Hypothesis Tests: One Sample Mean

Hypothesis Tests: One Sample Mean MAJOR POINTS
Sampling distribution of the mean revisited hypotheses: sigma known
Testing
An example
Testing
An hypotheses: sigma unknown example
Factors affecting the test
Measuring the size of the effect
Confidence intervals 2 REVIEW: HYPOTHESIS TESTING STEPS 1. 2. 3. 4. 5. 6. 7. State Null Hypothesis Alternative Hypothesis Decide on α (usually .05) Decide on type of test (distribution; z, t, etc.) Find critical value & state decision rule Calculate test Apply decision rule 3 1 SAMPLING DISTRIBUTIONS
In reality, we take only one sample of a specific size (N) from a population and calculate a statistic of interest.
Based upon this single statistic from a single sample, we want to know:
“How likely is it that I could get a sample statistic of this value from a population if the corresponding population parameter was ___” 4 SAMPLING DISTRIBUTIONS
BUT, in order to answer that question, we need to know what the entire range of values this statistic could be.
How can we find this out?
Draw all possible samples of size N from the population and calculate a sample statistic on each of these samples (Chapter 8)
Or we can calculate it 5 SAMPLING DISTRIBUTIONS
A distribution of all possible statistics calculated from all possible samples of size N drawn from a population is called a Sampling Distribution.
Three things we want to know about any distribution?
– Central Tendency, Dispersion, Shape 6 2 AN EXAMPLE – BACK TO IQPLUS
Returning to our study of IQPLUS and its affect on IQ
A group of 25 participants are given 30mg of IQPLUS everyday for ten days
At the end of 10 days the 25 participants are given the Stanford-Binet intelligence test. 7 IQPLUS
The mean IQ score of the 25 participants is 106 µ = 100, σ = 15
Is this increase large enough to conclude that IQPLUS was affective in increasing the participants IQ? 8 SAMPLING DISTRIBUTION OF THE MEAN
Formal solution to example given in Chapter 8.
We need to know what kinds of sample means to expect if IQPLUS has no effect.
i. e. What kinds of means if µ = 100 and σ = 15?
This is the sampling distribution of the 9 mean (Why?) 3 POPULATION DISTRIBUTION 10 SAMPLING DISTRIBUTION 11 What is the relationship between σ and the SD above? SAMPLING DISTRIBUTION OF THE MEAN
The sampling distribution of the mean depends on
Mean

St.
dev. of sampled population Why?
Size
of sampled population Why? of sample Why? 12 4 SAMPLING DISTRIBUTION OF THE MEAN
Shape of the sampling distribution
Approaches
normal Why? of approach depends on sample
Rate size
Why?
Basic theorem
Central limit theorem 13 CENTRAL LIMIT THEOREM
Central Tendency
The mean of the Sampling Distribution of the mean is denoted as µ X
Dispersion
The Standard Deviation of the Sampling Distribution of the mean is called the Standard Error of the Mean and is denoted as σ X 14 CENTRAL LIMIT THEOREM
Standard Error of the Mean
We defined this manually in Chapter 8
And it can be calculated as: σX =
Shape σ n
The shape of the sampling distribution of the mean will be normal if the original population is normally distributed OR 15
if the sample size is “reasonably large.” 5 DEMONSTRATION
Let a population be very skewed
Draw samples of size 3 and calculate means
Draw samples of size 10 and calculate means
Plot means
Note changes in means, standard deviations, and shapes 16 PARENT POPULATION Skewed Population 3000 Frequency 2000 1000 Std. Dev = 2.43 Mean = 3.0 N = 10000.00 0 20 18 .0 .0 .0 .0 0 .0 0 16 0 .0 8. 14 6. 0 12 4. 10 2. 0.0 X 17 Sampling Distribution Sample size = n = 3 1000 Std. Dev = 1.40 Mean = 2.99 N = 10000.00 0 0 .0 13 0 .0 12 0 .0 11 0 .0 10 0 9.0 0 8.0 0 7.0 0 6.0 00 5. 0 4.0 0 3.0 0 2.0 0 1.0 0 0.0 Sampling Distribution Sample Mean Sample size = n = 10 1600 1400 1200 Frequency Frequency 2000 1000 800 600 400 Std. Dev = .77 200 Mean = 2.99 N = 10000.00 0 0 6.5 0 6.0 0 5.5 0 5.0 50 4. 0 4.0 0 3.5 00 3. 0 2.5 0 2.0 0 1.5 0 1.0 Sample Mean 6 DEMONSTRATION
Means have stayed at 3.00 throughout
Except for minor sampling error
Standard deviations have decreased appropriately
Shape has become more normal as we move from n = 3 to n = 10
See superimposed normal distribution 19 for reference TESTING HYPOTHESES: µ AND σ KNOWN Called a 1-sample Z-test
H0: µ = 100
H1: µ ≠ 100 (Two-tailed)
Calculate p (sample mean) = 106 if µ = 100
Use z from normal distribution
Sampling distribution would be normal 20 USING Z TO TEST H0 → 2-TAILEDα = .05
Calculate zX = z X −µ σX = ___ − ___ __ = = ___ __ __ __ z > + 1.96, reject H0 (Why 1.96?)
