Find the critical values x² = x₁-a/2 and XR 2 = 13. 95% degree of confidence and the sample size n x² = X²/12 = = X²α/2 that correspond to
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- Find the critical values x2 and x²R for the given confidence level c and sample size n. c=0.95, n=22 L XL = (Round to three decimal places as needed.)Given two independent random samples with the following results: n1=297pˆ1=0.67n1=297p^1=0.67 n2=93pˆ2=0.41n2=93p^2=0.41 Use this data to find the 98%98% confidence interval for the true difference between the population proportions. Copy Data Step 1 of 3: Find the critical value that should be used in constructing the confidence interval.Find the critical values x1-a/2 and x- for a 95% confidence level and a sample size of n = 10. a/2 %3D X1-a/2 (Round to three decimal places as needed.)
- A 95% two-sided confidence interval for μ which has been calculated using R turns out to be (0, 1). A 90% two-sided confidence interval based on the same data will contain the value 0.9 (True, cannot tell, or False)Use a significance level of =α0.05 to test the claim that ≠μ39.2 . The sample data consists of 15 scores for which =x49 and =s6.9 .Find the critical value tcfor the confidence level c=0.80 and sample size n=28.
- The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65−74 years: n = 116, x = 12.5, and s = 6.35. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use ? = 0.05.)State the appropriate null and alternative hypotheses. H0: ? = 15Ha: ? ≠ 15H0: ? = 15Ha: ? > 15 H0: ? = 15Ha: ? ≤ 15H0: ? = 15Ha: ? < 15 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day.Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. Do not reject…A test is conducted for equality of two population means assuming equal population standard deviation. The square root of sum of squared deviations of data 1 and data 2 from their respective sample central tendencies are calculated as 20.23 and 19.83. If the sample variance used in test statistics is 34.89, what critical value is used in this test?What critical value of t* should be used for a 95% confidence interval for the population mean based on a random sample of 21 observations? Find the t-table here. t* = 1.721 t* = 1.725 t* = 2.080 t* = 2.086
- The ages of registered voters in Smith County are normally distributed with a population standard deviation of 3 years and an unknown population mean. A random sample of 18 voters is taken and results in a sample mean of 55 years. Find the margin of error for a 95% confidence interval for the population mean. z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.find out the confidence interval for population proportion for given data [ 4 decimal places ] population standard deviation =4 xbar = 123.5 sample size= 100 c= 0.95Find the critical value tc for the confidence level c=0.99 and sample size n=19. tc=
