Twelve different video games showing substance use were observed and the duration of times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the sample data to construct a 95% confidence interval estimate of o, the standard deviation of the duration times of game play. Assume that this sample was obtained from a population with a normal distribution. 4,512 4,630 3,913 4,857 4,898 3,936 4,034 4,572 4,563 4,305 3,902 4,075 Click the icon to view the table of Chi-Square critical values. The confidence interval estimate is sec
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- Before the furniture store began its ad campaign, it averaged 123 customers per day. The manager is investigating if the average is larger since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 110, 154, 108, 112, 151, 146, 144, 130, 138, 129, 142 Assuming that the distribution is normal, what can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: vSelect an answer H1: ?V Select an answer V c. The test statistic ? (please show your answer to 3 decimal places.) d. The p-value = e. The p-value is ?V a f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) the null hypothesis. O The data suggest the populaton mean is significantly more than 123 at a = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the…I need help on this practice problemInsurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂A study reported that finger rings increase the growth of bacteria on health-care workers’ hands. Research suggests that 31 percent of health-care workers who wear rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings is selected. What is the standard deviation of the sampling distribution of the sample proportions of health-care workers having bacteria on one or both hands? A. 100 B. 0.0462 C. 0.69A study is conducted to compare the performance of students with more than one personal electronic gadget and those with only one. A number of them were taken as subjects for the study. The mean grades of these students and the standard deviations are shown below: For one gadget: The mean is 83, standard deviation is 12, and sample size is 7. For more than one gadget: The mean is 79, standard deviation is 13, and sample size is 5. Is it possible to conclude that there is no significant difference in the mean grades of the two types of students at a 95% confidence interval? Note: This is a two-tailed test. Also take note that this is finding the difference between two population means.
- A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.10 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: Hq ZH₂ OC. Ho: H₁ H₂ H₁: Hy > H₂ The test statistic, t, is. (Round to two decimal places as needed.) (Round to three decimal places as needed.) The P-value is State the conclusion for the test. C... OB. Ho: H₁ H₂ H₁: Hy #H₂ OD. Ho: Hg #U2 H₁: HyGiven two dependent random samples with the following results: Population 1 35 34 45 30 49 25 34 Population 2 44 24 40 42 38 31 43 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that both populations are normally distributed. Step 2 of 4 : Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places. please highlight the answerInsurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$151 per month with a standard deviation of $16$16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$158 per month with a standard deviation of $19$19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: H₁ H₂ OC. Ho: H₁ H¹/₂ H₁: H₁A study is done to determine if students in the California state university (CSU) system take longer to graduate, on average, than students enrolled in private universities using the significant level of 5%. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years for CSU and 1 year for private universities. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. What is the decision rule of rejecting the null hypothesisIn a random sample of 38 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 61 months, with a standard deviation of 8 months. Construct and interpret a 95% confidence interval for the mean length of sentencing for this crime. Select the correct choice below and fill in the answer boxes to complete your choice.(Use ascending order. Round to one decimal place as needed.) A. 95% of the sentences for the crime are between ____ and ____ months. B. There is a 95% probability that the mean length of sentencing for the crime is between ____ and ____ months. C. We can be 95% confident that the mean length of sentencing for the crime is betweenRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. 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