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Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

Using sample data to draw conclusions about a population

Note:

Each sample may vary, but the population parameter doesn’t!

- If population is approximately normal, so is
- the sample distribution

- If population is skewed, the sample distribution
- is approximately normal if n30 by the central
- limit theorem

- If given sample data, look at the distribution to
- assess normality if needed. (Normal Prob. Plot)

- Uses the sample distribution to predict population parameter

- It is an interval of numbers above and below the sample statistic

The probability the interval will capture the true parameter value in repeated samples

Critical Value:

The probability p lying to its right under the standard Normal curve. ( Z* )

- How accurate our estimate is based on the variability of the sample distribution. We add and subtract this from our estimate.

estimate margin of error

Caution! Margin of error is only from random sampling errors. This does not include errors in collecting the data!

Confidence Level (C) Upper tail prob. Z* Value

90%

1.645

0.05

0.05

0.05

0.90

Z=?

Z=?

Confidence Level (C) Upper tail prob. Z* Value

95%

1.96

0.025

0.025

0.025

0.95

Z=?

Z=?

Confidence Level (C) Upper tail prob. Z* Value

99%

2.576

0.005

0.005

0.005

0.99

Z=?

Z=?

Critical Values

2nd Dist – invNorm( (1 + C)/2 )

OR: Look at the T-Tables for the most common ones!

(You will learn more about them later)

Confidence Interval for a Population mean ( known)

(Z-Interval)

estimate margin of error

estimate critical value standard error

Properties of Confidence Intervals

- The interval is always centered around the statistic

- The higher the confidence level, the wider the interval becomes

- If you increase n, then the margin of error decreases

Z-Interval

Stat – Tests – ZInterval

Data: If given actual values

Stats: If given summary of values

Interpreting a Confidence Interval:

What you will say:

I am C% confident that the true parameter is captured in the interval

What it means:

If we took many, many, SRS from a population and calculated a confidence interval for each sample, C% of the confidence intervals will contain the true mean

Never Say:

The interval will capture the true mean C% of the time.

It either does or does not!

(should say)

- SRS

(CLT or population approx normal)

2. Normality

(Population 10x sample size)

3. Independence

Steps to Construct ANY Confidence Interval:

PANIC

P:

Parameter of Interest (what are you looking for?)

A:

Assumptions (what are the conditions?)

N:

Name the type of interval (what type of data do we have?)

I:

Interval (Finally! You can calculate!)

C:

Conclusion in context (I am ___% confident the true parameter lies between ________ and _________)

Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is =13.4.

- Construct a 95% confidence interval for the mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients.

P:

The true mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients.

SRS:

Says randomly selected

Normality:

Approximately normal by the

CLT (n 30)

I am assuming that Dr. Oswick has 300 patients or more.

Independence:

N:

One sample Z-Interval

I am 95% confident the true mean serum HDL cholesterol for all of Dr. Oswick’s 20-29 year old female patients is between 45.875 and 55.465

Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is =13.4.

b. If the US National Center for Health Statistics reports the mean serum HDL cholesterol for females between 20-29 years old to be = 53, do Dr. Oswick’s patients appear to have a different serum level compared to the general population? Explain.

No,

53 is contained in the interval.

Serum Cholesterol-Dr. Paul Oswick wants to estimate the true mean serum HDL cholesterol for all of his 20-29 year old female patients. He randomly selects 30 patients and computes the sample mean to be 50.67. Assume from past records, the population standard deviation for the serum HDL cholesterol for 20-29 year old female patients is =13.4.

c. What two things could you do to decrease your margin of error?

Increase n

Lower confidence level

Suppose your class is investigating the weights of Snickers 1-ounce Fun-Size candy bars to see if customers are getting full value for their money. Assume that the weights are Normally distributed with standard deviation = 0.005 ounces. Several candy bars are randomly selected and weighed with sensitive balances borrowed from the physics lab. The weights are

0.95 1.02 0.98 0.97 1.05 1.01 0.98 1.00

ounces. Determine a 90% confidence interval for the true mean, µ. Can you say that the bars weigh 1oz on average?

P:

The true mean weight of Snickers 1-oz Fun-size candy bars

Says randomly selected

SRS:

Normality:

Approximately normal because the population is approximately normal

I am assuming that Snickers

has 80 bars or more in the 1-oz size

Independence:

N:

One sample Z-Interval

I am 90% confident the true mean weight of Snickers 1-oz Fun-size candy barsis between .9921 and .9979 ounces. I am not confident that the candy bars weigh as advertised at the 90% level.

Choosing a Sample Size for a specific margin of error

Note: Always round up! You can’t have part of a person! Ex: 163.2 rounds up to 164.

A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799.

- What is the sample mean income?

A statistician calculates a 95% confidence interval for the mean income of the depositors at Bank of America, located in a poverty stricken area. The confidence interval is $18,201 to $21,799.

b. What is the margin of error?

m

m = 21,799 – 20,000

m = 1,799

A researcher wishes to estimate the mean number of miles on four-year-old Saturn SCI’s. How many cars should be in a sample in order to estimate the mean number of miles within a margin of error of 1000 miles with 99% confidence assuming =19,700.

In the 10.1 we made an unrealistic assumption that the population standard deviation was known and could be used to calculate confidence intervals.

When the standard deviation of a statistic is estimated from the data

When we know we can use the Z-table to make a confidence interval. But, when we don’t know it, then we have to use something else!

Properties of the t-distribution:

- σ is unknown
- Degrees of Freedom = n – 1
- More variable than the normal distribution (it has fatter tails than the normal curve)
- Approaches the normal distribution when the degrees of freedom are large (sample size is large).
- Area is found to the right of the t-value

Properties of the t-distribution:

- If n < 15, if population is approx normal, then so is the sample distribution. If the data are clearly non-Normal or if outliers are present, don’t use!

