A random sample of 10 children found that their averagegrowth for the first year was 9.8 inches, with a standarddeviation of 0.96 inches. Find the 85% confidence intervalof the mean growth of all American children in the firstyear of life. Check which assumptions have not been met so we wouldsay "the assumption may not be met" or have to check thehistogram, box plot, and normal quantile plot.Fixed number of trialsThe proportion is constantIndependent trialsNormal Sampling Distribution: n,p, 2 5, n,(1 - p,) 2 5np, 2 5 & n,(1 - P,) 2 5Expected values 2 5Equal population variancesRandom sampleTwo outcomesIndependent samplesNormal Sampling Distribution: n 2 30 or Populationis normally distributedKnown population standard deviationNormal Sampling Distribution: np > 5 & n(1 - p) > 5Unknown population standard deviation

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A random sample of 10 children found that their average growth for the first year was 9.8 inches, with a standard deviation of 0.96 inches. Find the 85% confidence interval of the mean growth of all American children in the first year of life.

A random sample of 10 children found that their average growth for the first year was 9.8 inches, with a standard deviation of 0.96 inches. Find the 85% confidence interval of the mean growth of all American children in the first year of life.

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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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

A random sample of 10 children found that their average
growth for the first year was 9.8 inches, with a standard
deviation of 0.96 inches. Find the 85% confidence interval
of the mean growth of all American children in the first
year of life.

Check which assumptions have not been met so we would
say "the assumption may not be met" or have to check the
histogram, box plot, and normal quantile plot.
Fixed number of trials
The proportion is constant
Independent trials
Normal Sampling Distribution: n,p, 2 5, n,(1 - p,) 2 5
np, 2 5 & n,(1 - P,) 2 5
Expected values 2 5
Equal population variances
Random sample
Two outcomes
Independent samples
Normal Sampling Distribution: n 2 30 or Population
is normally distributed
Known population standard deviation
Normal Sampling Distribution: np > 5 & n(1 - p) > 5
Unknown population standard deviation

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