What is the null hypothesis for the ECDF-based tests? O Ho: M1 = μ2 ○ O Ho: F(X) = G (Y) H₁ : 0 = 02/2 The distribution of the data is Normal
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- A random sample of size n = 19 observations was drawn from a normal population with variance o? = 3.1. The sample mean was = 22.5. The researcher's interest was to defend that the population mean u > 25. What is the critical region (aka rejection region) for the test a = 0.05? Drawing a picture may be helpful. Draw your conclusion. Calculate the power of the test when the alternative value for u = 27.For mated pairs of gallinules (type of water bird), let X equal the weight in grams of the male and Y the weight in grams of the female. Assume that X and Y have a bivariate normal distribution with μx = 415, o = 611, uy = 347 and o=689. The correlation coefficient between X and Y is p = -0.25. (a) Explain, in context, what a correlation coefficient of -0.25 means.For Z-test used to analyze data on tablet weight in a batch with Ho: u = 300 mg and at a = 0.05. If a sample was taken from the batch and Z-score of the mean tablet weight of the sample was correctly calculated as 1.9. P-value is which suggest that, O 0.047, The mean tablet of the batch is not significantly different from 300 O 0.0574, The mean tablet of the batch is significantly different from 300 O 0.0287, The mean tablet of the batch is not significantly different from 300 O 0.047, The mean tablet of the batch is significantly different from 300 O0.9713, The mean tablet of the batch is significantly different from 300 0.9713, The mean tablet of the batch is not significantly different from 300 0.0287, The mean tablet of the batch is significantly different from 300 0.0574, The mean tablet of the batch is not significantly different from 300 Next pc
- Suppose we have a random sample X1,..., Xn that arise from a normal distribution with some mean u and unknown variance o?. A researcher is interested testing for the difference a mean 4o The hypothesis of interest in this case is given by Ho : µ= 68.6 vs Ha : µ + 68.6 Report an estimate for the parameter Report the value of the test statistic Report the value of the p-value (If you p-value < 0.001, report 0) Report a 94 % confidence interval for the mean ul If we test at the 6 % level we should + the null hypothesis 100.5 98.3 117.5 68.7 113.9 79.5 68.7 85.8 76.3 103.9 77 68.9 79.4 90.7 82.6 72.6 73.4 108.9 92.4 81.6 67.5 74.3 85 61.4 108 89.9 88.9 92.4 83.2 65.1 84.2 89.2 56.1 71.3 113.1 115.8 80 63.5 101.8 101.8 79.9 89.9 65 77.9 68.6 99.5 84.7 80.1 90.2 80.6 88.3 111.9 103.8 66.6 97.2 107.1 90.9 94.1 108 85.2 80.3 78.8 118.7 75.2 89 70.7 117.7 89.4 109.4 95 91.4 73.5 105.4 87.1 86.6The one-tailed test of hypothesis, H0: u = 100 versus Ha: u > 100.Suppose the test statistic is z = 2.71. Find the p-value of the test and the rejection region for thetest when alpha = 0.01. Draw the distribution and label rejection / fail-to-reject regions.A researcher was doing a study for a new Coronavirus Drug and conducted a hypothesis test like this one: H0: μμ = 15 Ha: μμ ≠≠ 15 a. What type of test is this? two-tailed left-tailed right-tailed b. She obtained a test statistic of -1.21, with a p-value of 0.245. So she concluded that her true mean must be equal to 15. Is this interpretation correct? Yes No Why or why not? (Will be graded manually!)v
- Suppose you run a regression y=alpha + beta*x + u. You know that the model sum of squares equal 2998.9, residual sum of squares equals 571,612.8, and residual mean squares (MS) squals 412.41. What is the number of observations in your sample?A 95% CI for true average serum-creatinine level was calculated as (3.81, 3.95), based on a sample size of 20 peoples (after they received a newly proposed antibiotic) and the assumption that serum -creatinine level is normally distributed.Suppose you want to test H0: μ = 4 versus Ha: μ 4 using 5% level of significance. Without testing the hypothesis, can you write down the statistical conclusion you would make about the hypothesis. If not, then give reasons.An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with o = 60 psi. A random sample of 12 specimens has a mean tensile strength of X = 3250 psi. (a) Test the hypothesis that mean strength is 3500 psi. Use a= 0.01. (b) What is the smallest level of significance at which you would be %3D willing to reject the null hypothesis? (c) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean tensile strength.
- A consumer advocacy group received a tip that an air conditioning company has been charging female customers more than male customers. The group's statistical expert decides examine this question at the alpha = 0.01 level of significance, by looking at the difference in mean charges between a random sample of female customers and a random sample of male customers. Let represent the average charges for female customers and mu M represent the average charges for male customers( Round your results to three decimal places) 1. Which would be correct hypotheses for this test? O H 0 : mu F - mu M =0 H 1 : mu F - mu M >0 O H 0 : mu F - mu M =0 , H 1 : mu F - mu K <0 O H 0 : mu F - mu M =0 , H 1 : mu F - mu M ne0 O H 0 : mu F - mu M >0,H 1 : mu F - mu M <=0 2. If we are going to test this using a confidence interval, which confidence interval should we construct? O 96 % 99 % 99.5 % 98 % A random sample of 46 female customers were charged an average of $917, with a standard deviation…Find the least square estimate for beta (the slope) given that its y-intercept is 0. Y_i = BetaX_i + e_i where e_i are independent and identically distributed N(0, variance_e) randon variables i = 1,...,2.If the coefficient ß₁ has a nonzero value, then it is helpful in predicting the value of the response variable. If B₁ = 0, it is not helpful in predicting the value of the response variable and can be eliminated from the regression equation. To test the claim that B₁ = 0 use the test statistic t = (b₁-0) /sp. Critical values or P-values can be found using the t distribution with n-(k+1) degrees of freedom, where k is the number of predictor (x) variables and n is the number of observations in the sample. The standard error sp, is often provided by software. For example, see the accompanying technology display, which shows that sp, = 0.076885101 (found in the column with the heading of "Std. Err." and the row corresponding to the first predictor variable of height). Use the technology display to test the claim that B₁ = 0. Also test the claim that B₂ = 0. What do the results imply about the regression equation? Click the icon to view the technology output. Test the claim that B₁ = 0.…