Confidence Intervals with Means

Confidence Intervals with

MeansChapter 9

What is the purpose of a confidence interval?

To estimate an unknown To estimate an unknown population parameterpopulation parameter

One-Sample z Confidence Interval for

2. The sample size n is large (generally n30), and

3. , the population standard deviation, is known then the general formula for a confidence interval for a population mean is given by

x z critical valuen

If

1. is the sample mean from a random sample, x

If

1. is the sample mean from a random sample, x

Formula:Formula:

nzx * :Interval Confidence

statistic

Critical value

Standard deviation of parameter

Margin of errorMargin of error

Find a 90% confidence interval estimate for the true mean fills of catsup from this machine.

ExampleA certain filling machine has a true population standard deviation = 0.228 ounces when used to fill catsup bottles. A random sample of 36 “6 ounce” bottles of catsup was selected from the output from this machine and the sample mean was 6.018 ounces.

The z critical value is 1.645

90% Confidence Interval (5.955, 6.081)

36n,228.0,018.6x 36n,228.0,018.6x

x (z critical value)n

0.2286.018 1.645 6.018 0.06336

Conclusion:We are 90% confident that the true mean fills of catsup from the machine is between 5.955 oz. and 6.081 oz.

In a randomized comparative experiment on the effects of calcium on blood pressure, researchers divided 54 healthy, white males at random into two groups. The participants either take calcium or a placebo. The paper reports a mean seated systolic blood pressure of 114.9 with standard deviation of 9.3 for the placebo group. Assume systolic blood pressure is normally distributed.

Can you find a z-interval for this problem? Why or why not?Can you find a z-interval for this problem? Why or why not?

We only know sample statistics! We do not know population standard deviation!

William S. Gossett

Quality control engineer for Guiness Brewery in Dublin, Ireland

Checked the stout’s quality by performing hypothesis tests

Figured out a new family of models

Student’s t distributions

Student’s t- distributionStudent’s t- distribution• Developed by William Gosset• Continuous distribution• Unimodal, symmetrical, bell-shaped

density curve• Above the horizontal axis• Area under the curve equals 1• Based on degrees of freedom

df = n - 1df = n - 1

t Distributions

How does the How does the tt-distributions -distributions compare to the standard compare to the standard normal distribution?normal distribution?

• Bell-shaped and centered at 0

• Shorter & more spread out

• More area under the tails

• As n increases, t-distributions become more like a standard normal distribution

Formula:Formula:

nstx * :Interval Confidence

statistic

Critical value

Standard deviation of statistic

Margin of errorMargin of error

Standard error – when you

substitute s for .

Since each t distribution would require a table similar to the standard normal table, we usually only create a table of critical values for the t distributions.

Appendix Table 3 in BOB

t Distributions

0.80 0.90 0.95 0.98 0.99 0.998 0.99980% 90% 95% 98% 99% 99.8% 99.9%

1 3.08 6.31 12.71 31.82 63.66 318.29 636.582 1.89 2.92 4.30 6.96 9.92 22.33 31.603 1.64 2.35 3.18 4.54 5.84 10.21 12.924 1.53 2.13 2.78 3.75 4.60 7.17 8.615 1.48 2.02 2.57 3.36 4.03 5.89 6.876 1.44 1.94 2.45 3.14 3.71 5.21 5.967 1.41 1.89 2.36 3.00 3.50 4.79 5.418 1.40 1.86 2.31 2.90 3.36 4.50 5.049 1.38 1.83 2.26 2.82 3.25 4.30 4.78

10 1.37 1.81 2.23 2.76 3.17 4.14 4.5911 1.36 1.80 2.20 2.72 3.11 4.02 4.4412 1.36 1.78 2.18 2.68 3.05 3.93 4.3213 1.35 1.77 2.16 2.65 3.01 3.85 4.2214 1.35 1.76 2.14 2.62 2.98 3.79 4.1415 1.34 1.75 2.13 2.60 2.95 3.73 4.0716 1.34 1.75 2.12 2.58 2.92 3.69 4.0117 1.33 1.74 2.11 2.57 2.90 3.65 3.9718 1.33 1.73 2.10 2.55 2.88 3.61 3.9219 1.33 1.73 2.09 2.54 2.86 3.58 3.8820 1.33 1.72 2.09 2.53 2.85 3.55 3.8521 1.32 1.72 2.08 2.52 2.83 3.53 3.8222 1.32 1.72 2.07 2.51 2.82 3.50 3.7923 1.32 1.71 2.07 2.50 2.81 3.48 3.7724 1.32 1.71 2.06 2.49 2.80 3.47 3.7525 1.32 1.71 2.06 2.49 2.79 3.45 3.7326 1.31 1.71 2.06 2.48 2.78 3.43 3.7127 1.31 1.70 2.05 2.47 2.77 3.42 3.6928 1.31 1.70 2.05 2.47 2.76 3.41 3.6729 1.31 1.70 2.05 2.46 2.76 3.40 3.6630 1.31 1.70 2.04 2.46 2.75 3.39 3.6540 1.30 1.68 2.02 2.42 2.70 3.31 3.5560 1.30 1.67 2.00 2.39 2.66 3.23 3.46

120 1.29 1.66 1.98 2.36 2.62 3.16 3.371.28 1.645 1.96 2.33 2.58 3.09 3.29

Central area captured:Confidence level:

Degrees of freedom

z critical values

How to find How to find tt**• Use Table B for t distributions• Look up confidence level at bottom &

df on the sides• df = n – 1

Find these t*90% confidence when n = 595% confidence when n = 15

t* =2.132t* =2.145

Can also use invT on the calculator!

