In the t-test of significance for the hypotheses Ho: μ = 11 and Haμ 11, the value of the test statistic was found to be t = 0.711 using a sample of size 18. Estimate the upper bound for the P-value associated with this test. That is, fill in the blank P<
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- The data in NutritionStudy include information on nutrition and health habits of a sample of 315 people. One of the variables is VitaminUse, indicating whether a person takes a multivitamin pill regularly or occasionally or not at all. Use technology to test whether the data provide evidence that the proportion taking a vitamin pill regularly is different from 35 % given. Clearly state the null and alternative hypotheses. : : 42 : Pi : P2 :: 0.35 : 2 :: P1 : p2 :: :: r Ho: vs Hg: ::Which of the following would probably NOT be a potential “cure” for non-normal residuals? Select one: a. Transforming two explanatory variables into a ratio. b. Removing large negative residuals. c. Removing large positive residuals. d. Using a procedure for estimation and inference which did not assume normality.Suppose μ1 and M₂ are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level 0.01 to test Ho: μ₁ −μ₂ = -10 versus H₂: M₁ M₂ < -10 for the following data: m = 8, x = 114.6, s₁ = 5.05, n = 8, y = 129.5, and s₂ = 5.33. USE SALT Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. Reject Ho. The data suggests that the difference between mean stopping distances is less than -10. Reject Ho. The data does not suggest that the difference between mean stopping distances is less than -10. Fail to reject Ho. The data suggests that the difference between mean stopping distances is less than -10. Fail to reject Ho. The data does not suggest that the difference between mean stopping distances is less than -10. You may need to use the…
- XYZ diet pills are supposed to cause significant weight loss. The following table shows the results of a recent study where some individuals took the diet pills and some did not. Diet Pills No Diet Pills No Weight Loss 80 20 Weight Loss 100 100 We are interested in determining if the proportions of individuals who experienced weight loss with and without Diet Pill usage are equivalent. Using the above information to perform a chi-square test of hypothesis with a level of significance of 0.05 A) Please develop the Expected frequencies table showing all the relevant calculations. B) Given that p1 represents the proportion of individuals who experienced weight loss with Diet Pills and p2 represents the proportion of individuals who experienced weight loss without Diet Pills, perform the Chi-square test of hypothesis (α = 0.05) for the following and also show all the relevant calculations. H0: p1 = p2 H1: p1 ≠ p2The following are measurements of the heat-producing capacity (in millions of calories per ton) of random samples of five specimens each of coal from two mines: Mine 1: AVOHA 8380 8210 8360 7840 7910 7540 7720 7750 8100 7690 Mine 2: Below are the EXCEL outputs of the data at 95% confidence level: t-Test: Paired Two Sample for Means t-Test: Two-Sample Assuming Equal Vari MINE 1 MINE 2 MINE 1 MINE 2 7760 Mean Variance Observations Pooled Variance Mean Variance Observations Pearson Correlation-0.69059 Hypothesized Mear df t Stat P(T0 B. H.: H1 – Hz = 0 vs HA: µ - Pz #0 C. H,: H1 - uz 20 vs HA: µ- 42 = 0 D. H,: H1 - 2 20 vs HA: 4 - 42 > 0 %3DSuppose µ, and µz are true mean stopping distances (in feet) at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level .01 to test: Họ: H1- H2 = -10 Car Type 1: m = 6, x = 115.7, s, = 5.03 Hạ: H1- H2 < -10 Car Type 2: n = 6, ỹ = 129.3, and s, = 5.38 Provide an explanation for the result of the test in terms of Car 1 and Car 2's stopping distances.
- Newspaper headlines recently announced a decline in science scores among high school seniors. In 2015, 15,109 seniors tested by The National Assessment in Education Program (NAEP scored a mean of 147 points. In 2000, 7537 seniors had averaged 150 points. The standard error of the difference in the mean scores for the two groups was 1.22. a) State the null and alternative hypotheses in terms of the appropriate parameter. Ho: p1 - µ2 HA: H1 - µ2 b) The point estimate is c) The test statistic is ? v . (If necessary, round your answer to the nearest hundredth.) d) The p-value is e) The correct decision is to Select an answer | the null hypothesis. f) There is Select an answer evidence that scores have decreased between 2000 and 2015. g) We are 95% confident that the mean score has decreased by between and points between 2000 and 2015. (If necessary, round your answers to the nearest hundredth.)the assumption that the t test for independent samples makes regarding the amount of variability in each of the two groups is called the?Suppose μ₁ and ₂ are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level 0.01 to test Ho: M₁ - μ₂ = -10 versus H₂: ₁-₂ < -10 for the following data: m = 8, x = 113.2, s₁=5.05, n = 8, y = 129.1, and s₂ = 5.35. USE SALT Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. O Reject Ho. The data suggests that the difference between mean stopping distances is less than -10. O Reject Ho. The data does not suggest that the difference between mean stopping distances is less than 10. O Fail to reject Ho. The data suggests that the difference between mean stopping distances is less than -10. O Fail to reject Ho. The data does not suggest that the difference between mean stopping distances is less than -10.
- A survey is conducted to determine if there is a The conditions for inference are met. The chi-square difference in the proportion of students, parents, and test statistic is 0.19, df = 2, and the P-value is greater than any reasonable significance level. What teachers who volunteer at least once a month. To investigate, a random sample of 45 students, 25 parents, and 12 teachers was selected from a large high school. The data are displayed in the table. conclusion should be made? There is convincing evidence that the proportions of these students, parents, and teachers who Student Parent Teacher volunteer at least once a month differs. Yes 24 12 O There is not convincing evidence that the Volunteer? No 21 13 proportion of these students, parents, and teachers who volunteer at least once a month differs. The researcher would like to test these hypotheses: O There is convincing evidence that the distribution of Họ: There is no difference in the distribution of responses for the…Suppose μ1 and μ2 are real average stopping distances at 50 mph of a certain type of car equipped with two different types of braking systems. Use the two-sample t test at the 0.01 significance level to test H0: (μ1 - μ2 = -10) vs. Ha: (μ1 - μ2 < -10) for the following data: m = 6 ; x̄ = 115,7 ; s1 = 5,03 ; n = 6 ; ȳ = 129,3 and s2 = 5,38 . After the calculations performed, it is possible to mark the alternative as correct: a) Test t = -1.25; the calculated degree of freedom v= 8.06; p-value = 0.03; therefore we reject H0 for the 0.05 significance level. b) Test t = 2.15; the calculated degree of freedom v= 9; p-value = 0.13; therefore we accept H0 for the 0.01 significance level. c) Test t = -1.20; the calculated degree of freedom v= 9.96; p-value = 0.130; therefore we accept H0 for the 0.01 significance level. d) Test t = -1.20; the calculated degree of freedom v= 9.96; p-value = 0.130; therefore we reject H0 for the 0.01 significance level. e) Test t = -1.25; the calculated…Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of u = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9 ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the Decision (step 5) for this problem? There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 7.3 ppb. O There is sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. O There is…