8) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1000 hours. A homeowner selects 40 bulbs and finds the mean lifetime to be 980 hours with a standard deviation of 80 hours. Test the manufacturer's claim. Use ? = 0.05. A) standardized test statistic z = -1.58; critical value z0 = ±1.96; fail to reject H0; There is not sufficient evidence to reject the manufacturer's claim. 9) You wish to test the claim that ? > 32 at a level of significance of ? = 0.05 and are given sample statistics n = 50, x? = 32.3, and s = 1.2. Compute the value of the standardized test statistic. Round your answer to two decimal places. A) 1.77 B) 2.31 C) 0.98 D) 3.11 10) Find the critical value for a two-tailed test with ? = 0.01 and n = 30. A) ±1.645 B) ±1.96 C) ±2.575 D) ±2.33 11) Given H0: ? ? 18 and P = 0.070. Do you reject or fail to reject H0 at the 0.05 level of significance? A) fail to reject H0 B) reject H0 C) not sufficient information to decide 12) Suppose you are using ? = 0.05 to test the claim that ? = 14 using a P-value. You are given the sample statistics n = 35, x? = 13.1, and s = 2.7. Find the P-value. A) 0.1003 B) 0.0488 C) 0.0591 D) 0.0244 13) Given H0: p = 0.85 and ? = 0.10, which level of confidence should you use to test the claim? A) 99% B) 80% C) 95% D) 90% 14) The owner of a professional basketball team claims that the mean attendance at games is over 30,000 and therefore the team needs a new arena. Determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. A) left-tailed B) two-tailed C) right-tailed 15) The mean age of bus drivers in Chicago is 53.7 years. State this claim mathematically. Write the null and alternative hypotheses. Identify which hypothesis is the claim. A) claim: ? = 53.7; H0: ? = 53.7; Ha: ? ? 53.7; claim is H0 16) You wish to test the claim that ? > 29 at a level of significance of ? = 0.05 and are given sample statistics n = 50, x? = 29.3, and s = 1.2. Compute the value of the standardized test statistic. Round your answer to two decimal places. A) 2.31 B) 0.98 C) 1.77 D) 3.11
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You need to test the claim that the mean lifetime of fluorescent bulbs is greater than 29 hours at a 0.05 level of significance using given sample statistics (n = 231, \( \bar{x} = 29.3 \), and \( \sigma = 1.2 \)). Show more…
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A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo. In a random sample of 37 floors of WaxWin and 30 of WaxCo, WaxWin had a mean lifetime of 26.2 and WaxCo had a mean lifetime of 21.9. The population standard deviation for WaxWin is assumed to be 9.1 and the population standard deviation for WaxCo is assumed to be 9.2. Perform a hypothesis test using a significance level of 0.10 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2. The correct hypotheses are: H0: μ1 = μ2 HA: μ1 ≠ μ2 (claim) Since the level of significance is 0.10, the critical value is ±1.645. The test statistic is: (round to 3 places) The p-value is: (round to 3 places) A random sample of 30 chemists from Washington state shows an average salary of $42,546. The population standard deviation for chemist salaries in Washington state is $868. A random sample of 39 chemists from Florida state shows an average salary of $48,395. The population standard deviation for chemist salaries in Florida state is $945. A chemist that has worked in both states believes that chemists in Washington make more than chemists in Florida. At α = 0.05, is this chemist correct? Let Washington be sample 1 and Florida be sample 2. The correct hypotheses are: H0: μ1 ≤ μ2 HA: μ1 > μ2 (claim) Since the level of significance is 0.05, the critical value is 1.645. The test statistic is: (round to 3 places) The p-value is: (round to 3 places) A researcher is interested in seeing if the average income of rural families is greater than that of urban families. To see if his claim is correct, he randomly selects 45 families from a rural area and finds that they have an average income of $66,299 with a population standard deviation of $668. He then selects 31 families from an urban area and finds that they have an average income of $67,979 with a population standard deviation of $534. Perform a hypothesis test using a significance level of 0.01 to test his claim. Let rural families be sample 1 and urban families be sample 2. The correct hypotheses are: H0: μ1 ≤ μ2 HA: μ1 > μ2 (claim) Since the level of significance is 0.01, the critical value is 2.326. The test statistic is: (round to 3 places) The p-value is: (round to 3 places)
Keondre P.
a. In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in a mean time between failures of 537.1 hours, with a standard deviation of 20.7 hours. At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours. Make sure you clearly show all steps of the hypothesis test. Show all of your work. b. Of 377 randomly selected medical students, 22 said that they planned to work in a rural community. Find a 92% confidence interval for the true proportion of all medical students who plan to work in a rural community. Give your answer(s) as decimal(s) rounded to 4 decimal places. Show all of your work. c. From previous studies, it is concluded that 70% of workers indicate that they are dissatisfied with their job. A researcher claims that the proportion is larger than 70% and decides to survey 400 working adults. Test the researcher's claim at the α=0.025 significance level. Preliminary: Have the requirements for this test been satisfied? Yes No Test the claim: The null and alternative hypotheses are H0: μ = 0.7 Ha: μ ≠ 0.7 H0: p ≤ 0.7 Ha: p > 0.7 H0: p = 0.7 Ha: p ≠ 0.7 H0: μ ≥ 0.7 Ha: μ < 0.7 H0: p ≥ 0.7 Ha: p < 0.7 H0: μ ≤ 0.7 Ha: μ > 0.7 The test is Select an answer: left-tailed, right-tailed, two-tailed Based on the sample of 400 people, 79% workers indicate that they are dissatisfied with their job. What is the test statistic? Round your answer to two decimal places. z = t = What is the p-value? Round your answer to four decimal places. Make a decision. Do not reject the null hypothesis. Reject the null hypothesis. Make a conclusion. There is not sufficient evidence that the proportion of people who workers indicate that they are dissatisfied with their job is larger than 70% at the α=0.025 significance level. There is sufficient evidence that the proportion of people who workers indicate that they are dissatisfied with their job is larger than 70% at the α=0.025 significance level.
Kari H.
Recall that "very satisfied" customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the random sample of 70 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. (a) Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42. H0: µ 42 versus Ha: µ 42. (b) The random sample of 70 satisfaction ratings yields a sample mean of x¯=42.800. Assuming that σ equals 2.65, use critical values to test H0 versus Ha at each of α = .10, .05, .01, and .001. (Round your answer z.05 to 3 decimal places and other z-scores to 2 decimal places.) z = Rejection points z.10 z.05 z.01 z.001 Reject H0 with α = .10, .05, .01, but not with α =.001 (c) Using the information in part (b), calculate the p-value and use it to test H0 versus Ha at each of α = .10, .05, .01, and .001. (Round your answers to 4 decimal places.) p-value = Since p-value = is less than .10, .05, .01; reject H0 at those levels of α but not with α = .001. (d) How much evidence is there that the mean composite satisfaction rating exceeds 42? There is evidence. Bayus (1991) studied the mean numbers of auto dealers visited by early and late replacement buyers. Letting μ be the mean number of dealers visited by all late replacement buyers, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence that μ differs from 4 dealers. A random sample of 100 late replacement buyers yields a mean and a standard deviation of the number of dealers visited of x¯ = 4.42 and s = .65. Using a critical value and assuming approximate normality to test the hypotheses you set up by setting α equal to .10, .05, .01, and .001. Do we estimate that μ is less than 4 or greater than 4? (Round your answers to 3 decimal places.) H0 : μ 4 versus Ha : μ 4. t tα/2 = 0.05 tα/2 =0.025 tα/2 =0.005 tα/2 =0.0005 There is evidence. μ is 4.
Sri K.
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