A random sample of size 20 from a normal population has a mean of x-bar = 62.8 and s = 3.55. Ifyou were to test the following hypothesis at the .05 level of significance the rejection regionwould best be defines asHo: H = 60Ho: H+60Ot> 2.093 or t < -2.093Ot> 1.729 or t < -1.729Ot> 1.725 or t < -1.725O t> 2.086 or t < -2.086

We have given that the random sample of size 20 from a normal distribution has mean 62.8 and…

Answered: A random sample of size 20 from a…

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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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…
A test of H₁: μ=7 versus H₁: u<7 is performed using a significance level of a=0.01. The P-value is 0.20. Part: 0/3 Part 1 of 3 (a) Is H rejected? Since P ≤ Part: 1 / 3 Part 2 of 3 (b) If the true value of u is 1, is the result a Type I error, a Type II error, or a correct decision? a, we do not reject Ho at the a= 0.01 level. The result is a Type I error Part: 2/3 Part 3 of 3 X 3 (c) If the true value of u is 7, is the result a Type I error, a Type II error, or a correct decision? The result is a [Choose one) ▾ Español
Below is the complete question Fail to reject / reject H0. There is not / is enough evidence at the 1% level of significance to support / reject the claim.
The average household size in a certain region several years ago was 4.3 people. A census worker wishes to test, at the 5% level of significance, whether it is different now. A random sample of 20 households will be used for the test. Which one of the following describes the correct rejection region to use for this test? Ot> 2.093 or t 2.093 Ot> 1.729 O t> 1.729 or t < -1.729
In a certain school district, it was observed that 24% of the students were classified as only children (only siblings). However, in the special program for talented and gifted children, 80 out of 286 are only children. The school administrators want to know if the proportion of only the children in the special program is significantly different from the proportion for the school district. Test at the a = 0.02 level of significance. c. Hypotheses for this situation Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for the sample based on the sample proportion? z = To 3 decimal places P-value = To 4 decimal places
Find the critical value(s) and rejection region(s) for a left-tailed chi-square test with a sample size n=22 and level of significance α=0.05. LOADING... Click the icon to view the Chi-Square Distribution Table. Find the critical value(s). enter your response here (Round to three decimal places as needed. Use a comma to separate answers as needed.) Find the Rejection Region(s).
Test the claim about the population variance o² at the level of significance oa. Assume the population is normally distributed. Claim: o? O B. x2 |
The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use a = 0.01 and assume normality. n- 40, Σd- 207, Σα? = 6199 (a) Find t. (Give your answer correct to two decimal places.) (ii) Find the p-value. (Give your answer correct to four decimal places.) (b) State the appropriate conclusion. O Reject the null hypothesis, there is significant evidence that the coating is beneficial. Reject the null hypothesis, there is not significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is not significant evidence that the coating is beneficial.
You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly different from 0.85. You use a significance level of α=0.002α=0.002. H0:p=0.85H0:p=0.85 H1:p≠0.85H1:p≠0.85You obtain a sample of size n=415n=415 in which there are 361 successes.What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =
For a fixed alpha level of significance used for a hypothesis test, the critical value for a chi-square statistic decreases as the size of the sample decreases. a. True b. False
We want to use a z-test to determine if the sample mean is greater than 20 at the alpha = 0.05. What is the critical value(s)?
please and thank you

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