The US vehicle sticker information states that you can expect 19 miles per gallon. It is desired to check the consistency of this information. Sixteen vehicles of the same model were tested. Assuming the miles per gallon X-N (μ, 0²) and we take a random sample of n=16. What is the p-value when testing = 19 Hop: H : μ # 19 We observe = 18. 16 and s-1.48 where s is the sample standard deviation. 0.038 1 0.019 0.037
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- Please solve this by hand and also with r-studio code not the excel output please. An experimenter is interested in the hypothesis testing problem with H0: µ = 420 versus Ha: µ <420, where µ is the average radiation level in a research laboratory. Suppose that a sample of 49radiation level measurements is obtained and that the experimenter wishes to use σ=10.0 for thestandard deviation of the radiation levels.(a) What is the rejection region when α=0.01?(b) Suppose that the sample mean is 415.7. What would be the experimenter’s decision of thehypothesis test based on the rejection region?(c) What is the p-value for this hypothesis test?The effectiveness of a new bug repellent is tested on 1616 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: ?⎯⎯⎯=92x¯=92, ?=13 s=13 The new repellent is considered effective if it provides a percent repellency of at least 9090. Using ?=0.05α=0.05, construct a hypothesis test with null hypothesis ?≤90μ≤90 and alternative hypothesis ?>90μ>90 to determine whether the mean repellency of the new bug repellent is greater than 9090 by computing the following: (a) the degree of freedom (b) the test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that ?≤90μ≤90. Our results do not provide enough evidence that the new bug repellent is effective. B. We can reject the null hypothesis that ?≤90μ≤90. Our results indicate that…A research center claims that 26% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 1000 adults in that country, 29% say that they would travel into space on a commercial flight if they could afford it. At α=0.05, is there enough evidence to reject the research center's claim? Complete parts (a) through (d) below. (a) Identify the claim and state H0 and Ha. "Recall that the claim is the percentage of adults in the country would travel into space on a commercial flight if they could afford it. Let a success be an adult in the country who would travel into space on a commercial flight if they could afford it. Translate the claim made about the population parameter from a verbal statement to a mathematical statement." (b) Use technology to find the P-value. (c) Decide whether to reject or fail to reject the null hypothesis and (d) interpret the decision in the context of the original claim.
- The research examining the effects of preschool childcare has found that children who spent time in daycare perform better on math and language tests than children who stay home with their mothers. The results show that a sample of n = 25 children who attended daycare before starting school had an average score of M = 87 with SS = 1536 on a standardized math test for which the population mean is μ = 81. The population distribution is normal. Use α= .01 a. Is this sample sufficient to conclude that the children with a history of preschool daycare are significantly better from the general population? b. Compute the 95% confidence interval. show all workThe research examining the effects of preschool childcare has found that children who spent time in day care perform better on math and language tests than children who stay home with their mothers. The results show that a sample of n = 25 children who attended day care before starting school had an average score of M = 87 with SS = 1536 on standardized math test for which the population mean is μ = 81. The population distribution is normal. Use α = .01. a. Is this sample sufficient to conclude that the children with a history of preschool day care are significantly better from the general population? b. Compute the 95% confidence interval.Molly works for a meat producer, and she needs to determine whether containers of ground beef have the correct fat content. She obtains a random sample of 120 containers of ground beef and finds that 84 percent have the correct fat content. Molly then conducts a hypothesis test of H0:p=0.80H0:p=0.80 versus Ha:p≠0.80Ha:p≠0.80 and calculates a test statistic of 1.10 with a pp-value of 0.273. Which of the following best represents the meaning of the pp-value? If the population proportion is 0.84, the probability of observing a sample proportion of 0.80 is 0.273. A If the population proportion is 0.84, the probability of observing a sample proportion of at least 0.04 less than 0.84 is 0.273. B If the population proportion is 0.80, the probability of observing a sample proportion within 0.04 of 0.80 is 0.273. C If the population proportion is 0.80, the probability of observing a sample proportion at least 0.04 greater than 0.80 is 0.273. D If the…
- Molly works for a meat producer, and she needs to determine whether containers of ground beef have the correct fat content. She obtains a random sample of 120 containers of ground beef and finds that 84 percent have the correct fat content. Molly then conducts a hypothesis test of H0:p=0.80H0:p=0.80 versus Ha:p≠0.80Ha:p≠0.80 and calculates a test statistic of 1.10 with a pp-value of 0.273. Which of the following best represents the meaning of the pp-value? If the population proportion is 0.84, the probability of observing a sample proportion of 0.80 is 0.273. A If the population proportion is 0.84, the probability of observing a sample proportion of at least 0.04 less than 0.84 is 0.273. B If the population proportion is 0.80, the probability of observing a sample proportion within 0.04 of 0.80 is 0.273. C If the population proportion is 0.80, the probability of observing a sample proportion at least 0.04 greater than 0.80 is 0.273. D If the…The result for a Pearson’s r between students’ anxiety and their achievement was, r(90) = - .75, p = .03. What can the researcher conclude?GLA2.4 Q1 For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples):
- A sample with M=8 has EX=56. How many scores are in the sample?Two separate samples receive two different treatments. The first sample has n = 9 with M = 50 and SS = 710. And the second sample has n = 6 with M = 37 and SS = 460. Are there significant treatment differences between these two samples? Use α = .05The result for a Pearson’s r for students’ perceived learning and their sense of community was, r(45) = .72, p = .02. What can the researcher conclude?