O d. 2.00Independent random sampling from two normally distributed populations gives thefollowing results: n,- 55, x= 520, o, - 30, n,= 45, y- 482, and o,-24.Find the 98% confidence interval for the difference in the means of the two populations.Lütfen birini seçin:O a 1002(+-)12.58Ob.38(+-)12.58c. 38(+-)19 86d. 1002(+)19.86The manufacturer of a new chewing gilm claims that at least 80% of dentists surveyed prefertheir type of gum and recommend it for their patients who chew gum An independentconsumer research firm decides to test their claim The findings in a sample of 200 dentistsindicate that 741% of the respondents do actually prefer their gum Which of the followingStatements IS most accurate?

Independent random sampling from two normally distributed populations gives the following results: n,- 55, x= 520, o, - 30, n, = 45, y- 482, and o,=24. Find the 98% confidence interval for the difference in the means of the two populations. Lütfen birini seçin. a 1002(+-)12.58 b. 38(+-)12.58 Oc 38(+-)19 86 Od 1002(+-)19 86

Independent random sampling from two normally distributed populations gives the following results: n,- 55, x= 520, o, - 30, n, = 45, y- 482, and o,=24. Find the 98% confidence interval for the difference in the means of the two populations. Lütfen birini seçin. a 1002(+-)12.58 b. 38(+-)12.58 Oc 38(+-)19 86 Od 1002(+-)19 86

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

Related questions

Q: Suppose that the number of students passed a Statistics class are normally distributed. A simple…

A: We have given that The number of students passed a Statistics class are normally distributed. A…

Q: A simple random sample of size n=40 is drawn from a population. The sample mean is found to be…

A: n=40 X-bar=121.2 S=13.2 99℅ confidence interval

Q: A random sample of ni = 233 people who live in a city were selected and 109 identified as a "dog…

Q: 6.06* From a random sample of n Canadians, 30% support the Liberal Party. The 95% confidence…

A: The formula for the confidence interval of population proportion is (p̂ ± E).

Q: Give a 99.8% confidence interval, for 41 - P2 given the following information. = 55, = 2.17, s =…

Q: A random sample of n measurements was selected from a population with unknown mean u and standard…

A: The random variable X follows normal distribution. We have to construct 95% confidence interval for…

Q: Construct a 95% confidence interval of the average.

A: Given : Y = 0.00039508 S = 0.0084646 n = 1,259

Q: Use the given confidence interval to find the margin of error and the sample mean. (67.6,82.0) The…

A: The confidence interval is 67.6,82.0. The margin of error is, Margin of error=Upper bound-Lower…

Q: Given a 99% confidence interval for mu1-mu2 as (5, 10). What can be concluded? A. nothing B.…

A: Given Information: 99% confidence interval for μ1-μ2 is (5, 10). Lower bound is 5 and upper bound is…

Q: Use the given confidence interval to find the margin of error and the sample mean. (15.8,23.6) The…

A: Given :Confidence interval (C.I)=(15.8 , 23.6)

Q: Last year a bill was passed. Of those who voted, 40 people gave two pieces of information: if they…

Q: Za/2 for the 99% confidence interval Za/2 Part 5 of 5 for the 94% confidence interval Za/2

A: Given99% confidence interval:-

Q: 9A. The time to failure of 12 batteries was measured. The average time to failure of the 12…

Q: A random sample of n1n1 = 272 people who live in a city were selected and 74 identified as a "dog…

Q: The mean replacement time for a random sample of 25 machines is 6.5 years and the standard deviation…

Q: e construct a confidence interval for the difference between two population means p, -, use the…

Q: Find t* for a 80% confidence interval based on N = 45 randomly selected observation

A: Given, Sample size n = 45 80% confidence interval

Q: A normally distributed population is having standard deviation 2.8, a sample of 28 is taken randomly…

A: Answer: From the given data, Sample size(n) = 28 mean (X¯) = 15.5 Standard deviation(σ) = 2.8 98%…

Q: 6.2.13 Use the given confidence interval to find the margin of error and the sample mean.…

A: We have to find sample mean and margin of error...

