Part 4 of 5for the 90% confidence intervalZa/2Za/2=%3DPart 5 of 5Za/2for the 85% confidence intervalZa/2= Find each of the following. Enter your answers rounded to at least two decimal places.Part 1 of 5for the 95% confidence intervalZ a/2%3DZa/2=Part 2 of 5for the 98% confidence intervalZ a/2Za/2=Part 3 of 5for the 99% confidence intervalZ a/2Za/2

We use standard normal Z distribution table to find the required critical value.

Answered: Part 4 of 5 Z/2 for the 90% confidence…

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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Can you solve this problem about the confidence interval?
7 FI A random sample of ni = 246 people who live in a city were selected and 70 identified as a "dog person." A random sample of n2 Find the 99% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person.' %3D 112 people who live in a rural area were selected and 62 identified as a "dog person." %3D Round answers to to 4 decimal places. Question Help: Message instructor Submit Question MacBook Pro %23 % 3. 00 R. S. H. K.
We would like to compute a 95% confidence interval for the difference of proportions of men and women who are frequent binge drinkers. Of 534 males, 26% admitted to frequent binge drinking as opposed to 20.6% of 847 women. The 95% confidence interval for the difference of proportions (proportion of males - proportion of females) is (0.0179, 0.1001). Based on this confidence interval, can we conclude that men are more likely to frequently binge drink than women? (4)
The CDC has asked for your help interpreting their data on COVID-19. From a sample of 64 patients, you find the average time for a person showing symptoms to recover is 21 days. The CDC reported a standard deviation of 6 days. For a 90% confidence level, what is the upper bound in days of the confidence interval for a sick patient to recover?
The 99% confidence interval on a population mean computed with a sample of size 36 and sample mean 1.4 is computed knowing that the standard deviation of this population is 0.3. The margin of error for this confidence interval is about
A random sample of ni = 270 people who live in a city were selected and 99 identified as a "dog person." A random sample of n2 = 108 people who live in a rural area were selected and 51 identified as a "dog person." Find the %3D 98% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. < pi – P2 < -
Solve Question 2 Please show work
A sample of size n=77 is drawn from a population whose standard deviation is 0= 33. Part 1 of 2 (a) Find the margin of error for a 95% confidence interval for µ. Round the answer to at least three decimal places. The margin of error for a 95% confidence interval for µ is Part 2 of 2 (b) If the confidence level were 90%, would the margin of error be larger or smaller? (Choose one) ▼ because the confidence level is (Choose one) ▼
A random sample of size 30 from a normal population yields = 39 and s = 4.9. The lower bound of a 95 percent confidence interval is (Round off upto 2 decimal places).
A sample of size n=89 is drawn from a population whose standard deviation is a 16. Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for μ. Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for μ is. Part 2 of 2 (b) If the sample size were n=87, would the margin of error be larger or smaller? (Choose one) , because the sample size is (Choose one)
We've seen how to compare the means of two samples with known population variances (refer, for instance, to Example 4 in Recitation 11), but when we need to compare the means of two samples with unknown (but assumed to be equal) population variances, a method called 'variance pooling' needs to be used to find the 'pooled sample variance' that should be used when calculating the confidence interval. The pooled variance is given by si (n1 - 1) + s5 (n2 - Spool = n, +n, -2 with v=n, +n, - 2 degrees of freedom (see page 308 of your e-book). Now look at the example below. Production engineers need to compare the average efficiencies of two production processes. Process 1 and Process 2 are sampled 11 and 15 times and they yielded average efficiencies of 80 and 75 (out of 100), respectively. The sample standard deviations are 6 and 3, again, respectively. Help them construct a 99% confidence interval, assuming that efficiency values are Normally distributed with equal variances. Click here to…
A sample of size n=79 Is drawn from a population whose standard deviation is o = 29, Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for l. Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for u is Part 2 of 2 (b) If the sample size weren=44, would the margin of error be larger or smaller? (Choose one) ▼ because the sample size is (Choose one) Larger Smaller

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Part 4 of 5 Z/2 for the 90% confidence interval Za/2= Part 5 of 5 Z2 for the 85% confidence interval Za/2