____ > 1.96
If
The difference is significant. 21 7 USING Z TO TEST H0 → 1-TAILEDα = .05
Calculate z (from last slide) z > + 1.64, reject H0 (Why 1.64?)
____ > 1.64
If
The difference is significant. 22 Z-TEST computed z to histogram of sampling distribution
The results should look consistent.
Logic of test
Compare Calculate probability of getting this mean if null true.
Reject if that probability is too small.
23 TESTING HYPOTHESES: µ KNOWN σ NOT KNOWN
Assume same example, but σ not known can make a guess at σ with s
But, unless we have a large sample, s is likely to underestimate σ (see next slide)
So, a test based on the normal distribution will lead to biased results (e.g. more Type 1 errors)
We 24 8 SAMPLING DISTRIBUTION OF THE VARIANCE 1400 1200 138.89 Frequency 1000 Let’s say you have a population variance = 138.89 800 If n = 5 and you take 10,000 samples 600 400 200 58.94% < 138.89 0 0 0. 0 0. 80 0 0. 75 0 0. 70 0 0. 65 0 0. 60 0 0. 55 0 0. 50 0 0. 45 0 0. 40 0 0. 35 0 0. 30 0 0. 25 0 0. 20 0 0. 15 0 0. 10 .0 50 Sample variance 25 TESTING HYPOTHESES: µ KNOWN σ NOT KNOWN s is the best estimate of σ;s X is the best estimate of σ X
Since Z does not work in this case we need a different distribution
Since
One that is based on s
Adjusts for the underestimation
And takes sample size (i.e. degrees of freedom) into account 26 THE T DISTRIBUTION
Symmetric, mean = median = mode = 0.
Asymptotic tails
Infinite family of t distributions, one for every possible df . For low df, the t distribution is more leptokurtic (e.g. spiked, thin, w/ fat tails)
For high df, the t distribution is more normal
With df = ∞, the t distribution and the z distribution are equivalent.
27 9 THE T DISTRIBUTION 28 DEGREES OF FREEDOM
Skewness of sampling distribution of variance decreases as n increases
t will differ from z less as sample size increases
Therefore need to adjust t accordingly
Degrees of Freedom: df = n - 1
t based on df 29 TESTING HYPOTHESES: µ KNOWN σ NOT KNOWN Called a 1-sample t-test
H0: µ = 100
H1: µ ≠ 100 (Two-tailed)
Calculate p (sample mean) = 106 if µ = 100
Use t-table to look up critical value using degrees of freedom
Compare tobserved to tcritical and make decision 30 10 USING T TO TEST H0 → 2-TAILEDα = .05
Same as z except for s in place of σ.
Let’s say for the 25, s = 7.78 tobserved = X − µ ___ − ___ __ = = = ___ ___ sX ___ __
With α = .05, df=24, 2-tailed tcritical = _____ (Table D.6; see next slide)
Since ____ > ____ reject H0 31 Psy 320 - Cal State Northridge df 1-tailed 2-tailed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 0.1 0.2 3.078 1.886 1.683 1.533 1.476 1.44 1.415 1.397 1.383 1.372 1.363 1.356 1.35 1.345 1.341 1.337 1.333 1.33 1.328 1.325 1.323 1.321 1.319 1.318 1.316 Critical Values of Student's t 0.01 0.005 0.0005 0.05 0.025 0.02 0.01 0.001 0.1 0.05 6.314 12.706 31.821 63.657 636.619 2.92 4.303 6.965 9.925 31.598 2.353 3.182 4.5415 5.841 12.941 8.61 2.132 2.776 3.747 4.604 6.859 2.015 2.571 3.365 4.032 5.959 1.943 2.447 3.143 3.707 5.405 1.895 2.365 2.998 3.499 5.041 1.86 2.306 2.896 3.355 1.833 2.262 2.821 3.25 4.781 1.812 2.228 2.764 3.169 4.587 4.437 1.796 2.201 2.718 3.106 4.318 1.782 2.179 2.681 3.055 2.65 3.012 4.221 1.771 2.16 4.14 1.761 2.145 2.624 2.977 4.073 1.753 2.131 2.602 2.947 2.583 2.921 4.015 1.746 2.12 2.567 2.898 3.965 1.74 2.11 3.922 1.734 2.101 2.552 2.878 4.883 1.729 2.093 2.539 2.861 3.85 1.725 2.086 2.528 2.845 2.518 2.831 3.819 1.721 2.08 3.792 1.717 2.074 2.508 2.819 2.5 2.807 3.767 1.714 2.069 3.745 1.711 2.064 2.492 2.797 2.485 2.787 3.725 1.708 2.06 T DISTRIBUTION 32 USING T TO TEST H0 → 1-TAILEDα = .05
