Three different assembly methods have been proposed for a new product. A completely randomized experimental design was chosen to deterr number of parts produced per hour, and 30 workers were randomly selected and assigned to use one of the proposed methods. The number o Method A в 95 94 100 75 100 93 92 94 88 99 56 66 73 77 60 92 91 74 100 85 83 87 72 74 91 89 86 93 82 85 Use these data and test to see whether the mean number of parts produced is the same with each method. Use a = 0.05. Compute the values identified below (to whole number).
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- Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of saturated fat for a sample of 6 brands of stick margarine (solid fat) and for a sample of 6 brands of liquid margarine and obtained the following results: Stick:[25.5,26.7,26.5,26.6,26.3,26.4] Liquid:[16.5,17.1,17.5,17.3,17.2,16.7] We want to determine if there a significant difference in the average amount of saturated fat in solid and liquid fats. What is the test statistic? a) z = 39.604 b) t = 25.263 c) t = 39.604 d) z = 39.104 e) t = 39.104A local nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For cost-estimating purposes, managers use 2.2 hours of labor time for planting of a medium-sized tree. A random sample of 36 tree plantings is selected and the planting times are recorded. The average planting time for the sample of 36 planting is 2.7 hours with a standard deviation of .9 hours. The researcher wants to test if the average planting time is significantly greater than 2.2 hours based on the data that was collected. Họ: H= 2.2 H1: u> 2.2 What is the value of the Test Statistic for this hypothesis test? Round Z to the 100ths place.A state fisheries commission wants to estimate the number of bass caught in a given lake during a season in order to restock the lake with the appropriate number of young fish. The commission could get a fairly accurate assessment of the seasonal catch by extensive “netting sweeps" of the lake before and after a season, but this technique is much too expensive to be done routinely. Therefore, the commission samples a number of lakes and record the seasonal catch (thousands of bass per square mile of lake area) and size of lake (square miles). A simple linear regression was performed and the following R output obtained. Estimate Std. Error t value Pr(>|t|) (Intercept) 2.5463 0.4427 5.7513 0.0000 size 0.0667 0.3672 0.1818 0.8578 If a scatterplot showed a non-linear relationship between the response and explanatory variables, what should be done? O Nothing. Continue the analysis as is. Stop the analysis. Perform a natural log transformation on the response variable first. O Perform a…
- Liquid suspensions are commonly used in children's medications. There is a concern about liquid suspensions around incorrect dosing, due to the imprecise nature of measuring out a dose. A sample of 350 parents are asked to pour out one dose (5 mL) of the medication in their usual way, and the amounts are measured. The sample mean is 7.4 mL, with a standard deviation of 0.79 mL. Calculate the margin of error and construct the 99% confidence interval for the true population mean dose that parents pour. We may assume that the sample standard deviation sis an accurate approximation of the population standard deviation o (i.e., s= 0), given that the sample size is so large (n> 200). Standard Normal Distribution TableThe General Social Survey (GSS) collects data on demographics, eduction and work, among many other characteristics of US residents. Suppose we want to estimate the difference between the average number of hours worked by all Americans with a college degree and those without a college degree. Is there sufficient evidence that there is a significant difference between the average number of hours worked by those Americans with a college degree vs. those Americans without a college degree? Use the following output: Welch Two Sample t-testdata: yes and not = 3.1181, df = 1098.5, p-value = 0.001867alternative hypothesis: true difference in means is not equal to 095 percent confidence interval: 1.011652 4.445822sample estimates:mean of x mean of y 42.81574 40.08701 Using the provided output, find the 95% confidence interval for the difference of the average amount of hours worked for those that have a college degree vs. those that do not have a college degree (ie: for the difference of two…Concerned about the initial monitoring data, the environmental action group decides to continue to monitor the plant, and try to gather more evidence for their case. A random sample of twenty-six hours is selected over a period of a week. The observations (gallons of wastewater discharged per hour) are 1036 1047 996 1052 1136 1131 958 1058 1087 1146 1111 1040 884 997 1130 994 963 1127 1136 1126 The output of a statistical analysis software on this dataset is shown below N Mean Std. Dev. 26 1081 89.6 Reading the output, we find that • The sample size is n = • The sample mean is a • The sample standard deviation is s = • From this we can calculate the standard error to be SE Note that the observed sample mean is is greater than 1000 gallons per hour. This could mean that the plant is discharging more wastewater than they promised, or the plant could be in compliance, and the large numbers were due to sampling variability. To see if this is the case, we will test the hypothesis that u…
- Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on arganisms, an expert claimed that the amount of gamma radiation needed to sterilize a colany of tardigrades no longer has a mean of 1200 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy.) A study was conducted on a sample of 21 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean af 1225 Gy, with a sample standard deviation of 65 Gy. Assume that the population of amounts of gamma radiation needed to sterilize a colony of tardigrades is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level af significance, to support the claim that H, the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is nat equal to 1200 Gy. (a) State the null hypothesis and the…An airline operates a call center to handle customer questions and complaints. The airline monitors a sample of calls to help ensure that the service being provided is of high quality. Ten random samples of 100 calls each were monitored under normal conditions. The center can be thought of as being in control when these 10 samples were taken. The number of calls in each sample of 100 not resulting in a satisfactory resolution for the customer is as follows: Excel File: data 19-09.xls 4 5 3 2 3 3 4 6 4 7 a. What is an estimate of the proportion of calls not resulting in a satisfactory outcome for the customer when the center is in control? Round your answers to three decimal places. 0.041 b. Construct the upper and lower limits for a p chart for the manufacturing process, assuming each sample has 100 calls. Round your answers to four decimal places. If answer of LCL is negative then enter "0". UCL 0.1005 LCL = c. With the results of part (b), what conclusion should be made if a sample…! Required information The lengths of two components will be measured several times. The uncertainty in each measurement of the length of the first component is 0₁ = 0.04 cm, and the uncertainty in each measurement of the length of the second component is σ2 = 0.09 cm. Let X denote the average of the measurements of the first component, and let y denote the average of the measurements of the second component. The total length of the two components will be estimated with the quantity X + Y. Find the uncertainty in the total length if the first component is measured 4 times and the second component is measured 12 times. The uncertainty in X + Y is cm.
- PETROGAS is testing new filters for its motorbikes. One brand of filter (Filter A) is placed in one motorbike, and the other brand (Filter B) is placed in the second motorbike. Random samples of air released from the motorbikes are taken at different times throughout the day. Pollutant concentrations are measured for both motorbikes at the same time. The following data attached represent the pollutant concentrations (in parts per million) for samples taken at 20 different times after passing through the filters. a. Test the hypothesis that the mean for the pollutant concentration for Filter B is greater than 30. Use the 1% level of significance. b. Construct a 95% confidence interval for the difference in mean pollutant concentration, where a difference is equal to the pollutant concentration passing through Filter A minus the passing through Filter B. c. Using the 5% significance level, determine whether there is evidence that mean for the pollutant concentration for Filter A…A researcher was interested in comparing the resting pulse rate of people who exercise regularly and people who do not exercise regularly. Independent random samples of 16 people aged 30-40 who do not exercise regularly (sample 1) and 12 people aged 30-40 who do exercise regularly (sample 2) were selected and the resting pulse rate of each person was measured. The summary statistics are as follows: Pulse Rate data Group 1 (no exercise) Group 2 (exercise) average 72.7 69.7 standard deviation 10.9 8.2 sample size 16 12 Test the claim that the mean resting pulse rate of people who do not exercise regularly is greater than the mean resting pulse rate of people who exercise regularly, use 0.01 as the significance level. Round you answer to 3 decimal places. Group of answer choices p-value=0.207, evidence not support claim p-value=0.267, evidence support claim p-value=0.414, evidence not support claim p-value=0.793, evidence not support claim p-value=0.207, evidence…