ean= 5.9 Sd= 3 N=200
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What are the confidence intervals for:
Sd= 3
N=200
α= .01
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- If the test statistic for a right sided test is z = 1.71, then the p-value isSelect one: A.0.9573 B.0.9564 C.0.0436 D.0.0427Step 3 The necessary value for Za/2 was determined to be 1.645. Recall the given information. Sample 1 n1 = 400 P1 = 0.56 P2 = 0.41 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) 1 72 lower bound = P₁-P₂-²a/2 = 0.560.41 Sample 2 n2 = 300 = 0.0559 = 0.56 0.41 = 02441 X Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) upper bound = P₁ P₂ + ²a/2√ n1 n2 X + ✔ - 1.645, + 0.56(1 0.56) 0.41(10.41) 300 +1.645. 400 0.56(1 0.56) + 400 + 0.41(1 0.41) 300find the t sub alpha/2 when n=23 for the 95% confidence interval.
- Give a 99.8% confidence interval, for µ1 - µ2 given the following information. 2.04, s1 = 0.39 20, ã1 60, x2 = 2.37, 82 - 0.7 n2 Use Technology Rounded to 2 decimal places.Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?Find the value for each ta/2 Prompts Submitted Answers n = 18 for the 99% confidence interval for the Choose a match mean. n = 10 for the 90% confidence interval for the Choose a match mean. n = 15 for the 98% confidence interval for the Choose a match mean.
- A researcher wishes to compare the mean of time spent watching TV in a day between men and women to test if the mean watching time for men equals the mean time for women. Assume that he performed the needed test and produce the following results: Gender N Mean Standard deviation Female 116 1.95 1.51 Male 59 2.37 1.85 95% confidence Interval for (µ,-µ,):(0.79,0.94) I. Which of the following is the correct conclusion about these results using a 5% significant level? OThe mean TV watching times of men and women are equal OThere is a statistically significant difference between the mean TV watching times of men and women. OThere is not a statistically significant difference between the mean TV watching tímes of men and women. OThere is not enough information to conclude II. Which statement is correct about p-value based on your conclusion about the test? Op-value0.05 Opvalue>0.005 OCannot be determine from the information givenn=10 b = 0.50 sb = 0.02 α=0.05 find the confidence interval for β.A sample of 1100 observations taken from a population produced a sample proportion of 0.38. Make a 90 % confidence interval for Round your answers to three decimal places. = ( ) to ( )
- You are going to create a 99% confidence interval for the mean amount ofsugar in coffee shop specialty drinks. You know the population standarddeviation is σ = 5.9 grams. You want your confidence interval to have amargin of error of 0.25 grams. How big should your sample size be to obtainthis margin of error?IID Suppose X1, ..., X, N(u, o?). Construct a 95% confidence interval for o? when u is known.The z confidence interval for the mean of a Normal population will have a specified margin of error m when the sample size is: a. n = (=*o/m). b. n = (z/m)². O c. n = (*o/m)². d. n ==*o/m.