Tests the claim that 41 2. Assume the samples are normally distributed, random and independent.of + 081 = 0.56 ; s2 = 0.51 = 44.37 ; 22 = 44.32n1 = 30 ; n2 = 30Standard Errort-Test StatisticDegrees of Freedom(#1 – #2) – (H – 42)Smaller of (n – 1) or (n2 – 1)S- =n2a. Calculate the Standard Error s;, -i, =(use 3 decimal places)(use 2 decimal places)b. Calculate the t-test statistic using the standard error from part a. t =c. What are the degrees of freedom? df ==d. At a 0.10, Use the distribution table to find the critical values for the rejection regionte = ±(use 4 decimal places)

Answered: Image /qna-images/answer/643b2cb8-c97e-44c6-8748-d39ce6aee083.jpg

Answered: 81 = 0.56 ; s2 = 0.5 1 = 44.37 ; #2 =…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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O Assighmant Score! < Question 12 of 16 Of the many species of oak trees in the United States, 28 grow on the Atlantic Coast and 11 grow in California. The back-to-back stemplot displays data on the average volume of acorns (in cubic centimeters) for these 39 oak species. Atlantie Const 998643 0 | Callifornia Which of the following are correct comparisons of the 4. 88864211111 1 50 2 6640 3 06 06 distributions of acorn volumes for the Atlantic Coast and California? 8 4 1 5 86 I. Both distributions have similar variability. II. Both distributions have similar medians. 59 1 8 1 9 10 11 III. Both distributions have possible outliers. O I and III only Key: 26 = An oak species whose acorn volume is 2.6 cm. O II and III only 15 16 17 O II only O I only O III only 1 From the Electronic Encyclopedia of Statistics Examples and Exercises (EESEE) story "Acorn Size and Oak Tree Range."
13. Two tail test with t-calc = 1.831 and degrees of freedom = 48. p-value of the test is: a) 0.05 < p-value < 0.1 b) 0.1 < p-value < 0.2 c) 0.01 < p-value < 0.02 d) 0.02 < p-value < 0.05
You are performing a two-tailed test with test statistic z=2.55z=2.55, find the p-value to 4 decimal places
17. please state all 6 steps of the p method clearly please! Strength of concrete: The compressive strength, in kilopascals, was measured for concrete blocks from five different batches of concrete, both three and six days after pouring. The data are as follows. Can you conclude that the mean strength after three days differs from the mean strength after six days? Let μ1 represent the mean strength after three days and =μd−μ1μ2 . Use the =α0.05 level and the P -value method with the TI-84 Plus calculator. Block 1 2 3 4 5 After 3 days 1371 1379 1344 1383 1386 After 6 days 1328 1337 1344 1321 1345
A previous study showed to determine whether a sales tax on soda will reduce consumption of soda in the US below the current per-capita level of about 50 gallons of soda per year. Hypotheses Test is Ho: m(mue) =50 vs. Ha: m<50 where m (mue) represents the average annual consumption of soda in communities where the sales tax is implemented. (b) Now supposed sample results give a p-value of 0.41. Interpret this p-value in terms of random chance and in the context of taxes and soda consumption.
I am confused. I do not understand how they got 0.0681 when using the standard normal Z table for 1.49. According to my table, that is for -1.49, and positive 1.49 has a z-score of 0.9319. Using the normalcdf function on my calculator, I can only get that result if I set 1.49 as my lower bounds and 10^99 as my upper bounds. Is this question asking for a right or left area? I thought the area to the left was supposed to be negative.
Use the information given in the table below to find a 95% Prediction Interval for y when x = 10. n = 20 x_bar = 15 %3D beta0_hat = 53 SSXX = 62 beta1_hat = - 4.5 S = 9 (- 14.79, 30.79) (-0.57, 16.57) (-10.25, 26.25) O (- 4.73, 20.73)

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