A study was carried out to evaluate the safety of motorcycle helmets. 40 randomly selected helmets were subiected to a certain impact test and 20 were found to show no damage. The study claimed that the proportion of all helmets that would show no damage when tested in the prescribed manner is smaller than 62.5% (i.e. 5/8). With what confidence can this be claimed? a. 0.91 b. 0.93 c.0.95
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- A dietician wishes to see if a person’s cholesterol level will change if the diet is supplemented by a certain mineral. Six randomly selected subjects were pretested, and then they took the mineral supplement of a 6 week period. The results are shown in the table below. Can it be concluded that the cholesterol level has changed at a significance level of 0.10? Use the P-value method with your calculator. Assume the variable is approximately normally distributed. You can use your calculator to answer this question but make sure that you clearly write the statements for the null and alternative hypothesis. Be sure that you thoroughly explain your final conclusion using the P-value method. Subject 1 2 3 4 5 6 Before 210 235 208 190 172 244 After 190 170 210 188 173 228How much customers buy is a direct result of how much time they spend in the store. A study of average shopping times in a large national houseware store gave the following information (Source: Why We Buy: The Science of Shopping by P. Underhill). Women with female companion: 8.3 min.Women with male companion: 4.5 min. Suppose you want to set up a statistical test to challenge the claim that a woman with a female friend spends an average of 8.3 minutes shopping in such a store. (a) What would you use for the null and alternate hypotheses if you believe the average shopping time is less than 8.3 minutes? Ho: ? = 8.3; H1: ? ≠ 8.3Ho: ? < 8.3; H1: ? = 8.3 Ho: ? = 8.3; H1: ? < 8.3Ho: ? = 8.3; H1: ? > 8.3 Is this a right-tailed, left-tailed, or two-tailed test? right-tailedtwo-tailed left-tailed (b) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from 8.3 minutes? Ho: ? = 8.3; H1: ? > 8.3Ho: ? ≠ 8.3; H1: ? = 8.3…In a study of minimizing tractor skidding distances along a new road in a European forest, the skidding distances were measured at 20 randomly selected road sites. The data are reported in S. the table below. A logger working on a road claims the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use a 0.10. (Yes) 488 350 457 199 285 409 435 574 439 546 385 295 184 261 273 400 311 312 141 425
- It has been claimed that at UCLA at least 40% of the students live on campus. In a random sample of 250 students, 90 were found to live on campus. Does the evidence support the claim at = .01?A researcher wanted to test whether there is a relationship between preferred sweet (Oreo cookies, Mars bars, or Twinkies) and age group (children, adolescents, or young adults). To test that, subjects from the three age groups were asked which of the three sweets they preferred. Which statistical test should be used to analyze the data? 1. chi square test of independece 2. chi square test for goodness of fit 3. one sample t test 4. one way anova testA magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 13% chose chocolate pie, and the margin of error was given as ±4 percentage poits. Describe what is meant by the statement that "the margin of error was given as ±4 percentage points." (...) Choose the correct answer below. OA. The statement indicates that the study is only 4% confident that the true population percentage of people that prefer chocolate pie is exactly 13%. OB. The statement indicates that the study is 100% -4% = 96% confident that the true population percentage of people that prefer chocolate pie is 13%. OC. The statement indicates that the interval 13% ± 4% is likely to contain the true population percentage of people that prefer chocolate pie. D. The statement indicates that the true population percentage of people that prefer chocolate pie is in the interval 13% ± 4%.
- Casey is a statistics student who is conducting a one-sample ?z‑test for a population proportion ?p using a significance level of ?=0.05α=0.05. Her null (?0H0) and alternative (??Ha) hypotheses are ?0:???:?=0.094≠0.094H0:p=0.094Ha:p≠0.094 The standardized test statistic is ?z = 1.20. What is the ?P-value of the test?An analysis of variances produces dftotal = 29 and dfwithin = 27. For this analysis, how many treatment conditions are being compared? a. 1 b. 2 c. 3 d. Cannot be determined without additional informationThe assumption of homogeneity of variances was tested, and the Levene’s test yielded a value of (p = .06). Was the assumption met?
- In a 4-week study about the effectiveness of using magnetic insoles to treat plantar heel pain, 58 subjects wore magnetic insoles and 40 subjects wore nonmagnetic insoles. The results are shown at the right. At a= 0.06, can you support the claim that there is a difference in the proportion of subjects who feel better between the two groups? Assume the random samples are independent. Complete parts (a) through (e). Do you Feel Better? Magnetic Insoles I Yes 20 I No 38 Nonmagnetic Insoles Yes 17 No 23 (Use a comma to separate answers as needed. Type an integer or a decimal. Round to two decimal places as needed.) Identify the rejection region(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) O A. z O B. z> O D. z< C. (c) Find the standardized test statistic. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis Choose the correct answer below.According to the NIMH website the average number of adult Americans diagnosed with a “serious mental illness” across all 50 states in 2019 was μ = 5.2 out of 100 residents (or 5.2%). Please test to see if the sample of states listed below (n=8) has a significantly higher level of mental illness than the population mean of 5.2. Sample data: Arkansas= 5.45%, Montana=5.34%, Vermont=5.32%, Oregon=5.19%, Utah=5.2%, Ohio=5.13, Kentucky=5.25% West Virginia=5.18%. Hint - drop the % signs and use 5.45, 5.32, etc as the X scores. If you want you can use JASP for help with the calculations. Use 3 decimal places for the M, s, SEM. The alpha level for the study was set to α=.05. a. State the hypotheses of the study. b. Find the critical t value for this studyc. Compute the one-sample t test. d. Compute the Cohen's d for the data (even if not significant) e. State the results - is the sample significantly higher? Do you reject the null hypothesis? Report the t test results in APA format…A researcher is studying how much electricity (in kilowatt hours) people from two different cities use in their homes. Random samples of 11 days from Tampa (Group 1) and 16 days from New York City (Group 2) are shown below. Test the claim that the mean number of kilowatt hours in Tampa greater than the mean number of kilowatt hours in New York City. Use a significance level of a = 0.10. Assume the populations are approximately normally distributed with unequal variances. Round answers to 4 decimal places. Tampa New York City 1015.4 1010 1010.3 1022.8 1001.7 1028.9 1001.7 1026.9 1022.6 1001.7 1020.2 1012.4 1021.9 1015.9 1018.9 1005.8 996.2 1006.1 1010.5 1012 1005.1 1025.9 1005.6 1020.2 985.3 994.9 1013.2 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols for each of the 6 spaces. Ho: Select an answer ✓ ? Select an answer ✓ H₁: Select an answer ✓ ? ✓ Select an answer ✓
