A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1500 hours. A homeowner...
A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1500 hours. A homeowner selects 40 bulbs and finds the mean lifetime to be 1480 hours with a population standard deviation of 80 hours. Test the manufacturer's claim. Use α = 0.05.
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A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1500 hours. A homeowner...
A manufacturer claims that the mean lifetime of its lithium batteries is 1101 hours. A homeowner...
A manufacturer claims that the mean lifetime of its lithium batteries is 1101 hours. A homeowner selects 25 of these batteries and finds the mean lifetime to be 1081 hours with a standard deviation of 81 hours. Test the manufacturer's claim. Use a = 0.05. OA. p-value = 0.114 > 0.05; do not reject the Ho; there is enough evidence to support the claim. OB. p-value = 0.114 > 0.05; reject the Ho; there is not enough evidence to support...
A manufacturer claims that the mean lifetime of its lithium batteries is 1100 hours but a...
A manufacturer claims that the mean lifetime of its lithium batteries is 1100 hours but a group of homeowners believe their battery life is different than this manufacturer's claim. A homeowner selects 35 of these batteries and finds the mean lifetime to be 1080 hours with a standard deviation of 80 hours. Test the manufacturer's claim. Use a 10% level of significance and the p-value approach. Round the test statistic to the nearest thousandth. Need hypothesis testing, values of symbols,...
A manufacturer claims that the mean lifetime of its lithium batteries is 1401 hours. A homeowner...
A manufacturer claims that the mean lifetime of its lithium batteries is 1401 hours. A homeowner selects 25 of these batteries and finds the mean lifetime to be 1380 hours with a standard deviation of 81 hours. Test the manufacturer's claim. Use a=0.01. p-value = 0.104 > 0.01; reject the Ho; there is not enough evidence to support the claim. ОА. p-value = 0.207 > 0.01; do not reject the Ho; there is enough evidence to support the claim. ОВ....
A manufacturer claims that the mean lifetime of its lithium batteries is less than 1095 hours....
A manufacturer claims that the mean lifetime of its lithium batteries is less than 1095 hours. A homeowner selects 27 of these batteries and finds the mean lifetime to be 1065 hours with a standard deviation of 76 hours. Test the manufacturer's claim. Use a = 0.02. P-value = 0.112 > 0.02; do not reject H0; There is not enough evidence support the claim that mean is less than 1500. ОА P-value = 0.025 0.02, do not reject HO, There...
1. A manufacturer claims that the mean lifetime of its lithium battery is 1000 hours. A...
1. A manufacturer claims that the mean lifetime of its lithium battery is 1000 hours. A homeowner selects 40 batteries and finds the mean lifetime to be 990 hours with a standard deviation of 80 hours. Test the manufacturer's claim. Use alpha = 0.05. a. Calculate the test statistics. test statistics= (Round your answer to the nearest hundredth) Answer is not 0.79 or 0.80 2. A local juice manufacturer distributes juice in bottles labeled 12 ounces. A government agency thinks that...
12) rind the critical value and rejection region for the type of t-test with level of...
12) rind the critical value and rejection region for the type of t-test with level of sign nance anu sa pe le n. z) Left-tailed test, a 0.01, n-36 A) to =-2.438; t <-2.438 C) to 1.306; t -1.306 B) to D) to 2.434; t <-2.434 2.438; t>2.438 Objective: (7.3) Find Critical Values in a t-distribution 13) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1400 hours. A homeowner 13)- selects 25 bulbs and finds the...
A bulb manufacturer claims that its compact fluorescent bulbs have an average of less than 3.5...
A bulb manufacturer claims that its compact fluorescent bulbs have an average of less than 3.5 mg of mercury. A sample of 25 bulbs showed a mean of 3.39 mg of mercury. The population standard deviation is 0.18 mg. Using α = 0.1, does the sample support the manufacturer’s claim? (use both methods)
A local retailer claims that the mean lifetime of its lithium batteries is normally distributed with...
A local retailer claims that the mean lifetime of its lithium batteries is normally distributed with a mean of 1400 hours. A homeowner randomly selects 29 of these batteries and finds the mean lifetime to be 1380 hours with a standard deviation of 50 hours. Test the manufacturer’s claim. At an ? = .1, test the retailer’s claim. a.) State the null and alternative hypotheses. b.) Verify that the requirements are met for conducting the hypothesis test. c.) Conduct the...
(SHOW YOUR WORK!!!) A local retailer claims that the mean lifetime of its lithium batteries is...
(SHOW YOUR WORK!!!) A local retailer claims that the mean lifetime of its lithium batteries is normally distributed with a mean of 1400 hours. A homeowner randomly selects 25 of these batteries and finds the mean lifetime to be 1340 hours with a standard deviation of 45 hours. Test the manufacturer’s claim. At an ? = .1, test the retailer’s claim. a.) State the null and alternative hypotheses. b.) Verify that the requirements are met for conducting the hypothesis test....
A manufacturer claims that the mean lifetime, it, of its light bulbs is 44 months. The...
A manufacturer claims that the mean lifetime, it, of its light bulbs is 44 months. The standard deviation of these lifetimes is 5 months. Fifty bulbs are selected at random, and their mean lifetime is found to be 45 months. Can we conclude, at the 0.05 level of significance, that the mean lifetime of light bulbs made by this manufacturer differs from 44 months? Perform a two-tailed test. Then fill in the table below Carry your intermediate computations to at...
A manufacturer claims that the mean lifetime of its lithium batteries is 1101 hours. A homeowner selects 25 of these batteries and finds the mean lifetime to be 1081 hours with a standard deviation of 81 hours. Test the manufacturer's claim. Use a = 0.05. OA. p-value = 0.114 > 0.05; do not reject the Ho; there is enough evidence to support the claim. OB. p-value = 0.114 > 0.05; reject the Ho; there is not enough evidence to support...
A manufacturer claims that the mean lifetime of its lithium batteries is 1401 hours. A homeowner selects 25 of these batteries and finds the mean lifetime to be 1380 hours with a standard deviation of 81 hours. Test the manufacturer's claim. Use a=0.01. p-value = 0.104 > 0.01; reject the Ho; there is not enough evidence to support the claim. ОА. p-value = 0.207 > 0.01; do not reject the Ho; there is enough evidence to support the claim. ОВ....
A manufacturer claims that the mean lifetime of its lithium batteries is less than 1095 hours. A homeowner selects 27 of these batteries and finds the mean lifetime to be 1065 hours with a standard deviation of 76 hours. Test the manufacturer's claim. Use a = 0.02. P-value = 0.112 > 0.02; do not reject H0; There is not enough evidence support the claim that mean is less than 1500. ОА P-value = 0.025 0.02, do not reject HO, There...
12) rind the critical value and rejection region for the type of t-test with level of sign nance anu sa pe le n. z) Left-tailed test, a 0.01, n-36 A) to =-2.438; t <-2.438 C) to 1.306; t -1.306 B) to D) to 2.434; t <-2.434 2.438; t>2.438 Objective: (7.3) Find Critical Values in a t-distribution 13) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1400 hours. A homeowner 13)- selects 25 bulbs and finds the...
A manufacturer claims that the mean lifetime, it, of its light bulbs is 44 months. The standard deviation of these lifetimes is 5 months. Fifty bulbs are selected at random, and their mean lifetime is found to be 45 months. Can we conclude, at the 0.05 level of significance, that the mean lifetime of light bulbs made by this manufacturer differs from 44 months? Perform a two-tailed test. Then fill in the table below Carry your intermediate computations to at...
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