34 0.59 0.34 0.84 0.50 0.34 0.28…
Given the two independent samples below, conduct a hypothesis test for the desired scenario. Assume all populations are approximately normally distributed. Sample 1 Sample 2 110 n2 : 179 28.9 23.6 81 = 3.8 82 = 3.9 Test the claim: Given the null and alternative hypotheses below, conduct a hypothesis test for a = 0.05. Ho: H1 = 42 Ha: H1 < 42 a. Given the alternative hypothesis, the test is Select an answer b. Determine the test statistic. Round to four decimal places. t = c. Find the p-value. Round to 4 decimals. с. p-value = d. Make a decision. O Fail to reject the null hypothesis. O Reject the null hypothesis.
Suppose that in a random selection of 100 colored candies, 20% of them are blue. The candy company claims that the percentage of blue candies is equal to 26%. Use a 0.01 significance level to test that claim. Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p +0.26 H₁: p=0.26 O B. Ho: p=0.26 H₁: p=0.26 O C. Ho: p = 0.26 H₁: p>0.26 O D. Ho: p=0.26 H₁: p<0.26
Suppose that in a random selection of 100 colored candies, 22% of them are blue. The candy company claims that the percentage of blue candies is equal to 30%. Use a 0.10 significance level to test that claim. Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.3 H₁: p=0.3 O B. Ho: p=0.3 H₁: p=0.3 C. Ho: p=0.3 H₁: p>0.3 O D. Ho: p=0.3 H₁: p<0.3 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test. A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 30% B. Reject Ho. There is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 30% O C. Fail to…
Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of blue candies is equal to 20%. Use a 0.10 significance level to test that claim. H, p#0.2 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.)
Suppose that in a random selection of 100 colored candies, 23% of them are blue. The candy company claims that the percentage of blue candies is equal to 30%. Use a 0.05 significance level to test that claim. Identify the null and alternative hypotheses for this test. Choose the correct answer below. OA. Ho: p=0.3 H₁: p0.3 Oc. Ho: p=0.3 H₁: p 0.3 OD. Ho: p*0.3 H₁: p=0.3 B
You suspect that an unscrupulous employee at a casino has tampered with a die; that is, he is using a loaded die. In order to test your suspicion, you rolled the die in question 200 times and obtained the following frequencies for each of the six possible outcomes of the die: Number Frequency 1 45 2 39 3 35 4 25 27 29 Can you conclude that the die is loaded? Use a=0.05 as the significance level and perform a hypothesis test. Remember to state the null and alternative hypothesis.
You are conducting a multinomial hypothesis test for the claim that the 4 categories occur with the following frequencies:HoHo : pA=0.1pA=0.1; pB=0.4pB=0.4; pC=0.25pC=0.25; pD=0.25pD=0.25Complete the table. Category ObservedFrequency ExpectedFrequency SquaredResidual A 4 B 39 C 29 D 25 What is the chi-square test-statistic for this data?χ2=Report all answers accurate to three decimal places.
A two-sample z-test for two population proportions is to be performed using the P-value approach. The null hypothesis is Hn: P, = p, and the altemative is H: P, #p2. Use the given sample data to find the P-value for the hypothesis test. Give an Interpretation of the p-value. A state university found it lost 25 students out of 352 in 2013 and 36 students out of 334 in 2014. O A P-value = 0. 0455; There is about a 4.55% chance that the two proportions are equal. O B. P-value = 0.091; If there is no difference in the proportions, there is about a 9.1% chance of seeing the observed difference or larger by natural sampling variation. OC. P-value = 0.0455; there is no difference in the proportions, there is about a 4.55% chance of seeing the observed difference or larger by natural sampling variation. O D. P-value = 0. 091; There is about a 9.1% chance that the two proportions are equal. O E. P-value = 0.9545: there is no difference in the proportions, there is about a 95.45% chance of…
Consider the hypothesis test below. Ho : P1 – P2 0 The following results are for independent samples taken from the two populations. Sample 1 Sample 2 100 n2 = 300 P1 = 0.27 P2 = 0.15 Use pooled estimator of p. a. What is the p-value (to 4 decimals)? Use Table 1 from Appendix B. b. With a = 0.05, what is your hypothesis testing conclusion? Select your answer -
You are conducting a multinomial hypothesis test for the claim that the 4 categories occur with the following frequencies:HoHo : pA=0.5pA=0.5; pB=0.3pB=0.3; pC=0.1pC=0.1; pD=0.1pD=0.1Complete the table. Category ObservedFrequency ExpectedFrequency SquaredResidual A 55 B 32 C 14 D 24 What is the chi-square test-statistic for this data?χ2=χ2= Report all answers accurate to three decimal places.

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