A sample of 900 computer chips revealed that 44 % of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed thatover 40% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim?State the null and alternative hypotheses for the above scenario.AnswerHo:Ha:KeypacKeyboard ShortcuTables

Sample size(n)=900Sample proportion()=0.44Population proportion(P)=0.40and =0.02

Answered: A sample of 900 computer chips revealed…

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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please solve question 4.19 attached 4.11 for your reference
A researcher is concerned with the effects of various courses that uses LMS and holding online exam with the academic standing. Exams results of students majoring in engineering that uses LMS are taken and analyzed. Is there a difference in the scores of students in various courses that uses LMS at 90% confidence level?Math 010 (Differential Calculus): 32 76 29 85 33 89 Math 013 (Linear Algebra): 72 81 98 21 63 13 Math 016 (Calculus 1): 98 21 34 88 38 54 Math 017 (Calculus 2): 72 73 26 93 42 63 Math 019 (EDA): 56 29 78 13 50 49 In your solution paper, show the null and alternative hypothesis, calculation of sum of squares and mean of square and the F values. Yes No
Based on information from a large insurance company, 66% of all damage liability claims are made by single people under the age of 25. A random sample of 54 claims showed that 41 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? State the null and alternate hypothesis. a) Ho: p = .66, Ha: p ≠ .66 b) Ho: p = .66, Ha: p > .66 c) Ho: p = .76, Ha: p < .76 d) Ho: p = .76, Ha: p > .76 e) Ho: p = .66, Ha: p < .66
Which of the following is a two-tailed test? O Ho: µ = 89, H1: u 84 O Ho: µ = 54, H1: u # 54
Do question 7,8 and 9 only The new study also carried out a test to determine whether the population proportion of unvaccinated school children contracting winter flu was higher than the population proportion of vaccinated school children. The Z test statistic to test this belief is found to be 1.874. The corresponding p-value is 0.0305 0.1212 0.3036 0.7724 Question 7 Suppose that the new study uses a level of significance of 0.05 to test the claim in Question 6. The probability of Type I error is 0.025 0.05 0.95 0.975 Question 8 Based on previous studies of school children who were vaccinated and contracted the flu, the time in hours that the flu symptoms last is assumed to follow a normal distribution with a mean of 20.7 hours and a standard deviation of 7.3 hours. The probability that a randomly selected school child has flu symptoms for more than 24 hours is 0.1628 0.3256 0.6744 0.8372 Question 9 Suppose that a random sample of 5 vaccinated school children is taken. Assuming the…
Question 3 Y Only about 12% of all people can wiggle their ears. Is this percent lower for millionaires? Of the 369 millionaires surveyed, 33 could wiggle their ears. What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ? Select an answer ✓ H₁: ? Select an answer (please enter a decimal) (Please enter a decimal) c. The test statistic?v= d. The p-value = e. The p-value is ? a f. Based on this, we should [Select an answer the null hypothesis. g. Thus, the final conclusion is that ... (please show your answer to 3 decimal places.) (Please show your answer to 3 decimal places.) O The data suggest the population proportion is not significantly lower than 12% at a = 0.05, so there is statistically significant evidence to conclude that the population proportion of millionaires who can wiggle their ears is equal to 12%. The data suggest the population proportion is not significantly…
Only 15% of registered voters voted in the last election. Will voter participation increase for the upcoming election? Of the 347 randomly selected registered voters surveyed, 56 of them will vote in the upcoming election. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use z-test for a population proportion b. The null and alternative hypotheses would be: Họ: ? Select an answer (please enter a decimal) H.: ?vSelect an answer v (Please enter a decimal) c. The test statistic ?v (please show your answer to 4 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is v f. Based on this, we should fail to reject g. Thus, the final conclusion is that ... O The data suggest the population proportion is not significantly higher than 15% at a = 0.01, so there is statistically significant evidence to conclude that the percentage of registered voters who will vote in the upcoming election will be equal to 15%. a…
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,476 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 153 of their customers and calculates that these customers used an average of 10,767 kWh of electricity last year. Assuming that the population standard deviation is 2478 kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 115 of their customers and calculates that these customers used an average of 11,425 kWh of electricity last year. Assuming that the population standard deviation is 3217 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
QUESTION 6 At-test is used to test hypotheses for the population mean u. A random sample of 25 is selected for the analysis. How many degrees of freedom would be used for the test? QUESTION 7

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