**Question**: Which of the below Z scores is the appropriate $ z^* $ when calculating a 98% confidence interval?**Options**:(a) $ Z = 2.05 $(b) $ Z = 1.96 $(c) $ Z = 2.33 $(d) $ Z = -2.33 $(e) $ Z = -1.65 $When calculating a 98% confidence interval, the correct $ z^* $ value represents the Z score that encompasses the central 98% of the standard normal distribution, leaving 1% in each tail. Therefore, the correct answer is:(c) $ Z = 2.33 $This Z score corresponds to the positive critical value needed for a 98% confidence interval. A negative value (like in options d and e) is incorrect in this context because confidence intervals conventionally use the positive Z score.

We hvae to find corrcet answer..

Answered: Which of the below Z scores is the…

Which of the below Z scores is the appropriate z* when calculating a 98% confidence interval? (a) Z = 2.05 (b) Z = 1.96 (c) Z = 2.33 (d) Z = -2.33 (e) Z = -1.65

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A survey of randomly selected university students found that 91 of the 101 first-year students and 96 of the 136 second-year students surveyed had purchased used textbooks in the past year. (a) Construct a 95% confidence interval for the difference the proportions of first-year and second-year university students who purchased used textbooks. (Round your answers to 4 decimal places, if needed.) Answer: (b) If a 98% confidence interval for the difference in proportion of first-year and second-year university students who purchased used textbooks is (0.0809, 0.3093). Give an interpretation of this confidence interval. O We are 98% confident that the first-year students at this university bought between 8.09% and 30.93% more used textbooks than second-year students. O We know that 98% of the first-year students bought between 8.09% and 30.93% higher than the proportion of second-year students who bought used textbooks. O We are 98% confident that, at this university, the proportion of…
Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from $3.37 to $3.60. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.) x = $ E = $
#2
During the 2009 holiday season, 436 randomly sampled American adults were surveyed regarding their average spending on Black Friday. A 95% confidence interval based on this sample is ($80.31,$89.11). (c) Determine the sample mean x and the sample standard deviation s of the sample. (d) Which would be the confidence level corresponding to an interval ($81.27,$88.15) for the data found in (c)? (e) If the values of the sample mean and sample standard deviation found in (c) do not change, find the sample size N for which a 99% confidence interval would have a margin of error smaller than 5.
Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from $3.37 to $4.18. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.) x=$ E=$
Lexi of Arm End insurance ,was hired to estimate the proportion of patients that had paid their Bill's on time. She took a random sample of 100 such patients and found that 85 of them had paid their Bill's on time. Help her find 96% confidence interval for population proportion. 1. What is the X and is the "n" sample size? What is the point estimate for the population proportion? 2 Estimate the 96% confidence interval for the proportion of people that paid their bills on time . 3 What is the standard error for the population proportion ? 4 The margin of error for the population proportion, is 4%
## In a survey of adults with diabetes, the average body mass index (BMI) in a sample of 1938 women was 32.2 with a standard deviation of 0.03. The BMI in a sample of 1555 men was 30.4, with a standard deviation of 0.06. Construct a 92% confidence interval for the difference in the mean BMI between women and men with diabetes, INTERPRET THE RESULTS. DOES THE CONFIDENCE INTERVAL CONTRADICT THE CLAIM THAT THE MEAN BMI IS THE SAME FOR BOTH MEN AND WOMEN WITH DIABETES?
Echinacea is widely used as an herbal remedy for the common cold, but does it work? In a double-blind experiment, healthy volunteers agreed to be exposed to common-cold-causing rhinovirus type 39 and have their symptoms monitored. The volunteers were randomly assigned to take either a placebo or an echinacea supplement daily for 5 days following viral exposure. Among the 103 subjects taking a placebo, 88 developed a cold, whereas 44 of the 48 subjects taking echinacea 1 developed a cold.
Hore are summary statistics for randomly selected weights of newborm girls: n 193, x 27.5 hg, s0.3 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very difforent from the confidence interval 26.4 hg
The interval [40 50] represents a 90% confidence interval for a population mean. Which of the following is most likely to be the corresponding 95% confidence interval? (Hint: in a normal distribution, close to 90% of the values are within 1.65 standard deviations of the mean and close to 95% of the values are within 2 standard deviations of the mean) O A. [42 48] В. [36 54] OC. [40 54] D. [30 60]
A flower shop owner wants to estimate the average retail value of the best-selling flower in his inventory. A random sample of 10 flowers indicates an average value of 15 pesos and a standard deviation of 3.5 pesos. What are the point estimate and the 99% confidence interval in the average value of the flower? What is the confidence interval estimate of the problem? a. (11.40 , 18.60) b. (7.29 , 12.71) c. (11.68 , 18.32) d. (7.06 , 12.94)
Suppose a new standardized test is given to 97 randomly selected third-grade students in New Jersey. The sample average score Y on the test is 62 points, and the sample standard deviation, sy, is 10 points. The authors plan to administer the test to all third-grade students in New Jersey. The 95% confidence interval for the mean score of all New Jersey third graders is (60.01, 63.99). (Round your responses to two decimal places.) Suppose the same test is given to 194 randomly selected third graders from lowa, producing a sample average of 66 points and sample standard deviation of 13 points. The 90% confidence interval for the difference in mean scores between lowa and New Jersey is (-6.26 -1.74). (Round your responses to two decimal places.) The p-value of the test of no difference in means versus some difference is (Round your response to four decimal places.)