A-E F-J K-O P-T U-Z Total Section 19 16 15 26 6. 82 004 Section 16 13 12 52 005 Section 21 10 8 80 006 Total 56 35 48 59 16 214. 2. 21 20 9,
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- Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. H0: μ=60 seconds H1: μ≠60 seconds B. H0: μ=60 seconds H1: μ<60 seconds C. H0: μ=60 seconds H1: μ>60 seconds D. H0: μ≠60 seconds H1: μ=60 seconds Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.)The label on a 4-quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. (a) Develop the appropriate null and alternative hypotheses. H0: ? > 1 Ha: ? ≤ 1 H0: ? < 1 Ha: ? ≥ 1 H0: ? ≤ 1 Ha: ? > 1 H0: ? = 1 Ha: ? ≠ 1 H0: ? ≥ 1 Ha: ? < 1 (b) What is the type I error in this situation? What are the consequences of making this error? It is claiming ? < 1 when it is not. This error would claim that the product is not meeting its label specification when it really is meeting its specification. It is claiming ? ≥ 1 when it is not. This error would miss the fact that the product is not meeting its label specification. It is claiming ? ≤ 1 when it is not. This error would miss the fact that the product is not meeting its label specification. It is claiming ? > 1 when it is not. This error would claim that the…An ad for a pest control company claims that 96% of their customers are 'extremely satisfied' with their service. A group of recent customers believes that figure is too high, and wish to conduct a hypothesis test at the α=0.05α=0.05 level of significance to determine if they are right. Which would be correct hypotheses for such a test? H0:p=0.96H0:p=0.96, H1:p<0.96H1:p<0.96 H0:p≠0.96H0:p≠0.96, H1:p=0.96H1:p=0.96 H0:p=0.96H0:p=0.96, H1:p>0.96H1:p>0.96 H0:p<0.96H0:p<0.96, H1:p=0.96H1:p=0.96 H0:p=0.96H0:p=0.96, H1:p≠0.96H1:p≠0.96 Which situation would represent a Type I error? Fail to reject the null hypothesis, which is actually false Reject the null hypothesis, which is actually false Reject the null hypothesis, which is actually true Fail to reject the null hypothesis, which is actually true If the resulting P-value is 0.024, which would be the appropriate conclusion? There is significant evidence to suggest that the actual proportion of satisfied…
- An ad for an artificial turf company claims that 92% of their customers are 'extremely satisfied' with their service. A group of recent customers believes that figure is too high, and wish to conduct a hypothesis test at the α=0.10α=0.10 level of significance to determine if they are right. Which would be correct hypotheses for this test? H0:p=0.92H0:p=0.92, H1:p>0.92H1:p>0.92 H0:p=0.92H0:p=0.92, H1:p<0.92H1:p<0.92 H0:p<0.92H0:p<0.92, H1:p=0.92H1:p=0.92 H0:p=0.92H0:p=0.92, H1:p≠0.92H1:p≠0.92 H0:p≠0.92H0:p≠0.92, H1:p=0.92H1:p=0.92 A random sample of 94 customers resulted in 85 that said they were 'extremely satisfied' with their service. Find the test statistic (Round to 2 decimals):Give the P-value (Round to 4 decimals - if less than 0.001 give answer as 0):The effectiveness of a new bug repellent is tested on 14 subjects for a 10 hour period. (Assume normally distributed population.) Based on the number and location of the bug bites, the percentage of surface area exposed protected from bites was calculated for each of the subjects. The results were as follows: x¯=93, ?=6 The new repellent is considered effective if it provides a percent repellency of at least 9090. Using ?=0.01, construct a hypothesis test with null hypothesis ?=90 and alternate hypothesis ?>90 to determine whether the mean repellency of the new bug repellent is greater than 9090 by computing the following: (a) the degrees of freedom (b) the test statisticThe effectiveness of a new bug repellent is tested on 25 subjects for a 10 hour period. (Assume normally distributed population.) Based on the number and location of the bug bites, the percentage of surface area exposed protected from bites was calculated for each of the subjects. The results were as follows: ?⎯⎯⎯=93, ?=12 The new repellent is considered effective if it provides a percent repellency of at least 91. Using ?=0.05, construct a hypothesis test with null hypothesis ?=91 and alternate hypothesis ?>91 to determine whether the mean repellency of the new bug repellent is greater than 91 by computing the following: (a) the degrees of freedom (b) the test statistic The final conclusion is A. We can reject the null hypothesis that ?=91. Our results indicate that the new bug repellent is effective.B. There is not sufficient evidence to reject the null hypothesis that ?=9. Our results do not provide enough evidence that the new bug repellent is effective.
- The label on a 3-quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. 1. What is the type I error in this situation? What are the consequences of making this error? A) It is claiming ? ≤ 1 when it is not. This error would miss the fact that the product is not meeting its label specification. B) It is claiming ? ≥ 1 when it is not. This error would miss the fact that the product is not meeting its label specification. C) It is claiming ? > 1 when it is not. This error would claim that the product is not meeting its label specification when it really is meeting its specification. D) It is claiming ? < 1 when it is not. This error would claim that the product is not meeting its label specification when it really is meeting its specification. #2 What is the type II error in this situation? What are the…Find the P-value for a left-tailed hypothesis test with a test statistic of z=−1.96. Decide whether to reject H0 if the level of significance is α=0.05.In 2002 a sta)s)cian named Fukuda found the following in Japan: Out of 565 births where both parents smoked more than a pack a day, 255 were boys. Out of 3602 births where both parents didn’t smoke, 1975 were boys. At ∝= 0.05, can we claim that there is a smaller propor)on of boys with parents who are smokers? Test using: a.) a hypothesis test b.) an appropriate CI
- An panelist on a morning news program recently claimed that 32% of drivers routinely drive while distracted by their smartphones. A statistician happened to be watching this program, and decided to conduct a hypothesis test at the α=0.01α=0.01 level of significance to determine if this is accurate. (Round your results to three decimal places)Which would be correct hypotheses for this test? H0:p≠0.32H0:p≠0.32, H1:p=0.32H1:p=0.32 H0:p<0.32H0:p<0.32, H1:p=0.32H1:p=0.32 H0:p=0.32H0:p=0.32, H1:p<0.32H1:p<0.32 H0:p=0.32H0:p=0.32, H1:p>0.32H1:p>0.32 H0:p=0.32H0:p=0.32, H1:p≠0.32H1:p≠0.32 A random sample of 176 drivers resulted in 151 that routinely drive while distracted by their smartphones. Find the test statistic (2 decimals):Give the P-value (4 decimals):The P-value for a hypothesis test is P = 0.034. Do you reject or fail to reject H 0 when the level of significance is α = 0.01?The P-value for a hypothesis test is 0.06. For each of the fol-lowing significance levels, decide whether the null hypothesis should be rejected. a. α = 0.05b. α = 0.10 c. α = 0.06