The birth weights of a sample of newborns were recorded for a group born to mothers who did not get regular prenatal care (Group 1) and for a group born to mothers who did (Group 2). A 93% confidence interval for $ \mu_1 - \mu_2 $ was computed as follows:(-2.12 pounds, -0.89 pounds)Interpret the endpoints of this interval in the context of this problem. Do not use the word "difference" in your answer.---This interval suggests that we can be 93% confident that the average birth weight for newborns whose mothers did not receive regular prenatal care is between 0.89 pounds and 2.12 pounds less than those whose mothers received regular prenatal care. The negative values indicate that the birth weights in Group 1 are likely lower than those in Group 2.

Solution; The 93% confidence interval for the difference in μ1-μ2 is ( -2.12 pounds , -0.89 pounds)…

The birth weights of a sample of newborns were recorded for a group born to mothers who did not get regular prenatal care (Group 1) and for a group born to mothers who did (Group 2). A 93% confidence interval for 41 - p2 was computed as follows: (-2.12 pounds, -0.89 pounds)

The birth weights of a sample of newborns were recorded for a group born to mothers who did not get regular prenatal care (Group 1) and for a group born to mothers who did (Group 2). A 93% confidence interval for 41 - p2 was computed as follows: (-2.12 pounds, -0.89 pounds)

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: Construct the indicated confidence interval for the difference between population proportions p1 -…

Q: A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Twelve…

A: The objective of this question is to construct a 99% confidence interval for the true mean…

Q: An environmental science teacher at a high school with a large population of students wanted to…

A: The 95% confidence interval for the proportion of all students at the school who would respond yes…

Q: In a study, 110 children of a certain race in a certain city were randomly assigned to one of two…

A: From the table contingency table we can see that there are 17+39 = 56 children who did not go to…

Q: In a sample of 40 water specimens taken from a construction site 26 contained detectable level of…

A: According to the provided information, sample size (N)=40 number of favorable cases (X) = 26 The…

Q: A recent survey of 54,000 randomly selected adults studied the religious affiliations of Americans.…

A: From the provided information, Sample size (n) = 645 From which 116 said they claimed no religious…

Q: The career center at a large university surveyed a random sample of students who graduated the…

A: First, we need to calculate the proportions of engineering and business majors who were able to find…

Q: A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Fourteen…

A: d = Initial level - level after 3 months

Q: A study of 420,024 cell phone users found that 140 of them developed cancer of the brain or nervous…

A: Given:

Q: Of 91 adults selected randomly from one town, 60 have health insurance. Find a 90% confidence…

A: From the provided information, Sample size (n) = 91 Out of which 60 have health insurance. Sample…

Q: A sleep disorder specialist wants to test the effectiveness of a new drug that is reported to…

A: From the provided information,n = 9Confidence level = 90%The difference d are as follow:d = x2 -…

Q: A study of 420,010 cell phone users found that 0.0319% of them developed cancer of the brain or…

A: Solution: Given information: n= 420010 Sample size p^=0.0319100= 0.000319 Sample Proportion

Q: An investigator computes a 95% confidence interval for a population mean on the basis of a sample of…

A: To find half of width, We need to calculate 4 times the sample Size.

Q: Independent samples from two different populations yield the following data. x1=200,x2=963,…

A: Given x1=200,x2=963, s1=19,s2=86 Find 90%CI for u1-u2

Q: Find the 98% confidence interval for the true difference in population proportions, pˆ A − pˆ B.

A: Given that For drug A, sample size n1 = 50, number of animals showed undesirable side effects, X1 =…

Q: In an engineering class, students want to determine the difference in the mean strength between two…

A: Here using the given information we give the answer such as

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

The birth weights of a sample of newborns were recorded for a group born to mothers who did not get regular prenatal care (Group 1) and for a group born to mothers who did (Group 2). A 93% confidence interval for $ \mu_1 - \mu_2 $ was computed as follows:

(-2.12 pounds, -0.89 pounds)

Interpret the endpoints of this interval in the context of this problem. Do not use the word "difference" in your answer.

