Chapter 7

Chapter 7

Section 7-1 Estimating a Population Proportion

Notations:

p=population proportion

p^= x/n  sample proportion of successes

x=number of successes

n=sample size

q^=1-p^   sample proportion of failures

Assumptions:

a) fixed number of trials

b) independent

c) success/failure (2 outcomes)

d) constant probabilities

    3)  np>5 and nq>5 (there are at least 5 successes and 5 failures)

Point estimate--A single value used to approximate a population parameter.

Critical value- cut off value(s) that separate the sample statistics that are likely to occur from those that are unlikely to occur.

Margin of Error

Confidence Interval for Population Proportion

Sample size requirement for estimating p:

Use p^ = q^ = .5 when p^ and q^ are unknown

ALWAYS ROUND n UP!!!

Interpreting a Confidence Interval (see page 301)

We are ___% confident that the interval from ____ to ___ actually does contain the true value of the population proportion p.

7-1 #13 In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders observed.  Construct a 95% confidence interval for the proportion of orders that are not accurate.

                                  a)  Find the best point estimate

d) Write a statement that correctly interprets the confidence interval.

“We are _95__% confident that the interval from __.0615__ to _.1209___ actually does contain the true value of the population proportion p.”

Video on using Statcrunch to perform confidence intervals for proportions.

7-1 #18 One of Mendel’s famous genetics experiments yielded 580 peas, with 428 of them green and 152 yellow.

A) Find a 99% confidence interval of the percentage of green peas.

B) Based on his theory of genetics, Mendel expected 75% of the offspring peas would be green. Given that the percentage of offspring peas is not 75%, do the results contradict Mendel’s theory? Why or why not?

no, the confidence interval contains 75%, so the true percentage could easily be 75%



7-1 #33 In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor’s degree in four years or less.  Find the sample size needed to estimate that percentage.  Use a 0.05 margin of error, and use a confidence level of 95%

This can be done by hand as below or on statcrunch.

Video on finding minimum sample size requirement for p on statcrunch

C) Does the added knowledge in part (b) have much of an effect on sample size?

no, sample size doesn't change much.

Video on how to use Statcrunch to find critical T values (This video starts with some general use of the T calculator and ends with a problem like the above) 

Confidence interval for a population mean,     ( Sigma KNOWN, population standard deviation known) (rare case)

Assumptions:

1) Sample is from a SRS

2) Sample is from a normal distribution or n > 30

3) Sigma known


Confidence interval for a population mean,  ( Sigma UNKNOWN) (This is the case we will be working with) 

Assumptions:

1) Sample is from a SRS

2) Sample is from a normal distribution or n > 30



Construct the confidence interval estimate for the mean.

7-2#9 Here are summary statistics for randomly selected weights of newborn girls: n=205,  x=30.4 hg, s=7.1 hg. Use a 95% confidence level.

Video directions for Statcrunch on finding confidence intervals like the below.


7-2#11 Data set 3 “Body Temperatures” in Appendix B includes a sample of 106 body temperatures having a mean of 98.20 F and a standard deviation of 0.62 F.  Construct a 95% (and 98%) confidence interval estimate of the mean body temperature for the entire population.  What does the result suggest about the common belief that 98.6 F is the mean body temperature? 

Note for below problem: A couple of friends and I do quite a bit of fishing off the coast of Northern California in the summertime.  We catch a variety of different fish and some have the characteristics that can be concerning for having high levels of mercury (predatory fish or slow growing).  The only data we could find for our area was severely lacking, so we decided to collect and analyze the data ourselves through a grant from Humboldt Bay Keeper. Below is a subset of our data from our catch of Lingcod ( a large predatory fish that lives on the bottom, the picture is of my son with a Lingcod).  Were our concerns of high mercury legit? 


The Office of Environmental Health Hazard Assessments (OEHHA) "do not consume" level of Methyl-Mercury is 0.44 ppm for women under 45 and children. Construct a 90% confidence interval for the levels of Methyl-Mercury in Lingcod.  Should women under 45 and children be eating Lingcod?


Methyl- Mercury (ppm) for Lingcod:

0.137  0.452  0.341  0.226  0.238  0.484  0.408  1.94  0.789  0.55  0.72  1.49  1.59  2.84  0.193  0.171  0.537  0.655 Note: The sample size is n=18 and data is only somewhat normally distributed.  It would be better if we had a few more in the sample to better meet the requirement for this T test.


Solution: Here raw data is given so the use of Statcrunch is a huge time and calculation saver.  Enter the data into a column of Statcrunch, click Stat, T-Stat, One-sample, with data.  Select the data set and enter in 90% for the confidence level.  You should get the following:

One sample T confidence interval:

μ : Mean of variable


90% confidence interval results:

Variable             Sample Mean      Std. Err.       DF       L. Limit          U. Limit

Methyl-Mecury         0.7645         0.17275369     17       0.46397651     1.0650235


The 90% confidence interval for the levels of Methyl-Mercury in Lingcod is (.46397651, 1.0650235) ppm.


Should women under 45 and children be eating Lingcod based on OEHHA's "do not consume" level set at .44 ppm? NO!  The interval of (.46397651, 1.0650235) ppm suggests that the level of Methyl-Mercury is above .44 ppm.  Lingcod appear to have high levels of Methyl-Mercury!  More to come on this topic as the course progresses, what else did we learn about Lingcod in our study?

Statcrunch example with raw data

7-2#30 Assume that  sigma = 15 for the IQ scores. Attorneys are a group with IQ scores that vary less than the IQ scores of the general population.  Find the sample size needed to estimate the mean IQ of attorneys, given that we want 98% confidence that the sample mean is within 3 IQ points of the population mean. Does the sample size appear to be practical?

This sample size requirement can be done by hand as it is below or using statcrunch.

Video on finding the sample size requirement for mean using statcrunch

Sample size requirement formula