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Mathematics, 15.02.2022 22:30 aashya16

Please help me i have no idea im so confused


Please help me i have no idea im so confused

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Answers: 3

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Overproduction of uric acid in the body can be an indication of cell breakdown. this may be an advance indication of illness such as gout, leukemia, or lymphoma.† over a period of months, an adult male patient has taken nine blood tests for uric acid. the mean concentration was x = 5.35 mg/dl. the distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.87 mg/dl. (a) find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. what is the margin of error? (round your answers to two decimal places.) lower limit upper limit margin of error (b) what conditions are necessary for your calculations? (select all that apply.) σ is unknown n is large σ is known normal distribution of uric acid uniform distribution of uric acid (c) interpret your results in the context of this problem. there is not enough information to make an interpretation. the probability that this interval contains the true average uric acid level for this patient is 0.05. the probability that this interval contains the true average uric acid level for this patient is 0.95. there is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. there is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. (d) find the sample size necessary for a 95% confidence level with maximal margin of error e = 1.10 for the mean concentration of uric acid in this patient's blood. (round your answer up to the nearest whole number.) blood tests
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Mathematics, 22.06.2019 03:40, cherylmorton7302
Anull hypothesis states that the difference between 8 hours of sleep and 3 hours of sleep has no effect on the number of pictures a student will remember during a picture memory interface test (pmit). examining the mit database, an extremely small t-value of 0.1611 is found. this t-value is much smaller than the smallest t-value on the chart for p-value reference. from this information we can: select one: a. accept the null hypothesis because the p-value obtained shows that the difference between the two groups being tested is not statistically significant b. accept the null hypothesis because the p-value obtained was statistically significant c. reject the null hypothesis because of the p-value obtained d. reject the null hypothesis because the data obtained is statistically significant
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