2. Suppose that we are testing Ho = o versus H₁>po. Calculate the p-value for the following observed values of the test statistic: (a) Zo=2.45 (c) Zo= 2.15 (e) Zo= -0.35 (b) Zo= -1.53 (d) Zo=1.95
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- a. Report and interpret the P-value for Fisher's exact test with (i) Ha: 0 > 1 and (ii) Hạ: 0 # 1. Explain how the P-values are calculated. b. Find and interpret the mid P-value for Ha: 0 > 1. Summarize advantages and disadvantages of this type of P-value. Table 3.14 Data for Exercise 3.18 on Therapy for Cancer of Larynx Cancer Controlled Cancer Not Controlled Surgery Radiation therapy 21 15 23A study was done to look at the relationship between number of lovers college students have had in their lifetimes and their GPAs. The results of the survey are shown below. Lovers 0 3 1 2 5 1 8 GPA 3.3 2.9 3.4 3.3 2.5 2.8 0.7 1. The p-value is: _____ (Round to four decimal places) 2. Use a level of significance of α=0.05 to state the conclusion of the hypothesis test in the context of the study (pick one). a. There is statistically significant evidence to conclude that there is a correlation between the number of lovers students have had in their lifetimes and their GPA. Thus, the regression line is useful. b. There is statistically significant evidence to conclude that a student who has had more lovers will have a lower GPA than a student who has had fewer lovers. c. There is statistically insignificant evidence to conclude that a student who has had more lovers will have a lower GPA than a student who has had fewer lovers. d. There is statistically insignificant…The price X (dollars per pound) and consumption y (in pounds per capita) of beef were samples for 10 randomly selected years. The following data should be used to answer the question that follows. n = 10 Ex = 36.19 Ix² = 134.17 2.9 < x s 6.2 Ey = 774.7 Iy² = 60739.23 Exy = 2832.21 Using this data, a student calculated SSy = 28.43 SSx = 3.2 SSy= 717. Calculate the %3D %3D value of the standard error or regression, Se , and enter you answer accurate to the nearest hundredth (2 decimal places).
- You wish to test the following claim (Ha) at a significance level of a = 0.10. H.: µ1 – µ2 = 0 Ha: µ1 – µ2 > 0 You obtain a sample of size nị the first population. You obtain a sample of size of s2 = 54.9 and a standard deviation of s1 = 78 with a mean of 2 65 with a mean of x1 17.1 from 46.7 and a standard deviation n2 7.2 from the second population. You will need to determine whether or not to pool so you will use the correct df. What is the critical value for this test? Report answer accurate to three decimal places. critical valuea. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: uv (Please enter a decimal) c. The test statistic z v (please show your answer to 3 decimal places.) %3D d. The p-value = (Please show your answer to 4 decimal places.) %3D e. The p-value is> va f. Based on this, we should fail to reject vthe null hypothesis. g. Thus, the final conclusion is that ...Calculate the missing values in the Stata output (SKIP E!!!). Two-sample t test with A variances X y combined diff Obs Ho: diff = 0 49 30 79 diff = mean(x) Ha: diff |t|) = Std. Dev. 6.5 7.4 6.935314 G E [95% Conf. Intervall 40.33298 39.62126 C F degrees of freedom = 44.86702 Pr{T > t) 42.72811 1.6996 77 Ha: diff > 0 H
- Fluoride Exposure in Drinking Water Exercise 2.250 introduces a study showing that fluoride exposure might have long-term negative consequences for the offspring of pregnant women. Part of the study examines the effect of adding fluoride to tap water on mean fluoride concentration in women. Summary statistics for fluoride concentration (measured in mg/L) for the two groups are given in the table below. Tap water Fluoridated Non-fluoridated Conclusion: Sample size 141 228 Mean 0.69 0.40 St.Dev. 0.42 0.27 Find and interpret a 99% confidence interval for the mean increase in fluoride concentration for those with fluoridated tap water. Let Group 1 represent those with fluoridated tap water and Group 2 represent those without fluoridated tap water. Confidence interval: i to i (round to three decimal places)Some frozen food dinners were randomly selected from this week's production and destroyed in order to measure their actual calorie content. The claimed calorie content is 200. Here are the calorie counts for each frozen dinner selected: 191 189 198 210 207 203 209 215 209 204 200 194 Assume the distribution of calories is normal. (a) The test statistic (z/t) is Use two decimals. (b) Does the sample indicate that the mean calorie content is 200? Set a = (c) Find the P-value of the test in (a)-(b). 0.07 ? Yes NoUsing a significance level of α = 0.05, state your conclusion in terms of H0.?
- Answers: 1.a) rXY = 0.461, t = 1.272, Do not reject H0: no evidence for a significant correlation between the two variables b) 43.66You wish to test Ho: Md = 0 versus Ha:μd 0 at a significance level of 0.10. For the context of this problem, d = μ₂ μ₁ where the first data set represents a pre-test and the second data set represents a post-test. You obtain pre-test and post-test samples for na 25 subjects. The average difference (post-pre) is d 7 with a standard deviation of the differences of sd = 11.6. = Round your answers to three decimal places, and round any interim calculations to four decimal places. What is the test statistic? =The following data represent the results from an independent-measures study comparing two treatment conditions. Treatment I II 10 7 N = 16 8 4 G = 120 7 9 ∑X² = 1036 9 3 13 7 7 6 6 10 12 2 M = 9.000 M = 6.000 T = 72 T = 48 SS = 44 SS = 56 Use an independent-measures t test with α = .05 to determine whether there is a significant mean difference between the two treatments. (Use three decimal places; subtract MIIII from MII.) t-critical = ± t = t Distribution Degrees of Freedom = 21 -3.0-2.0-1.00.01.02.03.0t Conclusion: Fail to reject the null hypothesis; there are significant differences between the two treatments. Reject the null hypothesis; there are significant differences between the two treatments. Fail to reject the null hypothesis; there are no significant differences between the two treatments. Reject the null hypothesis; there are…