A lot size of 500 cartons of food A is to be inspected using ANSI/ASQ Z1.4-2003 (refer to Appendix), with an AQL of 1.5% for critical defect and 2.5% for major defect of a double normal level II plan. Explain how acceptance and rejection can be done on a critical and a major defect.
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- How many stomach cancer patients and controls would be required to estimate the difference in proportion of patients with sufficient concentrations of plasma antioxidant vitamins with a margin of error not exceeding 5% using a 95% CI? The data in table 8-2 were reported recently from a similar study investigating the sufficiency of plasma antioxidant vitmians in throat cancer patients.The new version of the Standard requires the measurement uncertainty (MU) contributions of sampling to be considered Comment on whether this is still applicable if the facility is not accredited for sampling or is not responsible for sampling.Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on organisms, an expert claimed that the amount of gamma radiation needed to sterilize a colony of tardigrades no longer has a mean of 900 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy .) A study was conducted on a sample of 27 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean of 907 Gy , with a sample standard deviation of 17 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 0.05 level of significance, to support the claim that μ , the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is not equal to 900…
- Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on organisms, an expert claimed that the amount of gamma radiation needed to sterilize a colony of tardigrades no longer has a mean of 900 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy .) A study was conducted on a sample of 27 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean of 907 Gy , with a sample standard deviation of 17 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 0.05 level of significance, to support the claim that μ , the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is not equal to 900…Perform an ANOVA using Company as the independent variable, and Driver_Height as the dependent variable use alpha = 0.05. What is your p-value? Can someone confirm my coding is correct?Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on organisms, an expert claimed that the amount of gamma radiation needed to sterilize a colony of tardigrades no longer has a mean of 1350 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy .) A study was conducted on a sample of 18 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean of 1375 Gy , with a sample standard deviation of 75 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 0.05 level of significance, to support the claim that μ , the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is not equal to 1350 Gy .
- In this study, the protection of aquatic ecosystems is an important goal of Ville de Montréal. In particular, Thomas wants to see whether the contamination of creeks (ruisseaux in French) by Escherichia coli differs among the months of May, June, July, and August. Therefore, he randomly samples six creeks in different areas. For each creek, he collects water samples from May to August, and measures the amount of E. coli, expressed in colony forming units (CFU) per 100 mL of water. For each combination of a month and a creek, he thus obtains an E. coli contamination value, i.e., 24 observations in total. A preliminary statistical analysis produces the following results. 13-a) Does the mean E. coli contamination differ significantly among the four months? Justify your answer to this question with the result of a statistical test. Use α = 0.05. 13-b) Which pairs of months show significant differences in mean E. coli contamination? Explain your reasoning and justify your answer…Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 part per million (ppm). M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Boletus pinicola and published the results in the paper “Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain” (Journal of Environmental Science and Health, Vol. B33(4), pp. 439–455). A hypothesis test is to be performed to decide whether the mean cadmium level in Boletus pinicola mushrooms is greater than the government’s recommended limit. Hypothesis tests are proposed. For each hypothesis test,a. determine the null hypothesis.b. determine the alternative hypothesis.c. classify the hypothesis test as two tailed, left tailed, or right tailed.What would be the minimum sample size for an independent samples t-test with a small effect size where eta squared = .01, alpha = .05, and statistical power = .70?
- Seedlings of understory trees in mature tropical rainforests must survive and grow using inter- mittent flecks of sunlight. How does the length of exposure to these flecks of sunlight (fleck duration) affect growth? Leakey et al. (2005) experimentally irradiated seedlings of the Southeast Asian rainforest tree Shorea leprosula with flecks of light of varying duration while maintaining the same total irradiance to all the seedlings. Their data for 21 seedlings are listed in the following table. i. What is the rate of change in relative growth rate per minute of fleck duration? provide a standard error for your estimate. ii. using these data, test the hypothesis that fleck duration affects seedling growth rate. iii. Calculate a 99% confidence interval for the slope of the population regression.Part of the ANOVA summary table for using efficiency ratio (X1) and total risk-based capital (X2) to predict ROAA is shown to the right. The results also state that SSRX1=3.7902 and SSRX2=3.7615. a. At the 0.05 level of significance, determine whether each independent variable makes a significant contribution to the regression model. On the basis of these results, indicate the most appropriate regression model for this set of data. Write the hypotheses for the contribution of the variable efficiency ratio.In this study, the protection of aquatic ecosystems is an important goal of Ville de Montréal. In particular, Thomas wants to see whether the contamination of creeks (ruisseaux in French) by Escherichia coli differs among the months of May, June, July, and August. Therefore, herandomly samples six creeks in different areas. For each creek, he collects water samples from May to August, and measures the amount of E. coli, expressed in colony-forming units (CFU) per 100 mL of water. For each combination of a month and a creek, he thus obtains an E. coli contamination value, i.e., 24 observations in total. A preliminary statistical analysis produces the following results. Which pairs of months show significant differences in mean E. coli contamination? Explain your reasoning and justify your answer to this question with the results of a statistical procedure. Use α = 0.05.