---

This interval suggests that we can be 93% confident that the average birth weight for newborns whose mothers did not receive regular prenatal care is between 0.89 pounds and 2.12 pounds less than those whose mothers received regular prenatal care. The negative values indicate that the birth weights in Group 1 are likely lower than those in Group 2.

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

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

The number of cell phones per 100 residents in countries in Europe is given in table #9.3.9 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 ("Population reference bureau," 2013). Find the 98% confidence interval for the different in mean number of cell phones per 100 residents in Europe and the Americas. (Show all work) Table #9.3.9: Number of Cell Phones per 100 Residents in Europe 100 100 76 100 130 75 84 112 112 84 138 133 118 134 126 126 188 129 93 64 128 124 124 122 109 121 127 152 96 96 63 99 95 151 147 123 123 95 67 67 118 125 110 110 115 140 115 141 77 98 98 102 102 112 118 118 54 54 23 121 126 47 Table #9.3.10: Number of Cell Phones per 100 Residents in the Americas 158 117 106 159 53 50 78 66 88 92 42 3…
PT. “ABC” in 2018 offers a new portfolio for investment. For this new product, companies need to know the ability of investors to face risk. For this purpose, samples were taken from two groups, namely the old and young groups, each of which was taken as many as 160 old and young investors. The survey results showed that as many as 90 older people and 70 young people agreed to accept greater risk. Make a confidence interval for the difference in the proportion of the ability to face the risk.
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 226 milligrams withs = 17.9 milligrams. Construct a 98% confidence interval for the true mean cholesterol content of all such eggs
If n=280 and ˆpp^ (p-hat) = 0.39, construct a 95% confidence interval.Give your answers to three decimals.
Dr. Moore wanted to estimate the mean cholesterol level for all adult men living in Hartford. He took a sample of 25 adult men from Hartford and found that the mean cholesterol level for this sample is 186 mg/dL with a standard deviation of 12 mg/dL. Assume that the cho- lesterol levels for all adult men in Hartford are (approximately) normally distributed. Con- struct a 95% confidence interval for the population mean u.
A group of 40 bowlers showed that their average score was 192 with a standard deviation of 8. Find the 95% confidence interval of the mean score of all bowlers.
If n = 420 and ˆpp^ (p-hat) = 0.7, construct a 90% confidence interval.Give your answers to three decimals
A researcher was interested in the mean corn yeild for farmers in Illinois. She took an SRS of 28 farmers from Illinois and survey their corn yeild. The sample mean was 136.25 bushels per acre and the standard deviation was 13.25 bushels per acre assuming the conditions are met how many farmers in Illinois should be sampled in order to get a 95% confidence interval with a margin of error of 3.5?
If n = 500 and ˆpp^ (p-hat) = 0.2, construct a 99% confidence interval.
If n=360 and ˆpp^ (p-hat) = 0.6, construct a 90% Confidence Interval.In order to use technology, you need to compute the number of successes:x = n*ˆpp^Give your answers to three decimals.
A random sample of 42 KU college students gave a mean age of their vehicles of 11.3 years and a standard deviation of 3.4 years. A separate random sample of 38 KU college employees gave a mean vehicle age of 9.2 years with a standard deviation of 3.1 years. Find a 90% confidence interval for the difference in the population mean of vehicle ages between students and employees, u1-u2. Use students as the first population.
The mean noise level of 19 randomly selected areas designated as emergency rooms was 63.1 dBA and the sample standard deviation is 5.0 dBA. The mean noise level for 24 randomly selected areas designated as operating theaters was 56.3 dBA and the sample standard deviation was 7.0 dBA. An engineer reported a confidence interval of (1.8454, 11.409) for the difference in the mean noise level between areas designated as emergency rooms and areas designated for operating theaters, but neglected to specify the significance level. Estimate the significance level of the confidence interval if the population standard deviations, 01 and oz of the two populations are unknown and unequal. At the estimated significance level, could it be concluded that there is a difference in the means?