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Good Evening, I need Answers for some questions please. 1.) Part 3? (The options in the first box are the same options for the second
Good Evening, I need Answers for some questions please.
1.) Part 3? (The options in the first box are the same options for the second box.)
Type I error: A company that manufactures steel wires guarantees that the mean breaking strength (in kilonewtons) of the wires is greater than 50. They measure the strengths for a sample of wires and test H0: p. = 50 versus H1 : p. 22} 50. If a Type I error is mader what conclusion will be drawn regarding the mean breaking strength? The conclusion will be that the mean breaking strength is greater than 'l' 50- If a Type [I error is made, what conclusion will be drawn regarding the mean breaking strength? The conclusion will be that the mean breaking strength is |equal to 'l' | 50. This test uses a onetailed alternative hypothesis. Explain why a onetailed hypothesis is more appropriate than a twotailed hypothesis in this situation. With a onetailed hypothes'w we can conclude that the mean breaking strength is 50. With a twotailed hypothesis, we will not know whether - the mean breaking strength is 50. not equal to rnative hypothesis. Explain why a onetailed hypothesis is more appropriate than a twotailed less than greater than \"355 than or areater than we can conclude that the mean breaking strength is 50. With a twotailed hypothesis, we will not know whether _ the mean breaking strength is equal to g 50. A test is made of H\": p. = 19 versus H]: 11> 19. A sample of size n =49 is drawn, and E = 22. The population standard deviation is (F = 8. (a) Compute the value of the test stastic z. (b) Is HI] rejected at the u = 0.01 level? (c) Is H0 rejected at the a: 0.10 level? (a) Compute the value of the test statistic. Round the answer to at least two decimal places. Choose the correct type of hypothesis test. Then nd the critical values for o: = 0.01 and a = 0.10. Round your answers to three decimal placesr if necessary. H] Critical Values for a: 0.01 Critical Valut for o. = 0.10 Lefttailed D D O Righttailed D D O Twotailed D and D D and D Facehook: A study showed that two years ago, the mean time spent per visit to Facebook was 20.2 minutes. Assume the standard deviation is o = 10.0 minutes. Suppose that a simple random sample of 103 visits was selected this year and has a sample mean of E = 19.2 minutes. A social scientist is interested to know whether the mean time of Facebook visits has decreased- Use the [1: 0.10 level of signicance and the Pvalue method with the 1184 calculator. (a) State the appropriate null and alternate hypotheses. (b) Compute the Fvalue. (c) State a conclusion. Use the :1: 0.10 level of signicance. (a) State the appropriate null and alternate hypotheses. H0: p = 20.2 H1: p.Step by Step Solution
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