How does 0.4 - 0.453 = 0.247 and how does 0.4 + 0.453 = 0.553?
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How does 0.4 - 0.453 = 0.247 and how does 0.4 + 0.453 = 0.553?
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- · A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard devia- tion of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time sho- uld be normal with the same standard devia- tion, but the mean effective time may be lo- wer. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sam- ple of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. Test hypothesis in level of significance = 0.10If the variance of a national accounting examination is 900, how large a sample is needed to estimate the true mean score within 5 points with 99% confidence? Z score is Use an Excel Function to calculate Z sample size is Show the Formula that you use to calculate it.The standard error of the mean (Qx) is a(n) estimate of population means which make up the sampling distribution. indication of variability in the distribution of sample means indication of variability in the raw data derived from the sample measure of variability in the population
- In a fish restaurant, population variance for fish to go bad is at least 4 days. After buying a new cooling system, its expected to be less than 4 days. After buying the new cooling system, 10 fish are tested and with an average of 8 days without going bad with a population variance of 3 days. Test with 95 confidince if the population variance is really less than 4 days. (use chi square please)Scientists need to be able to detect amounts of pollutants in the environment. Measurements were made on n 32 test specimens with a known mean concentration 2.35 ug/l of lead. It n was found that the > x; = 79.456 and > (x; – a)² = 1.0767. i=1 i=1 (a) If the population distribution of lead concentration is normal, state the minimum variance unbiased estimator of the mean. Then, compute the corresponding point estimate. Round your answer to 3 decimal places. (b) Find the estimated standard error of X. Round your answer to 3 decimal places. (c) Compute and interpret a 90% confidence interval for the true average lead concentration. (d) Did the interval you compute contain the true average? If it does not, is this a surprising result? Discuss. (e) Would you expect a wider or narrower interval if you were to compute a 95% confidence interval? Explain your answer without performing any computations.The sample mean was given to be x = $253.45 spent per day and the sample standard deviation was given to be s = $75.50. The value of ta/2 was determined to be 1.998. All the values are now known to construct the confidence interval. S The lower bound of the confidence interval is calculated using the expression x-ta/2 √n Find this lower bound, rounding the result to two decimal places. lower bound = x-ta/2 = 253.45 1.998 Ei S upper bound = x + ta/2 √n The upper bound of the confidence interval is calculated using the expression x + ta/2 to two decimal places. 75.50 = 253.45 + 1.998 √64 75.50 √64 S √n Find this upper bound, rounding the result 74°F Sunny 4:18 PM 4/17/2023 Ę
- An economic instructor at UCF is interested in the relationship between hours spent studying and total points earned in a course. Data collected on 19 students who took the course last semester follow # of observation(s) n = 19 # of independent variable(s) = 1 SSR = 3,882 SSE = 256 Find the Value for MSE (Use two decimals)confidence interval = [0.3652, 0.4596] find out the margin of error for given dataconfidence interval = [0.3652, 0.4596] find out the margin of error and population proportion for given data
- As the director of a research laboratory you are paid to decide if there is a significantdifference between the mean values of two sets of data obtained by two different scientists, asenior scientist and one recently hired. Use 95% CL. (t at df = 10 is 2.23)• Data of Senior Scientist: x mean = 24.66% with s = 0.06% for n = 5• Data of the New Kid: x mean = 24.55% with s = 0.10% for n = 7X~N(12,3) Find the 87.7 th percentile X.= = A dietician selected a random sample of n 50 male adults and found that their average daily intake of dairy products was x 756 grams per day with a standard deviation of s 35 grams per day. Use this sample information to construct a 95% confidence interval for the mean daily intake of dairy products for men.