Q: A random sample of 25 individuals finds a sample mean value of 18 and sample standard deviation of…

A: Given information- Sample size (n) = 25 Mean x̅ = 18 Standard deviation (s) = 6 Significance…

Q: An Independent sample from two different populations yields the following data. x̄1= 947,x̄2 =…

Q: Two random samples are selected from two independent populations. A summary of the samples sizes,…

Q: A researcher collected a random sample of n = 46 squirrels and found 6 with a broken tail. Find a…

A: From the provided information Sample size (n) = 46, x = 10 The sample proportion can be obtained as:

Q: 6.2.14 Use the given confidence interval to find the margin of error and the sample mean.…

Q: Use the given confidence interval to find the margin of error and the sample mean. (15.3,24.1) The…

Q: 7. Bob draws a simple random sample from an original population. The sample size is n = 51, the…

Q: A student conducted a test to see if people preferred one of two possible new logos for UF, one with…

A: Confidence interval is a technique of estimation.

Q: A random sample of n1 = 246 people who live in a city were selected and 70 identified as a "dog…

A: Given problem is :

Q: Q.3/ Assume that the following five numbers are sampled from a normal distribution: 2, 3. 5, 6 and 9…

A: Solution: x (x-x) (x-x)2 2 -3 9 3 -2 4 5 0 0 6 1 1 9 4 16 n= 5 Sample size ∑x=…

Q: A random sample of 78 measurements was selected from a population with ?=2.76σ=2.76. The sample mean…

A: We have given that.Sample size (n) = 78 Population Mean(µ) = 2.77Sample mean (x̅) = 122.5Standard…

Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

O d. 2.00
Independent random sampling from two normally distributed populations gives the
following results: n,- 55, x= 520, o, - 30, n,= 45, y- 482, and o,-24.
Find the 98% confidence interval for the difference in the means of the two populations.
Lütfen birini seçin:
O a 1002(+-)12.58
Ob.38(+-)12.58
c. 38(+-)19 86
d. 1002(+)19.86
The manufacturer of a new chewing gilm claims that at least 80% of dentists surveyed prefer
their type of gum and recommend it for their patients who chew gum An independent
consumer research firm decides to test their claim The findings in a sample of 200 dentists
indicate that 741% of the respondents do actually prefer their gum Which of the following
Statements IS most accurate?

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.

Similar questions

Use the sample information x⎯⎯x¯ = 37, σ = 6, n = 15 to calculate the following confidence intervals for μ assuming the sample is from a normal population.
A random sample of n1n1 = 236 people who live in a city were selected and 90 identified as a "dog person." A random sample of n2n2 = 103 people who live in a rural area were selected and 56 identified as a "dog person." 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." Round answers to to 4 decimal places. ___________< p1−p2p1-p2 <_____________
A sample of 36 breastfed infants found that 2 of them developed iron deficiency by age 5.5 months. Using the plus-four method, what is the 95% confidence interval for the population proportion of breastfed infants who have iron deficiency? Choose from available choices. a. (-0.0193, 0.1304) b. (-0.0072, 0.1184) c. (0.0071, 0.1929)
6.2.13 Use the given confidence interval to find the margin of error and the sample mean. (14.2,24.0) The sample mean is (Type an integer or a decimal.)
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.
random sample of 25 business students required an average of 50.7 minutes to complete a statistics exam. Assume that the population standard deviation to complete the exam was 10.4 minutes. The 99% confidence interval around this sample mean is
Use the given confidence interval to find the margin of error and the sample mean. (13.2,20.6) The sample mean is The margin of error is (Type an integer or a decimal ) (Type an integer or a decimal)
A random sample of n1n1 = 235 people who live in a city were selected and 79 identified as a "dog person." A random sample of n2n2 = 93 people who live in a rural area were selected and 67 identified as a "dog person." 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." Round answers to to 4 decimal places. < p1−p2<
The z confidence interval for the mean of a Normal population will have a specified margin of error m when the sample size is: a. n = (=*o/m). b. n = (z/m)². O c. n = (*o/m)². d. n ==*o/m.