- If n > 15, sample distribution is normal, except if population has outliers or strong skewness

- If n 30, sample distribution is normal, even if population has outliers or strong skewness

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

10

1.093

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

0.15

10

1.093

Determine the degrees of freedom and use the t-table to find probabilities for each of the following:

0.85

10

1.093

23

0.685

0.25

23

0.685

0.25

23

-0.685

10

0.70

1.093

10

0.1

0.70

1.093

.25 – .15 = 0.1

One sample t-Interval

Stat – Tests – TInterval

Data: If given actual values

Stats: If given summary of values

- SRS

(should say)

(population approx normal and n<15, or moderate size (15≤ n < 30) with moderate skewness or outliers, or large sample size n ≥ 30)

2. Normality

3. Independence

(Population 10x sample size)

The probability calculations remain fairly accurate when a condition for use of the procedure is violated

The t-distribution is robust for large n values, mostly because as n increases, the t-distribution approaches the Z-distribution. And by the CLT, it is approx normal.

As part of your work in an environmental awareness group, you want to estimate the mean waste generated by American adults. In a random sample of 20 American adults, you find that the mean waste generated per person per day is 4.3 pounds with a standard deviation of 1.2 pounds. Calculate a 99% confidence interval for and explain it’s meaning to someone who doesn’t know statistics.

P:

The true mean waste generated per person per day.

Says randomly selected

SRS:

Normality:

15<n<30. We must assume the population doesn’t have strong skewness. Proceeding with caution!

It is safe to assume that there are more than 200 Americans that create waste.

Independence:

N:

One Sample t-interval

df = 20 – 1 = 19

I am 99% confident the true mean waste generated per person per day is between 3.5323 and 5.0677 pounds.

Subjects are matched according to characteristics that affect the response, and then one member is randomly assigned to treatment 1 and the other to treatment 2. Recall that twin studies provide a natural pairing. Before and after studies are examples of matched pairs designs, but they require careful interpretation because random assignment is not used.

Apply the one-sample t procedures to the differences

Archaeologists use the chemical composition of clay found in pottery artifacts to determine whether different sites were populated by the same ancient people. They collected five random samples from each of two sites in Great Britain and measured the percentage of aluminum oxide in each. Based on these data, do you think the same people used these two kiln sites? Use a 95% confidence interval for the difference in aluminum oxide content of pottery made at the sites and assume the population distribution is approximately normal. Can you say there is no difference between the sites?

1.7

3.2

1.3

-2.5

.6

μn = New Forrest percentage of aluminum oxide

μa = Ashley Trails percentage of aluminum oxide

μd = μn - μa = Difference in aluminum oxide levels

The true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails.

Says randomly selected

SRS:

Normality:

Says population is approx normal

It is safe to assume that there are more than 50 samples available

Independence:

N:

Matched Pairs t-interval

df = 20 – 1 = 19

I am 95% confident the true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails is between –1.754 and 3.4743.

Can you say there is no difference between the sites?

zero is in the confidence interval, so it is safe to say there is no difference.

Yes,

The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language teachers. One institute hosted 20 Spanish teachers for four weeks. At the beginning of the period, the teachers took the Modern Language Association’s listening test of understanding of spoken Spanish. After four weeks of immersion in Spanish in and out of class, they took the listening test again. (The actual spoken Spanish in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Below is the pretest and posttest scores. Give a 90% confidence interval for the mean increase in listening score due to attending the summer institute. Can you say the program was successful?

μB = Pretest score

μA = Posttest score

μd = μB - μA = Difference in test scores

The true mean difference in test scores between the Pretest and Posttest

We must assume the 20 teachers are randomly selected

SRS:

Normality:

15<n<30 and distribution is approximately normal, so safe to assume

It is safe to assume that there are more than 200 Spanish teachers

Independence:

N:

Matched Pairs t-interval

df = 20 – 1 = 19

I am 90% confident the true mean difference in test scores between the Pretest and Posttest

is between –2.689 and –0.2115.

Can you say the program was successful?

zero is not in the confidence interval, so the pretest score is lower than the posttest score.

Yes,

Confidence Interval for a Population Proportion:

Notice! We use Z* and not t*

A news release by the IRS reported 90% of all Americans fill out their tax forms correctly. A random sample of 1500 returns revealed that 1200 of them were correctly filled out. Calculate a 92% confidence interval for the proportion of Americans who correctly fill out their tax forms. Is the IRS correct in their report?

P:

The true percent of Americans who fill out their tax forms correctly

Says randomly selected

SRS:

Normality:

0.80

Yes, safe to assume an approximately normally distribution

It is safe to assume that there are more than 15,000 people who file their taxes

Independence:

I am 92% confident the true percent of Americans who fill out their tax forms correctly is between 78.19% and 81.8%

Is the IRS correct in their report?

No,

90% is not in the interval!

Sample size for a Desired Margin of Error

If we want the margin of error in a level C confidence interval for p to be m, then we need n subjects in the sample, where:

p* =

An estimate for

n

Note: If p is unknown use the most conservative value of p = 0.5. Since n is the sample size, it must be a whole number!!! Round up!

You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.

a. Find the sample size needed if a prior study found that 19% of computers needed repairs or had problems by the time the product as three years old.

You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.

b. If no preliminary estimate is available, find the most conservative sample size required.

You wish to estimate with 95% confidence; the proportion of computers that need repairs or have problems by the time the product is three years old. Your estimate must be accurate within 3.5% of the true proportion.

c. Compare the results from a and b.

Using 0.5 makes the sample size very large, ensuring that enough people will be surveyed.

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