Consider the following data for two independent random samples taken from two normal populations.Excel File: data10-11.xlsxSample 110 7 13 7 9 8Sample 28 7 8 4 6 9a. Compute the two sample means. (to nearest whole number)9.7b. Compute the two sample standard deviations. (to 2 decimals)S1 =2.28 O82 =1.79 Oc. What is the point estimate of the difference between the two population means?2d. What is the 90% confidence interval estimate of the difference between the two population means? (to 2 decimals and enter negative value as negative number).053.95

From the given information we calculate confidence interval for two sample mean t test.

Answered: cel File: data10-11.xlsx Sample 1 10 7…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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6. In year 2022 a random sample of 35 households in Ottawa found that the average annual internet service cost per household was $1755 with a sample standard deviation of $600.00. (a) Construct a 90% confidence interval for the average annual internet service cost per household. (b) In year 2018 the average annual cost of internet services per household was $2000. Does the sample data show that these average annual costs have declined since 2018? Use a - 0.I to do the test.
U21T. Construct a 99% confidence interval estimate for the mean u using the given sample information. (Give your answers correct to two decimal plac n = 20, x = 16.8, and s = 2.1 |3D 15.6 X to 18 You may need to use the appropriate table in Appendix B to answer this question. Need Help? Read It Watch It Talk to a Tutor
Find the 98% confidence interval for the difference between two means based on this information about two samples. Assume independent samples from normal populations. (Use conservative degrees of freedom.) (Give your answers correct to two decimal places.) Sample Number Mean Std. Dev. 1 16 38 29 2 29 28 29 Lower Limit Upper Limit
n1=16 x‾1=124 s1=35 n2=12 x‾2=86 s2=31 Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. Step 3 of 3 : Please also find the high and low endpoints and round to four decimal points.
just the last part thank you!
Q-3) Show all the steps for the problem.
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Also included in question find estimated variance of errors, estimated variance of slope, and 90% and 95% confidence intervals for slope. Thank you so much
Based on a sample of size 49, a 95% confidence interval for the mean score of all students, μ, on an aptitude test is from 59.2 to 64.8. Find the margin of error. Question content area bottom Part 1 A. 2.8 B. 0.05 C. 5.6 D. 0.78 E. There is not enough information to find the margin of error.
please answer the question below to 4 decimal places
12. A random sample of 16 individuals finds a sample mean value of 32 and sample standard deviation of 12. What is the 90% confidence interval for the population mean? A random sample of 16 individuals finds a sample mean value of 32 and sample standard deviation of 12. What is the 90% confidence interval for the population mean? C. [27.06 , 36.94] B. [27.91 , 36.09] D. [25.40 , 38.60] A. [26.61 , 37.39] Previous QuestionPreviousNext QuestionNext
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