Consider two independent random samples with the following results: n, = 573 P = 0.63 %3D n2 = 604 %3D = 0.49 Use this data to find the 95 % confidence interval for the true difference between the population proportions. Step 2 of 3: Find the margin of error. Round your answer to six decimal places.
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- A student who took Stat 2 last semester learns that their overall course score was at the 85th percentile. Assuming that a histogram of all the Stat 2 course scores from last semester follows the normal curve, this means that this student had a score that was how many SDs above the average score? Choose the answer that is closest. Group of answer choices 1.15 1.05 1.10 1.0 PreviousNextTina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.Aleks spins a 9 sided spinner 56 times. Each of the nine sides has a different colour on it. He records the number of times that he spins the colour purple. What values need to be determined in order to use a normal approximation? Determine these values.
- Find the value:Solve only the part a, b and c for question 10.A medical doctor is interested in determining whether mean body temperatures are different in the morning and at night. Five patients were recruited and their body temperatures were measured first at 8 AM and then again at 10 PM. The data is in the following table: Patient 1 2 4 Morning Night 98.0 97.6 97.2 97.0 98.0 97.0 98.8 97.6 97.7 98.8 For this matched pairs experiment, you should take the differences by calculating: Morning - Night. Also, you may use the fact that the sample standard deviation of 44/V5 the differences is 0.844 and that 0.844 0.3774. What is the test statistic for the test? Round your final answer to two decimals. Next Page Page 5 of 15 126 53 1. NOV 18
- Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. a. Identify the independent and dependent variables. b. In a sentence, state the null hypothesis being tested. c. Using symbols, state the null and alternative hypotheses. d. Determine the critical region using a one-tailed test with alpha = .05. e .Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). f. State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). g. Based on your analysis, is there a relationship between the drug and attention span?Beth wants to determine a 99% confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.04? Assume we have no prior estimate of the proportion and want a conservative choice for the sample size. n =Use the data to calculate the sample variance, s. (Round your answer to five decimal places.) n = 7: 1.6, 3.5, 1.6, 2.1, 3.1, 2.9, 3.0 %3D Construct a 95% confidence interval for the population variance, of. (Round your answers to two decimal places.) to Test Ho: = 0.8 versus H: of + 0.8 using a = 0.05. State the test statistic. (Round your answer to two decimal places.) x2 State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to two decimal places.) x2 > x² State the conclusion. H, is rejected. There is sufficient evidence to indicate that the population variance is different from 0.8. H, is rejected. There is insufficient evidence to indicate that the population variance is different from 0.8. H, is not rejected. There is sufficient evidence to indicate that the population variance is different from 0.8. O Ho is not rejected. There is insufficient evidence to indicate that the population variance is different from 0.8.