There is a p(x1) = 0.4 probability to observe x1 = 10 and p(x2) = 0.6 probability to observe x2 = 5. Find the mean and the variance of x. Show work.
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There is a p(x1) = 0.4
x2 = 5. Find the
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- Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. (a) What is the level of significance?State the null and alternate hypotheses. Ho: ?2 = 47.1; H1: ?2 < 47.1Ho: ?2 = 47.1; H1: ?2 > 47.1 Ho: ?2 = 47.1; H1: ?2 ≠ 47.1Ho: ?2 < 47.1; H1: ?2 = 47.1 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)What are the degrees of freedom?What assumptions are you making about the original distribution? We assume a uniform population distribution.We…Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. ased on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis.Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient…A correlation of +.20 means that [blank] of the variance in the outcome variable is accounted for by the predictor variable.
- Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker's times have a variance of 12.2. The second worker's times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times and that his commute time is shorter. Test the claim at the 10% level.What is the F statistic? (Round your answer to two decimal places.)How did you get the variance for this problem ?A researcher takes sample temperatures in Fahrenheit of 16 days from Miami and 14 days from Atlanta. Use the sample data shown in the table. Test the claim that the mean temperature in Miami greater than the mean temperature in Atlanta. Use a significance level of α=0.10α=0.10.Assume the populations are approximately normally distributed with unequal variances.Note that list 1 is longer than list 2, so these are 2 independent samples, not matched pairs. Miami Atlanta 68.3 73.6 83.1 69.3 79.1 54.9 72 81.1 72.8 78.6 83.3 54 82.7 36.1 80.7 44.3 87 58.4 83.1 50.8 77.4 60.5 86.1 61.2 76.3 46.8 74.5 54.4 83.3 78.5 The Null Hypotheses is: H0: μ1 - μ2 = 0 What is the alterative hypothesis? Select the correct symbols for each space. (Note this may view better in full screen mode.)HA: μ1 - μ2 Based on these hypotheses, find the following. Round answers to 4 decimal places. Test Statistic = p-value = The p-value is The correct…
- A company is doing a hypothesis test on the variation in quality from two suppliers. Both distributions are normal, and the populations are independent. Use a = 0.05. A sample of 31 products were selected from Supplier 1 and a standard deviation of quality was found to be 3.77. A sample of 16 products were selected from Supplier 2 and a standard deviation of quality was found to be 3.5659. Test to see if the variance in quality for Supplier 1 is larger than Supplier 2. What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: 0.² H₁: 0.² Based on the hypotheses, compute the following: Round answers to at least 4 decimal places. The test statistic is = The p-value is = The decision is to [Fail to reject the null hypothesis The correct summary would be: There is not enough evidence to support the claim in quality for Supplier 1 is larger than Supplier 2. o that the varianceA researcher found a significant correlation of .12 between gender (coded as 0 = ‘male’ and 1 = ‘female’) and number of friends at university. If the assigned values of gender were changed to 1 = ‘female’ and 2 = ‘male’, how would this affect the correlation? The strength of the relationship would become weaker The amount of variance explained by the relationship would increase It would become nonsignificantThe sign would change to become negative A researcher is trying to predict anxiety (measured on a 7-point scale) from exposure to stress (operationalised as the number of stressful life events participants can recall in the last week). She has found a correlation coefficient of r = .54 between anxiety and stress, with a standard deviation of 4.45 for anxiety and a variance of 17.21 for stress. What other information is needed to calculate b? The sums of squares of stressThe sums of products of anxiety and stress The covariance of anxiety and stressNo other information is…Explain whether homogeneity of variance is satisfied for this analysis, and be clear on how you decided this.
- Records from previous years for a casualty insurance company show that its clients average a combined total of 1.9 auto accidents per day, with a variance of 0.31. The actuaries of the company claim that the current variance, oʻ, of the number of accidents per day is not equal to 0.31. A random sample of 17 recent days had a mean of 2 accidents per day with a variance of 0.62. If we assume that the number of accidents per day is approximately normally distributed, is there sufficient evidence to conclude, at the 0.05 level of significance, that the actuaries are correct? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round…Suppose that the blood pressure for a randomly selected patient has an average of 22 and variance of 441. If a sample of 220 patients is selected, then what is P(-1.2 < X< 1.4) ?A certain IQ test is known to have a population mean of 100 and standard deviation of 15 in the general population. You want to test whether psychology majors have a different average IQ than the population as a whole. Assume the variance of IQ is the same for Psych majors as it is in the general population. Suppose that Psychology majors actually have an average IQ of 108. If you do a 2-tailed test at α= .05 with a sample of 56 Psychology majors, you will be able to reject the null hypothesis if the mean IQ of your sample is below [L] or above [H]. Find L and H values. Options listed below. [L] answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92. [H] Answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92.