A factory produces a component used in manufacturing computers, each of which is tested prior to shipment to determine whether or not it is defective. In a random sample of 300 units, 12 were found to be defective. Estimate, with 99% confidence, the true proportion of defective components produced by this factory.
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- Anterior cruciate ligament (ACL) injuries occur most frequently in planting and cutting sports such as basketball, soccer, and volleyball. National collegiate athletic association injury data show that female athletes injure the ACL more frequently than their male counterparts do. A sports medicine professor wants to test this claim. He surveyed 5 universities that had both men's and woman's basketball, soccer, and volleyball and asked them how many grade 3 ACL tears they had during the past 5 seasons. Using the following results, are women at greater risk than men for ACL injuries? (use a standard deviation of 0.01, one tail test). Women- 4,3, 5, 2,4 Men- 2, 0, 1 ,2, 1 What is the difference between a repeated measure study and a matched pairs study?A recent study looked at the relationship between condom use and the spread of AIDS, by following heterosexual couples in which one partner was infected with HIV. The couples were classified by their condom use. Of the 98 couples who always used condoms, 4 partners became infected with HIV. Of the 67 couples who did not always use condoms, 11 partners became infected. Test whether the proportions of partners who became infected with HIV are different for each group, using a 5% level of significance. Give each of the following: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your workA physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 75 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. N Mean StDev SE Mean Men Women 95% Cl for mu Men - mu Women - 9.81, - 0.79) T-Test mu Men = mu Women (vs H2 Oc. Họ: H1 = H2; Hai H1In a random sample of males, it was found that 29 write with their left hands and 212 do not. In a random sample of females, it was found that 70 write with their left hands and 444 do not. Use a 0.01 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below.Anterior cruciate ligament (ACL) injuries occur most frequently in planting and cutting sports such as basketball, soccer, and volleyball. National collegiate athletic association injury data show that female athletes injure the ACL more frequently than their male counterparts do. A sports medicine professor wants to test this claim. He surveyed 5 universities that had both men's and woman's basketball, soccer, and volleyball and asked them how many grade 3 ACL tears they had during the past 5 seasons. Using the following results, are women at greater risk than men for ACL injuries? (use a standard deviation of 0.01, one tail test). Women- 4,3, 5, 2,4 Men- 2, 0, 1 ,2, 1 Calcualte the sum of squares of injuries for men.The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = IIIn college, talented runners may join a cross-country team. Runners tend to run their best times when they run even splits. Even splits occur when the runners maintain an even pace throughout the race. The cross-country coach wants to estimate the typical variability in his best runner's 1-mile splits. He takes a random sample of 25 of this runner's mile splits and finds that this runner's mean 1-mile split is 5.44 minutes per mile, with a standard deviation of 0.14 minutes per mile. This runner's 1-mile splits follow a normal distribution. (a) Find the chi-square critical values XL² and Xu² to be used in constructing a 95% confidence interval for the true population standard deviation. (Round your answers to two decimal places.) XL²= XU²= (b) Find the 95% confidence interval for the true variability in his best runner's 1-mile splits. (Round your answers to three decimal places.) lower bound and the upper boundA researcher believes that among the residents of a remote community, 30% have type O blood, 50% have type A blood, 10% have type B blood and 10% have type AB blood. She takes a random sample of 80 residents and finds that 20 had type O blood, 40 had type A blood, 12 had type B blood and 8 had type AB blood. Do the data follow the hypothesized distribution of blood type for this community? Test at a = .05. Round your answers to three decimal places, if necessary. Null hypothesis: Ho: Po = 0.3, PA = 0.5, PB = 0.1, PAB = 0.1 20 40 12 8 80 80 80 80 1 1 1 20 40 8 Ho: Po = Ho: Po = Alternative hypothesis: Ho: Po = PA = PB = PAB = Ha: po # Type O 1 Ha: Po PA PB = PAB = 4 20 › PA = PA = Type 0 20 24 40 80 80 Ha: Po 0.3, PA # 0.5, PB At least one of the proportions is different. Complete the table of observed counts: Type O , PA # 0.667 Type A PB = PB = 40 Type A 40 What is the test statistic? 2.667 1 ,PB # 0 , PAB = 1 12¹ PAB Complete the table of expected counts under the null hypothesis: 12…Gary has discovered a new painting tool to help him in his work. If he can prove to himself that the painting tool reduces the amount of time it takes to paint a room, he has decided to invest in a tool for each of his helpers as well. From records of recent painting jobs that he completed before he got the new tool, Gary collected data for a random sample of 7 medium-sized rooms. He determined that the mean amount of time that it took him to paint each room was 3.4 hours with a standard deviation of 0.3 hours. For a random sample of 6 medium-sized rooms that he painted using the new tool, he found that it took him a mean of 3.2 hours to paint each room with a standard deviation of 0.2 hours. At the 0.05 level, can Gary conclude that his mean time for painting a medium-sized room without using the tool was greater than his mean time when using the tool? Assume that both populations are approximately normal and that the population variances are equal. Let painting times without using…Clothes dried outside smell better than those dried in a dryer because of a process called photolysis. Sunlight breaks down compounds that cause odor. A trained smeller is asked to smell a random sample of 20 towels. The smeller does not know it, but all of the 20 towels have been dried in the sun. The smeller is blindfolded and is not allowed to touch the towels but leans forward to smell the towel then rates the towel on a smell-scale of 1 to 10, where 1 is the an extremely unpleasant smell and 10 is a very good smell. A smell rating of 8 is typically attributed to a towel that is dried in the dryer. The mean smell rating for these 20 towels is approximately normal with a mean of 8.45 and the standard deviation is 0.887. A statistician performs a test of Ho μ = 8 versus H₁ μ> 8 where is the true mean smell rating for all towels dried in the sun. This test yields a P-value of 0.0176. What conclusion would you make at the significance level a = 0.05? Because the P-value of 0.0176 < α =…The plant-breeding department at a major university developed a new hybrid boysenberry plant called Stumptown Berry. Based on research data, the claim is made that from the time shoots are planted 90 days on average are required to obtain the first berry. A corporation that is interested in marketing the product, tests 60 shoots by planting them and recording the number of days before each plant produces its first berry. The sample mean is 92.3 days. The corporation will not market the product if the mean number of days is more than the 90 days claimed. The hypotheses are: H0:μ = 90 H1:μ > 90 What is a type I error in the context of this problemA researcher claims that more than 38% of US adults have been a victim of identity theft.ln a random sample of 485 US adults,194 of them have been a victim of identity theft.Use a 5% significance level to determine if there is enough evidence to support the researcher's claim. Ho:- HA:- Z stat = ( Round to 2 decimal places ) P- value = Decision: Type either Reject H0 or Retain H0. Write a summary statement for the hypothesis.
