12. Mr. Sy asserts that fewer than5% of the bulbs that he sells aredefective. Suppose 300 bulbs arerandomly selected, each are testedand 10 defective bulbs are found.Does the data provide sufficientevidence for Mr. Sy to conclude thatthe fraction of defective bulbs isless than 0.05 using a = 0.01? Whichof the following is the critical value?O A. ± 1.96B. ± 1.645O C. ± 2.58O D. ± 2.33 *13. Mr. Sy asserts that fewer than5% of the bulbs that he sells aredefective. Suppose 300 bulbs arerandomly selected, each are testedand 10 defective bulbs are found.Does the data provide sufficientevidence for Mr. Sy to conclude thatthe fraction of defective bulbs isless than 0.05 using a = 0.01? Whatis the value of p hat (p^)?O A. 0.03OB. 0.97O C. 0.30D. 0.05

Answered: Image /qna-images/answer/e27a5ddb-8d8d-4356-ad3f-7df84f50bd2b.jpg

Answered: * 13. Mr. Sy asserts that fewer than 5%…

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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Assume we wanted to show that the true proportion of virginica iris seeds that fail to germinate is less than 20%. Let's say we tested at an alpha of 0.05, and found a p-vale of 0.082. Which of the following would be an appropriate conclusion and interpretation? A) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state we have insufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. B) With an alpha of 0.05, and a p-value of 0.0082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. C) With an alpha of 0.05, and a p-value of 0.082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. D) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state…
The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO2 in a sample is normally distributed with ? = 0.32 and that x = 5.21. (Use ? = 0.05.) (a) Does this indicate conclusively that the true average percentage differs from 5.5?State the appropriate null and alternative hypotheses. H0: ? = 5.5Ha: ? ≠ 5.5H0: ? = 5.5Ha: ? ≥ 5.5 H0: ? = 5.5Ha: ? < 5.5H0: ? = 5.5Ha: ? > 5.5 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.Reject the null hypothesis. There is sufficient evidence…
The test statistic of z=−2.07 is obtained when testing the claim that p<0.85. a. Using a significance level of α=0.05, find the critical value(s). b. Should we reject H0 or should we fail to reject H0?
A survey of 90 recently delivered women on the rolls of a county welfare department revealed that 27 had a history of intrapartum or postpartum infection. What is the critical value of z if we need to conclude that the population proportion with a history of intrapartum or postpartum infection is less than 0.25.
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Dr. Chapman conducted an experiment in which participants watched paint dry for 30 minutes twice, once being paid $1 and once being paid $30. When comparing the samples, he calculated t = 3.57. He assumed α = .01 with df = 5, so the tcv = ±4.032. Because the calculated value was: A. greater than the critical value, Dr. Chapman can reject the null hypothesis. B. greater than the critical value, Dr. Chapman failed to reject the null hypothesis. C. less than the critical value, Dr. Chapman failed to reject the null hypothesis. D. less than the critical value, Dr. Chapman can reject the null hypothesis.
If I have a sample mean less than 30 at alpha= 0.05. What is the critical value using the Z table
The test statistic of z=−2.31 is obtained when testing the claim that p=1/2. a. Using a significance level of α=0.05,find the critical value(s). b. Should we reject H0 or should we fail to reject H0?
Suppose a technician claims that her monthly repairs are less than 50 units, on average. Several of her coworkers do not believe her, so the technician decides to do a hypothesis test, at a 10% significance level, to persuade them. She records data from 16 previous months and works through the testing procedure: H0: μ=50; Ha: μ<50 x¯=41 σ=6 α=0.1 (significance level) The test statistic is z0=x¯−μ0σn√=41−50616√=−6 The critical value is −z0.1=−1.28. Conclude whether to reject or not reject H0, and interpret the results.
Suppose a technician claims that her monthly repairs are less than 50 units, on average. Several of her coworkers do not believe her, so the technician decides to do a hypothesis test, at a 10% significance level, to persuade them. She records data from 16 previous months and works through the testing procedure: H0: μ=50; Ha: μ<50 x¯=41 σ=6 α=0.1 (significance level) The test statistic is z0=x¯−μ0σn√=41−50616√=−6 The critical value is −z0.1=−1.28. Conclude whether to reject or not reject H0, and interpret the results. Select the correct answer below: Reject H0. At the 10% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis. Reject H0. At the 10% significance level, the data provide sufficient evidence to conclude that the mean monthly repairs is less than 50 units. Do not reject H0. At the 10% significance level, the test results are not statistically significant and at best, provide…
2. The critical value of t is 1.383, the value of the test statistic t = -1.383; therefore, we do not reject H0

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