H0: µ ≤ 100
H1: µ > 100 (One-tailed)
The tobserved value is the same → _____
With α = .05, df=24, 1-tailed tcritical = ____ (Table D.6; see next slide)
Since _____ > _____ reject H0 33 11 df 1-tailed 2-tailed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 0.1 0.2 3.078 1.886 1.683 1.533 1.476 1.44 1.415 1.397 1.383 1.372 1.363 1.356 1.35 1.345 1.341 1.337 1.333 1.33 1.328 1.325 1.323 1.321 1.319 1.318 1.316 Critical Values of Student's t 0.01 0.005 0.0005 0.05 0.025 0.02 0.01 0.001 0.1 0.05 6.314 12.706 31.821 63.657 636.619 2.92 4.303 6.965 9.925 31.598 2.353 3.182 4.5415 5.841 12.941 8.61 2.132 2.776 3.747 4.604 6.859 2.015 2.571 3.365 4.032 5.959 1.943 2.447 3.143 3.707 5.405 1.895 2.365 2.998 3.499 5.041 1.86 2.306 2.896 3.355 1.833 2.262 2.821 3.25 4.781 1.812 2.228 2.764 3.169 4.587 4.437 1.796 2.201 2.718 3.106 4.318 1.782 2.179 2.681 3.055 2.65 3.012 4.221 1.771 2.16 4.14 1.761 2.145 2.624 2.977 4.073 1.753 2.131 2.602 2.947 2.583 2.921 4.015 1.746 2.12 2.567 2.898 3.965 1.74 2.11 3.922 1.734 2.101 2.552 2.878 4.883 1.729 2.093 2.539 2.861 3.85 1.725 2.086 2.528 2.845 2.518 2.831 3.819 1.721 2.08 3.792 1.717 2.074 2.508 2.819 2.5 2.807 3.767 1.714 2.069 3.745 1.711 2.064 2.492 2.797 2.485 2.787 3.725 1.708 2.06 T DISTRIBUTION 34 CONCLUSIONS
With n = 25, tobserved(24) = _____
Because _____ is larger than both _____ (1-tailed) and _____ (2tailed) we reject H0 under both 1and 2-tailed hypotheses
Conclude that taking IQPLUS leads to a higher IQ than normal. 35 FACTORS AFFECTING…
t test between sample and population means
Magnitude of sample variance
Sample size
Difference
Decision
Significance level α
One-tailed versus two-tailed test 36 12 SIZE OF THE EFFECT
We know that the difference is significant.
That doesn’t mean that it is important.
Population mean = 100, Sample mean = 106
Difference is 6 words or roughly a 6% increase.
Is this large? 37 EFFECT SIZE
Later we develop this more in terms of standard deviations.
For Example:
In our sample s = 7.78 Effect size =
over X − µ 106 − 100 6 = = = .77 s 7.78 7.78 3/4 of a standard deviation 38 CONFIDENCE INTERVALS ON MEAN
Sample mean is a point estimate
We want interval estimate
Given the sample mean we can calculate an interval that has a probability of containing the population mean
This can be done if σ is known or not 39 13 CONFIDENCE INTERVALS ON MEAN
If σ is known than the 95% CI is CI.95 = X ± 1.96(σ X )
If σ is not known than the 95% CI is CI.95 = X ± ( t(2−tailed ,α =.05) * s X ) 40 FOR OUR DATA ASSUMING σ KNOWN CI = X ± (1.96 * σ X ) .95 = 106 ± (1.96* ___) = 106 ± ____ = ____ ≤ µ ≤ ____ 41 FOR OUR DATA ASSUMING σ NOT KNOWN = X ±  t CI * s X  .95 (2 − tailed , α = .05)   = 106 ± (____* ____) = 106 ± ____ = ____ ≤ µ ≤ ____ 42 14 CONFIDENCE INTERVAL
Neither interval includes 100 - the population mean of IQ
Consistent with result of t test.
Confidence interval and effect size tell us about the magnitude of the effect.
What else can we conclude from confidence interval? 43 15