It is of interest to determine if the average drying time of a certain brand of latex paint is not 4.5 hours. To do this, 15 cans of the branded latex paint were selected. Normality test of drying time of the branded latex paint: Shapiro-Wilk normality test data: drying time W = 0.9691, p-value = 0.1396 Which of the following is(are) TRUE at 5% level of significance? I. The assumption on the normality of drying time of the branded latex paint is NOT SATISFIED. II. To determine if the average drying time of a certain brand of latex paint is not 4.5 hours, a NONPARAMETRIC TEST is more appropriate to use than a parametric test. Choices: I only, II only, Both I and II, Neither I nor II
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It is of interest to determine if the average drying time of a certain brand of latex paint is not 4.5 hours. To do this, 15 cans of the branded latex paint were selected.
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- The score on a standardized test for a certain year is modeled using the normal distribution shown below. The mean of the distribution is 56.7 points and the standard deviation is 4.8 points. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) v 68% 95% 99.7% 40 45 50 55 O 60 65 | 70 75 Test score200 people were randomly sampled and asked what they regularly eat for breakfast or lunch. Each person was identified as either a consumer or a non consumer of high-fiber cereals, and the number of calories consumed at lunch was measured and recorded. These data are summarized below; Consumer of high fiber cereals Non consumer of high fiber cereals η1 =41 η2 = 159 Mean 1 =603 Mean 2 =639 Stanadard deviation 1 = 110 Standard deviation 2 = 141 If the scientist claims that people who eat high fiber cereals for breakfast do consume on average fewer calories for lunch than people who don’t eat high fiber cereals for breakfast, and if it is true, high fiber cereal manufacturer will be able to claim another advantage of eating their products-potential weight reduction for dieter.Use the following data to conduct a one-sample t test (with alpha = 0.05) to determine if the score on a standardized test (for scores that are normally distributed) with a known population mean of 100 is influenced by a new instructional method: 95, 105, 110, 90, 120, 110, 100, 95, 105, 125, 80, 100, 120 What is your critical t value? What is your sample standard deviation? What is your obtained t value? What do you conclude?
- Suppose the mean retail price oer gallon of regular gasoline int the United States is $3.42 with a standard eviation of $0.40 and the retail price per gallon has a normal bell shaped distribution. Use the Empirical Rule to answer the following quesitona. What percentage of regular gasoline sold between $2.62 and $3.82 per gallon? what percentage of regualr gasoline sold for more the $3.82 per gallon?The height of an 8th grader is modeled using the normal distribution shown below. The mean of the distribution is 59.1 in and the standard deviation is 1.2 in. In the figure, Vis a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) V 56 58 60 62 Height (in inches)A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage x¯1x¯1 = $ 1,101 x¯2x¯2 = $ 1,766 Sample Standard Deviation s1 = $ 696 s2 = $ 838 Repair Incidents n1 = 12 n2 = 9 (a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round answers to 3 decimal places. Negative values should be indicated by a minus sign.) (b) Repeat part (a), using the assumption of unequal variances with Welch's formula for d.f. (Round answers to 2 decimal places. Negative values should be indicated by a minus sign.) (d) Construct separate…
- You are conducting a study to see if the accuracy rate for fingerprint identification is significantly different from 0.51. Thus you are performing a two-tailed test. Your sample data produce the test statistic - 2.153. Find the p-value accurate to Z = - 4 decimal places. p-valueA test was conducted on the differences in academic performance of three groups of students subjected to different teaching strategies. Below is the ANOVA table of the results of data analysis. Source F- MS Ratio critical value F- P- of df SS variation Among 3 2100 MST groups ??? 3.18 0.036 Within 45 10000 MSE groupAccording to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 21 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, u, is now less than 21 kilometers. The administrator chooses a random sample of 85 visitors. The mean distance hiked for the sample is 20.5 kilometers. Assume the population standard deviation is 8.3 kilometers. Can the administrator conclude that the mean distance hiked by each visitor is now less than 21 kilometers? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H and the alternative hypothesis H . H: 0 OThe herbal supplement ginkgo biloba is advertised as producing an increase in physical strength and stamina. To test this claim, a sample of n = 120 adults is obtained and each person is instructed to take the regular daily dose of the herb for a period of 30 days. At the end of the 30 days each person is tested on a standard treadmill task for which the average age-adjusted score is μ = 15. The individuals in the sample produce a mean score of M = 17.4 with SS = 1260. Are these data sufficient to conclude that the herb had a statistically significant effect? Conduct as a two-tailed test, using all hypothesis steps with p < .05. BONUS POINTS: Calculate the 95% confidence intervalA study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment The previous treatment regime had produced an average increase of μ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of μ in comparison to the previous treatment. The sample data yielded ?̅ = 2.17 and ? = 1.05: Using α = 05, what conclusions can you draw about the research hypothesis?A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 50 men and 74 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. Two sample T for Men vs Women Mean N StDev SE Mean Men Women 95% Cl for mu Men - mu Women (- 10.67, - 1.13) T-Test mu Men = mu Women (vs H2 B. Ho: H1 = H2;Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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