escSay we have a population parameter u, and the 95% confidence interval for it is (-3, 10).What can you say to someone who claims that u = 11 ?7O You would accept their claim.QAO You would reject their claim.F1Z2F2WSJ#X380F3ED$4F4RF%5F5TMacBook Air6GF6Y&7HF7U00 *8CV BNF8J9F9KMOF10P+F12 escSay we have a population parameter u, and the 95% confidence interval for it is (-3, 10).What can you say to someone who claims that u = 11 ?7O You would accept their claim.QAO You would reject their claim.F1Z2F2WSJ#X380F3ED$4F4RF%5F5TMacBook Air6GF6Y&7HF7U00 *8CV BNF8J9F9KMOF10P+F12

For the population distributed with mean μ and variance σ2, the sampling distribution of the sample…

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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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If n=16, ¯xx¯(x-bar)=35, and s=17, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. I used the formula 35-NORM.INV(1-2.0/2,0,1)*((17)/SQRT(16)) for the lower and the same but with a plus sign for the upper, I got it wrong and dont understand why. We have to use Excel so please explain what excel formula to use.
A researcher wants to estimate the true proportion of Americans who suffer side-effects after taking a particular medication. A 99% confidence interval for the true proportion of Americans who suffer side-effects after taking the particular medication was found to be (0.024, 0.160). If a hypothesis test is conducted with the hypotheses, H,: p = 0.10 vs H: p# 0.10 , what would be the most likely conclusion? Select one: a. There is very strong evidence that p 0.10. Ob. There is little to no evidence that p = 0.10. c. There is little to no evidence that p # 0.10. d. Cannot tell.
Please answer this question for me because i do not understand how to work it.
Say that on average teenagers eat 10 slices of pizza each week, with a 95% confidence interval of (8-12). We take a sample of teenagers and find that this group eats an average of 6 slices each week. What can we say about the teenagers in the sample in relation to the confidence interval?
You have taken a random sample of sizen & 95of a normal population that has a population mean ofμ = 140and a population standard deviation ofo = 23. Your sample, which is Sample 1 in the following table, has a mean ofx = 141.3. (In the table, Sample 1 is indicated by "M1", Sample 2 by "M2", and so on.) (to) Based on Sample 1, plot the confidence intervals of80%and95%for the population mean. Use1,282as the critical value for the confidence interval of 80%and use1960 as the critical value for the confidence interval of95%. (If necessary, you can refer to a list of formulas .) • Write the upper limit and the lower limit on the graphs to indicate each confidence interval. Write the answers with one decimal place. • For the points (♦and ◆), write the population mean, μ = 140. 128.0 128.0 80% confidence interval 139.0 X Ś 150.0 150.0 128.0 128.0 95% confidence interval 139.0 X Ś 150.0 150.0
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, ×, is found to be 110, and the sample standard deviation, s, is found to be 8. (a) Construct a 99% confidence interval about u if the sample size, n, is 24. (b) Construct a 99% confidence interval about u if the sample size, n, is 13. (c) Construct a 96% confidence interval about u if the sample size, n, is 24. (d) Should the confidence intervals in parts (a)(c) have been computed if the population had not been normally distributed? An example was provided with additional pictures.
- Construct the 99% confidence interval for the difference μ₁ −μ₂ when x₁=476.88, x₂=318.32, s₁=42.11, s₂=26.80, n₁=19, and n₂=22. Use tables to find the critical value and round the answers to at least two decimal places. A 99% confidence interval for the difference in the population means is
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 104, and the sample standard deviation, s, is found to be 8. (a) Construct a 95% confidence interval about u if the sample size, n, is 22. (b) Construct a 95% confidence interval about u if the sample size, n, is 16. (c) Construct a 90% confidence interval about u if the sample size, n, is 22. (d) Should the confidence intervals in parts (a)-(c) have been computed if the population had not been normally distributed? (a) Construct a 95% confidence interval about u if the sample size, n, is 22. Lower bound: 100.5; Upper bound: 107.5 (Round to one decimal place as needed.) (b) Construct a 95% confidence interval about u if the sample size, n, is 16. Lower bound: Upper bound: (Round to one decimal place as needed.)
You are conducting a study to see if the proportion of voters who prefer the Democratic candidate is significantly smaller than 58% at a level of significance of αα = 0.01. According to your sample, 43 out of 85 potential voters prefer the Democratic candidate. For this study, we should use The null and alternative hypotheses would be: Ho: (please enter a decimal) H1: (Please enter a decimal) The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly smaller than 58% at αα = 0.01, so there is sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is equal to 58%. The data suggest the populaton proportion is significantly smaller than 58% at αα = 0.01, so there is…

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Say we have a population parameter μ, and the 95% confidence interval for it is (-3, 10). What can you say to someone who claims that μ = 11 ? O You would accept their claim. O You would reject their claim.