In a test of the hypothesis H0: μ ≥ 40 versus Ha: μ < 40, a sample of n = 50 observations possessed the mean x = 39.3 and standard deviation s = 4.1. Find the P-value for this test.
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In a test of the hypothesis H0: μ ≥ 40 versus Ha: μ < 40, a sample of n = 50 observations possessed the mean x = 39.3 and standard deviation s = 4.1. Find the P-value for this test.
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- A sample of 100 observations from a normal population has a mean of 50 and a standard deviation of 10. The null hypothesis is that the population mean is equal to 48, and the alternative hypothesis is that the population mean is greater than 48. Using a significance level of 0.05, what is the decision and conclusion of the hypothesis test?In a right-tailed test comparing two proportions, the test statistic was zcalc = +1.5. The p-value is:A researcher is testing a hypothesis of a single mean. The critical t value for a = .05 and a one-tailed test is 2.0639. The observed t value from sample data is 1.742. The decision made by the researcher based on this information is to the null hypothesis. reject not reject redefine change the alternate hypothesis into restate
- Scores on a certain "IQ" test for 18-25 year olds are normally distributed. A researcher believes that the average IQ score for students at a certain NJ college is less than 110 points, and so wants to test this hypothesis. The researcher obtain a SRS of 45 student IQ scores from school records and found the mean of the 45 results was 108 with a sample standard deviation of 21. The level of significance (alpha) used for this problem is 0.05. What is the appropriate test statistic (Student must complete by showing by formula using the ap- propriate values in that formula "showing work" and the final answer and appropriate label)? O T test score = (108-110)/(21/sqrt(45)) = -.639 T test score (108-110)/(21/sqrt(45)) = .639 %3D OT test score = (110-108)/(21/sqrt(45)) = .639 %3D T test score (108-110)/(45/sqrt(21)) =-.2037 %3D 素The average test score at a university is 78 with a standard deviation of 5. A professor believes the average test score of students in his class is different form 78. If a sample of 102 students was selected with a sample mean score of 79.1 and our level of significance is set at .05: What is the standard error of the mean? What is the calculated test statistic (zobt). Compute effect size using Cohen’s dA random sample of 25 people has a sample mean height of 70 inches and a sample standard deviationof 4 inches. Confirm or reject the hypothesis H0: = 68 inches (H1: ≠ 68 inches) using = 0.05.
- A sample of 50 earthquakes was found to have a sample mean of x ¯ = 1.184 and a sample standard deviation of s = 0.5864. Test the claim that the population of earthquakes has a mean magnitude greater than 1.00. Use a 0.01 significance level. Find the test statistic and critical value used for this test. Round to three decimal places.A simple random sample of 80 adults is obtained, and each person’s red blood cell count is measured. The sample mean is 6.25 and a standard deviation of .65. Use a .01 significance level to test the claim that the sample is from a population with a mean less than 6.4, which is a value often used for the upper limit of the range of normal values. (For each hypothesis test state H0 and H1)The percentage of titanium in an alloy used in aerospace castings is measured in 51randomly selected parts. The sample standard deviation is σx̅= 0.37Test the hypothesis H0: σ = 0.25 versus Ha: σ ≠ 0.25 using α = 0.05.
- Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is 0x = 8.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is dy = 11.5. The true (unknown) mean 6AEQ for vaccine X recipients is μx, while the true (unknown) mean 6AEQ for vaccine Y recipients is y. 6AEQ measurements are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation Vaccine X 78 Vaccine Y 93 151 148 8.7 11.5 a) Calculate the variance of the random variable X which is the mean of the 6AEQ measurements of the 78 vaccine X recipients.…Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is o, = 9.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is o, = 10.5. The true (unknown) mean SAEQ for vaccine X recipients is ly, while the true (unknown) mean 6AEQ for vaccine Y recipients is 4,. 6AEQ measurements are known to be a normally distributed. summary: Sample Size Sample Mean Standard Deviation 9.7 Гуре Vaccine 78 151 Vaccine Y 93 148 10.5 What is the length of your 96% confidence interval for uy Hy ? ) If we used this data to test Hg: Hy - Hy =0 against the alternative H: uy…
