work together harmoniously so that the process is always smooth but for this particular patient it did not seem to happen so smoothly. I had every confidence in my performance and also in the responsibility of my patient but the next department, the pre-operative evaluation center, who actually confirms the surgery schedules did not have the same confidence and there was some friction. They attempted to contact my patient to confirm her date and time to report for her procedure and became irritated
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7a) Width of the confidence interval=0.8349-0.6651=0.1698 (2x1.96)((0.75)(1-0.75)/n)^1/2=0.1698 n=100 7b) (2z) )((0.75)(1-0.75)/100)^1/2=0.1299 Z=1.499955 P(z<1.50)=0.9332 So the new confidence level is 93.32% 8a) Yes, the manufacturer should assume that the average would be 21714 miles driven. It is because the sample mean(21714miles) is usually equal to the population mean. Sample mean is the unbiased estimator of the population mean
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| 9 | 14.8 | 19 | 15.8 | 29 | 14.8 | 10 | 15.2 | 20 | 14.5 | 30 | 14.6 | Write a two to three (2-3) page report in which you: 1. Calculate the mean, median, and standard deviation for ounces in the bottles. 2. Construct a 95% Confidence Interval for the ounces in the bottles. 3. Conduct a hypothesis test to verify if the claim that a bottle contains less than sixteen (16) ounces is supported. Clearly state the logic of your test, the calculations, and the conclusion of your test
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found to be 0.2. Give an interval based on your data so that you are 95% confident that the true value of the unknown proportion lies inside it. How would you explain 95% confidence to a layman? Suppose a professor of IIMA thinks that true proportion is 0.3. Are you ready to accept the professor’s perception based on your data at 99% confidence level? Solution – 1 Sample Size n = 100 (male smokers) p = 0.2 Sd (P) = √(pq / n) = .04 95% confidence interval of p = 0.2 ± 2 x 0
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I. NORMALITY IN THE TESTPAPER II. EQUAL VARIANCES | Ho: | The two data have equal variances. | | | | | | | | | Ha: | The two data did not have equal variances. | | | | | | | | | | | | | | | | | | | | | Since the p value (0.006035) is lesser than alpha (0.05), then Ho is rejected. | | | | | | | | | | | | | | | | | | | Therefore, the two data did not have two equal variances. Meaning the two data comes from the different population. |
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as I have done in the past. My fears almost caused me to give up during my LPN to RN bridge program. I felt overwhelmed and lost confidence in myself. Thankfully I had a strong support system. I was able to refocus on my goals, prioritize and utilize my resources to overcome my fears of failure and succeed. With hard work and determination I regained my confidence and drive to become a professional and a scholar. From that point there was no turning back. I have too much at stake to allow self-doubt
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verify my standard process is operating in a way that allows me to complete all of the tasks and affords me additional time. This paper will also outline the control limits of my morning process, the effects of any seasonal factors, and the confidence intervals involved. Statistical Process Control Data was recorded for a period of two weeks on how long it took me to complete my morning routine and get my kids off to school. I tracked the minutes it took me to complete each step of my morning
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Process Plan Sims OPS/571 Operations Management December 12, 2011 Manuel Gonzalez Process Plan This paper will analyze the process of driving to an interview it will define a strategy, which can be applied to improve of the method. With building on week three’s task, which was recognizing the bottlenecks, I constant gather data that concerning the time it takes to drive to an interview during the week. The data is studied plus examined to decide what the lower and upper control limits
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2011). Confidence Interval In addition to the Consad Research, information includes addressing data analysis using descriptive statistics, which included central tendency, dispersion, and skew data and statistical data using graphic and tabular techniques, was provided. A confidence interval was also computed to support the team’s conclusion. From our research the confidence interval is M 73.0 – 4.8 = 69.1 < M 73.9 > 73.9+ 4.8= 78.7Females
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Managerial Report 1. Use appropriate descriptive statistics to summarize the data on assets and yields for the money market funds. 2. Develop a 95% confidence interval estimate of the mean assets, mean 7-day yield, and mean 30-day yield for the population of money market funds. Provide a managerial interpretation of each interval estimate 3. Discuss the implication of your findings in terms of how Lisa could use this type of information in preparing her weekly newsletter. 4. What other
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