2. A researcher is testing a null hypothesis that states, HO: µ= 50. A sample of 25scores is selected and M= 55. Assuming the sample variance is s² = 100,compute the estimated standard error and the t-statistic. Is this samplesufficient to reject the null hypothesis using a two-tailed test with a= .05? one-tailed or two-tailed test:State the Hypotheses;df=ta or t value for the critical regionsM =%3Dt DDecision:

Answered: 2. A researcher is testing a null…

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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A random sample of 46 adult coyotes from Northern Minnesota showed the average age to be sample mean of 2.05 years. The population standard deviation is known to be o = 1.13 years. However, it is thought that the average age of coyotes is u = 1.75 years. Test to see if the coyotes in Northern Minnesota live longer than the national average of 1.75 years. Use 0.03 level of significance. What are the hypotheses? A: Ho µ= 1.75 years vs HA µ> 1.75 years B: Ho µ= 1.75 years vs HA H 2.05 years D: Ho µ = 2.05 years vs HA u< 2.05 years O A D O O
a. What are the sample estimates of β0,β1,and β2? b. What is the least squares prediction equation? c. FindSSE,MSE,and standard deviation . Interpret the standard deviation in the context of the problem. d. Test H0: β1=0 against Ha: β1≠0.Use α=0.01. e. Use a 95% confidence interval to estimate β2. f. Find R2 and R^2_a and interpret these values. g. Find the test statistic for testing H0: β1=β2=0. h. Find the observed significance level of the test in part g.interpret the result.
The environmental research scientists report that the exhaust emissions for a certain car model has a normal distribution with a mean of 1.45 grams of nitrous oxide per mile and a standard deviation of 0.4. The car manufacturer claims their new process reduces the mean level of exhaust emitted for this car model. A Simple random sample of 36 cars is taken and the mean level of exhaust emitted for this sample is 1.21 grams. (a) State the null and alternative hypotheses. (b) Calculate the test statistic. (c) Calculate the Zc and p-value. Draw the picture. (d) What is the decision at the 0.05 significance level?
A sample of n = 18 data values randomly selected from a normally distributed population has variance s = 24.8. Construct a 95% confidence interval for the population variance. Round your endpoints to one decimal place. < o? <
A sample of n=41n=41 data values randomly selected from a normally distributed population has variance s2=30s2=30. Construct a 95% confidence interval for the population variance. Round your endpoints to one decimal place.
A random sample of 23 bags of apples (marked as 10 pounds each) received by a large grocery chain tested out as having a mean of 9.2 pounds with a variance of 2.56 pounds. Test whether the true mean of all bags is under 10 pounds. Set up hypotheses. Perform the appropriate test by showing your formula. Interpret the results using a Type I (alpha) error of .05. Also provide the p value here. Also construct a 95% confidence interval around your sample mean (X bar) that should contain the true mean (mu). Also interpret this interval.
For each of the tests, what would be the conclusion if the test were carried out using the α = 0.05 level of significance? H0 : µ = 120, H1:µ ≠ 120, n = 36, S.D of population= 10.3, sample mean = 117.0 Should the null be rejected?
ou wish to test the following claim (Ha) at a significance level of α=0.005. Ho:μ1=μ2Ha:μ1≠μ2 You believe both populations are normally distributed, but you do not know the standard deviations for either, and you have no reason to believe the variances of the two populations are equal. You obtain a sample of size n1=18 with a mean of M1=81.6 and a standard deviation of SD1=8.2 from the first population. You obtain a sample of size n2=23 with a mean of M2=88.9 and a standard deviation of SD2=5.8 from the second population. What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =
A sample of n = 24 data values randomly selected from a normally distributed population has variance s = 21.7. Construct a 90% confidence interval for the population variance. Round your endpoints to one decimal place. < o?
An independent measures design study with n= 16 in each sample produces a sample meam difference of 4 points and a pooled variance of 24. Wht is the value of t statistic?
Bob selects independent random samples from wo populations and obtains the values pi 0.700 and p2 confidence interval for p1 – P2 and gets: 0.500. He constructs the 95% 0.200 ± 1.96(0.048) = 0.200 ±0.094. %3| Note that 0.048 is called the estimated standard error of pi – P2 (the ESE of the estimate). Tom wants to estimate the mean of the success rates: Pi + p2 (a) Calculate Tom's point estimate. (b) Given that the estimated standard er- ror of (p1 + p2)/2 is 0.024, calculate the 95% confidence interval estimate of (Pi + P2)/2. Hint: The answer has our usual form: Pt. est. ±1.96 × ESE of the estimate.
Two separate samples, each with 16 individuals, receive two different treatments. After treatment, the first sample has SS = 1540 and the second has SS = 1460. a. Find the pooled variance, s2p, for the two samples b. The estimated standard error is c. If the sample mean difference is 10 points (i.e., M1 - M2 = 10), compute the t-statistic. The t-statistic is:

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