Need upper t* value with 5% is above – so 95% is below

invT(p,df)

Let’s do some comparing:Finding probabilities

Student’s t models:Normal model:

Normalcdf(1.645, ∞) tcdf(1.645, ∞, 4)

tcdf(1.645, ∞, 9)

tcdf(1.645, ∞, 14)

tcdf(1.645, ∞, 29)

tcdf(1.645, ∞, 99)

Student’s t models resemble normal models as sample size gets bigger….

.04998

.05157

.05538

.06111

.06719

.08766

Finding critical values:

Student’s t models:Normal model:

95% invT(.975, 4)

invT(.975, 9)

invT(.975, 14)

invT(.975, 29)

invT(.975, 99)

T critical values approach z critical values as sample size increases….

1.96

1.984

2.045

2.145

2.262

2.776

z critical values: t critical values:

invNorm(.975)

Steps for doing a confidence Steps for doing a confidence interval:interval:1) Identify by name or formula

One-sample confidence interval for means

2) Assumptions

3) Calculate the interval

4) Write a statement about the interval in the context of the problem.

nstx * :Interval Confidence

Assumptions for Assumptions for tt-inference-inference

• Have an SRS from population (or randomly assigned treatments)

• unknown

• Normal (or approx. normal) distribution– Given– Large sample size– Check graph of data

Use only one of these methods to check normality

Statement: Statement: (memorize!!)(memorize!!)

We are ________% confident

that the true mean context is

between ______ and ______.

Ex. 1) Find a 95% confidence interval for the true mean systolic blood pressure of the placebo group.

Assumptions:

• Have randomly assigned males to treatment

• Systolic blood pressure is normally distributed (given).

• is unknown

We are 95% confident that the true mean systolic blood pressure is between 111.22 and 118.58.

)58.118,22.111(273.9056.29.114

Ex. 2) A medical researcher measured the pulse rate of a random sample of 20 adults and found a mean pulse rate of 72.69 beats per minute with a standard deviation of 3.86 beats per minute. Assume pulse rate is normally distributed. Compute a 95% confidence interval for the true mean pulse rates of adults.

One-sample confidence interval for means

Assumptions:

• random sample of adults

• Pulse rate is normally distributed (given).

• is unknown

3.8672.69 2.093 (70.883, 74.497)20

We are 95% confident that the true mean pulse rate of adults is between 70.883 & 74.497.

Ex 2 continued) Another medical researcher claims that the true mean pulse rate for adults is 72 beats per minute. Does the evidence support or refute this? Explain.

The 95% confidence interval contains the claim of 72 beats per minute. Therefore, there is no evidence to doubt the claim.

Ex. 3) Consumer Reports tested 14 randomly selected brands of vanilla yogurt and found the following numbers of calories per serving:160 200 220 230 120 180 140130 170 190 80 120 100 170Compute a 98% confidence interval for the average calorie content per serving of vanilla yogurt.

We are 98% confident that the true mean calorie content per serving of vanilla yogurt is between 126.16 calories & 189.56 calories.

Ex 3 continued) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate?

Since 120 calories is not contained within the 98% confidence interval, the evidence suggest that the average calories per serving does not equal 120 calories.

Note: confidence intervals tell us if something is NOT EQUALNOT EQUAL

– never less or greater than!

RobustRobust• An inference procedure is ROBUST if the

confidence level or p-value doesn’t change much if the normality assumption is violated.

• t-procedures can be used with some skewness, as long as there are no outliers.

• Larger n can have more skewness.

Since there is more area in the tails in t-distributions, then, if a distribution has

some skewness, the tail area is not greatly affected.

CI & p-values deal with area in the tails – is the area changed greatly

when there is skewness

Find a sample size:Find a sample size:

n

zm *

• If a certain margin of error is wanted, then to find the sample size necessary for that margin of error use:

Always round up to the nearest person!

Ex 4) The heights of SHS male students is normally distributed with = 2.5 inches. How large a sample is necessary to be accurate within + .75 inches with a 95% confidence interval?

n = 43

Some Cautions:Some Cautions:

• The data MUST be a SRS from the population (or randomly assigned treatment)

• The formula is not correct for more complex sampling designs, i.e., stratified, etc.

• No way to correct for bias in data

Cautions continued:Cautions continued:

• Outliers can have a large effect on confidence interval

• Must know to do a z-interval – which is unrealistic in practice

Confidence Interval ExampleTen randomly selected shut-ins were each asked

to list how many hours of television they watched per week. The results are82 66 90 84 7588 80 94 100 91

Find a 90% confidence interval estimate for the true mean number of hours of television watched per week by shut-ins.

t-critical value of 1.833 by looking on the t table at 90% confidence with df = 9 or invt(.95, 9).

Calculating the sample mean and standard deviation we have n = 10, = 85, s = 9.843x 86

ns*tx

Can we meet the assumptions for a t –confidence interval?

9.84385 1.833( ) 85 5.70510

(79.295, 90.705)

We are 90% confident that the true mean number of hours of television watched per week is between 79.295 hours and 90.705 hours.