What should their null hypothesis be if they want to collect evidence against the manufacturer's claim? What form should their p-value have?
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I bought a new set of headphones, and it failed after two days. The manufacturer claims that this happens only in less than 1 percent of cases.
The Bureau of Consumer Affairs decides to investigate my complaint and tests 120 sets of headphones.
What should their null hypothesis be if they want to collect evidence against the manufacturer's claim? What form should their p-value have?
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- I am having difficulty finding the correct test statistic. I tried 2.2987, which was incorrect. I would be grateful for any help at all, thanks so much! :)It has been claimed that 71% of all adults use their cell phones within an hour of going to bed. Believing this claimed value is too low, a researcher surveys a large random sample of adults and obtains a test statistic of 0.5. What will the P-value be? Between 0.01 and 0.05 O There is not enough information available in the problem to determine the P-value. Larger than 0.10 Between 0.05 and 0.10 O Less than 0.01Suppose are running a study/poll about the proportion of women over 40 who regularly have mammograms. You randomly sample 138 people and find that 91 of them match the condition you are testing.Suppose you are have the following null and alternative hypotheses for a test you are running:H0:p=0.67Ha:p>0.67Calculate the test statistic, rounded to 3 decimal places z=
- The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65-74 years: n = 117, x = 12.2, and s = 6.60. Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use a = 0.05.) State the appropriate null and alternative hypotheses. Ο H, μ 15 Ha: u > 15 Ho:H = 15 Ha: u < 15 Ο H: μ= 15 H: u + 15 Ο H: μ= 15 H3: Hs 15 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.) P-value =The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65-74 years: n = 113, x = 12.4, and s = 6.63. Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use a = 0.05.) State the appropriate null and alternative hypotheses. O Ho: H = 15 H: u > 15 O Ho: H = 15 H: u < 15 Ο H: μ15 H:us 15 Ο Η: μ- 15 H: u = 15 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. O Reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There…Let's examine the mean of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 by drawing samples from these values, calculating the mean of each sample, and then considering the sampling distribution of the mean. To do this, suppose you perform an experiment in which you roll an eight-sided die two times (or equivalently, roll two eight-sided dice one time) and calculate the mean of your sample. Remember that your population is the numbers 1, 2, 3, 4, 5, 6, 7, and 8. The true mean (p) of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 is The number of possible different samples (each of size n = 2) is the number of possibilities on the first roll (8) times the number of possibilities on the second roll (also 8), or 8(8) = 64. If you collected all of these possible samples, the mean of your sampling distribution of means (HM) would equal , and the standard deviation of your sampling distribution of means (that is, the standard error or GM) would be The following chart shows the sampling distribution of the…
- Suppose are running a study/poll about the proportion of women over 40 who regularly have mammograms. You randomly sample 117 people and find that 92 of them match the condition you are testing.Suppose you are have the following null and alternative hypotheses for a test you are running:H0:p=0.83Ha:p<0.83Calculate the test statistic, rounded to 3 decimal places z=z=Generally speaking, would you say that most people can be trusted? A random sample of n1 = 257 people in Chicago ages 18-25 showed that r1 = 40 said yes. Another random sample of n2 = 281 people in Chicago ages 35-45 showed that r2 = 70 said yes. Does this indicate that the population proportion of trusting people in Chicago is higher for the older group? Use α = 0.01. (a) What is the level of significance? (b) What is the value of the sample test statistic? (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)A certain magazine stated that the mean time for a Ford Mustang to go from 0 to 60 miles per hour is 4.1 seconds. (a) If you want to set up a statistical test to challenge the claim of 4.1 seconds, what would you use for the null hypothesis? 04.1 O = 4.1 (b) The town of Leadville, Colorado, has an elevation over 10,000 feet. Suppose you wanted to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer Leadville (because of less oxygen). What would you use for the alternate hypothesis? OH 4.1 OM 4.1 O = 4.1 (c) Suppose you made an engine modification and you think the average time to accelerate from 0 to 60 miles per hour is reduced. What would you use for the alternate hypothesis? OH> 4.1 O μ< 4.1 OH 4.1. (d) For each of the tests in parts (b) and (c), would the P-value area be on the left, on the right, or on both sides of the mean? O left; both O right; left O left; right O both; left
- Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?Bags of a certain brand of tortilla chips claim to have a net weight of 14 oz. Net weights actually vary slightly from bag to bag and are Normally distributed with mean u. A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised, so he intends to test the hypotheses Ho: μ = 14, Ha: μ < 14. To do this, he selects 16 bags of this brand at random and determines the net weight of each. He finds the sample mean to be x = 13.88 and the sample standard deviation to be s = 0.24. Suppose in a similar test of 16 bags of these tortilla chips the P-value is 0.001. Further suppose that a is chosen to be 0.001. In that case, we would conclude that: Othere is not significant evidence that the mean net weight of the bags of chips is less than the advertised 14 oz. O there is not significant evidence that the mean net weight of the bags of chips is not less than the advertised 14 oz. O there is significant evidence that the…Suppose you had obtained a test statistic of 2.45 (this is incorrect). What would the P-value be? Draw a conclusion based on a = 0.01. Select one: O a. 0.01 < P-value < 0.02, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O b. 0.005 < P-value < 0.01, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O c. 0.01 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O d. 0.005 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet…
