A genetic experiment with peas resulted in one sample of offspring that consisted of 407 green peas and 151 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? ..... a. Construct a 95% confidence interval. Express the percentages in decimal form.

A genetic experiment with peas resulted in one sample of offspring that consisted of 407 green peas and 151 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? ..... a. Construct a 95% confidence interval. Express the percentages in decimal form.

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

Please help!
If n= 100 and x= 80, construct a 99% confidence interval. Give your answers to three decimals
Given a confidence interval {21.98, 28.64 and a mean of 25.31, the margin of error is 6.66. True or False
A survey of 240 teachers found that two-thirds of those surveyed do not live in Sydney. Can you conclude statistically that the population proportion who do not live in Sydney is less than 0.75? Use a 95% confidence interval. What is the confidence interval and is 0.75 within the interval?
At one large college, 660 out of the 1500 students surveyed said they plan to take another science class.Find the 95% confidence interval for the proportion of students that said they plan to take another science class.
2. In a survey participated by 100 college students, it was found out that 60% of them have personal computers at home. Construct a 95% confidence interval estimate of the true population proportion of all college students who have personal computers.
In a sample of 500 Canadians, it was found that 400 people preferred apple products. In a sample of 700 Americans it was found that 600 preferred apple products. Is the proportion Canadians who prefer apple products less than Americans. 1. Construct 95% confidence interval for difference of proportions. 2. From the confidence interval, do you think that the proportion of Canadians who prefer apple products is different from Americans.
According the the U.S. Census bureau, the percentage of Santa Ana residents living in poverty is 17.7% and the percentage of Irvine residents living in poverty is 13.3%. Suppose that the sample sizes were 1200 for the Santa Ana group and 1500 for the Irvine group. Use the survey data to calculate and interpret a 90% confidence interval for the difference in the proportion living poverty between Santa Ana and Irvine
If n=470 and p̂ (p-hat) = 0.65, construct a 95% confidence interval.
In a survey of 1798 US adults, 1365 said they had paid a bill online within the past year. What is the LOWER endpoint of a 90% confidence interval for the population proportion? What is the UPPER endpoint of a 90% confidence interval for the population proportion?
You want to estimate the average difference in the number of complaints filed by women at work and the number of complaints filed by men at work. You took an SRS of 7 businesses and collected the following data for number of complaints in a year: men 1 3 2 4 6 3 5 women 8 7 6 3 5 12 15 Create a 90% confidence interval for the average difference in the number of complaints filed by women at work and the number of complaints filed by men at work (in a year). You must do this using table C. If this is a 2 sample interval, be sure to specify who is population 1 and who is population 2. If this is a matched pairs interval, be sure to state the order you subtracted your data. Once you create your interval, give a sentence explaining the meaning of the interval (what the numbers in the interval mean, NOT the meaning of 95%) in context of this problem. Round to any appropriate number of decimals.
A survey of 32,782 adults in 2004 indicated that 24.6% of adults were current smokers. A similar study conducted in 1995 of 40,000 adults indicated that 29.7% were current smokers. a. Find and interpret a point estimate of the difference between the proportion of current smokers in 1995, and the proportion of current smokers in 2004. b. A 95% confidence interval for the true difference is (-0.058,-0.045). Interpret. c. What assumptions must you make for the confidence interval i part b to be valid? (…) a. Find a point estimate of the difference between the proportion of current smokers in 1995 and the proportion of current smokers in 2004. (Round to three decimal places as needed.)