A manufacturer of rechargeable laptop batteries claims its batteries have, on average, 500 charges. A consumer group decides to test this claim by assessing the number of times 30 of their laptop batteries can be recharged and finds the average is 497, with a standard deviation of 10. The resulting p-value is .1111; thus, the null hypothesis is not rejected. What is the Type II error? -The consumer group concludes that the manufacturer's claim that its laptop batteries can be recharged, on average, 500 times is inaccurate, when in fact the claim is accurate. -The correct decison was made, there was no error possible. -The consumer group concludes that the manufacturer's claim that its laptop batteries can be recharged, on average, 500 times is accurate, when in fact the claim is inaccurate.

A manufacturer of rechargeable laptop batteries markets its batteries as having, on average, 500 charges. A consumer group decides to test this claim by assessing the number of times 30 of their laptop batteries can be recharged and finds the average is 497, with a standard deviation of 10. The resulting p-value is 0.1111; thus, the null hypothesis is not rejected. The consumer group concludes that the manufacturer’s claim that its laptop batteries can be recharged, on average, 500 times is accurate. What type of error is possible in this situation?   type I type II neither both

The breaking strengths of cables produced by a certain manufacturer have a mean, u, of 1875 pounds, and a standard deviation of 100 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 50 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1912 pounds. Can we support, at the 0.1 level of significance, the claim that the mean breaking strength has increased? (Assume that the standard deviation has not changed.) Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round to three or more decimal…

A credit bureau analysis of undergraduate students' credit records found that the average number of credit cards in an undergraduate's wallet was 4.03. It was also reported that in a random sample of 136 undergraduates, the sample mean number of credit cards that the students said they carried was 2.5. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.1. Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.03? (Use a = 0.05. Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.) n USE SALT t = P-value = State your conclusion. O Do not reject H.. We have convincing evidence that the mean number of credit cards carried by undergraduates is less than the credit bureau's figure of 4.03. O Reject Ha. We do not have convincing evidence that the mean…

A company provides a full range of insurance and financial services. Suppose the company reported the results of a survey of 1,045 identity theft victims. The survey shows victims spend an average of 79 hours trying to resolve their cases. If the true average time spent was 79 ​hours, determine the probability that a test of hypothesis designed to test that the average was less than 83 hours would select the research hypothesis. Use α=0.05 and a standard deviation of 50.   The probability that the hypothesis test would select the research hypothesis is

A highly anticipated scl-fl movie opened in theaters with a special midnight showing this past Friday. For the Saturday edition of his town's newspaper, Alan wrote a column about local interest In the movie. He found that the attendance for the midnight showing at his town's theater was 288 people. For all the theaters in the state, attendance at the midnight showing had a mean of 245 people with a standard deviation of 17 people. Find the z-score of the attendance at the theater in Alan's town relative to the attendances among all the theaters in the state. Round your answer to two decimal places. Submit Assignment Continue Privacy Accessibility 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Next » · Previous

A survey of nonprofit organizations showed that online fundraising has increased in the past year. Based on a random sample of 55 nonprofit organizations, the mean one- time gift donation in the past year was $ 75, with a standard deviation of $ 9. If you test the null hypothesis at the 0.01 level of significance, is there evidence that the mean one- time gift donation is greater than $ 70? Interpret the meaning of the p- value in this problem.

Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at a university used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in a paper are shown. Use these data with the resulting sample mean 25.750000 and sample standard deviation 1.870829, to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. (Use x = 0.05.) 27 22 OH₂ μ = 20 Hà U = 20 Ho: μ = 20 H₂ μ 20 State the appropriate null and alternative hypotheses. Ο Μ μ 20 ⒸH: μ> 20 Hàu<20 25 24 P-value = 26 26 28 27 25 27 24 28 24 31 26 26 26 24 25 24 25 27 27 24 Find the test statistic and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = State your conclusion. O We reject Ho. We do not have convincing evidence that the mean wrist…

2) A restaurant association claims that the average amount a family spends per year eating out is $2700. Suppose you are a consumer reporter and want to test this claim. You select a random sample of 32 homes and obtain that their mean is $2850 with a standard deviation of $270. using α= 0.05, do you have enough evidence to reject the association's claim?