hypothesis testing comparing one sample to its population

Hypothesis Testing

Comparing One Sample to its Population

Hypothesis Testing w/ One Sample

If the population mean (μ) and standard deviation (σ) are known: Testing if our sample mean ( ) is significantly

different from our sampling distribution of the mean

Similar to testing if how different an individual score is from other scores in the sample What is this test called?

X


z-score formula [for an individual score (x)] =

z-score formula [for means ( )] =

X

z

X

N

Xz


Testing score versus standard deviation for an distribution of scores

Testing mean versus standard deviation for distribution of sample means I.e. standard error

X

z

N

Xz


Two implications of this formula: 1. Because we are dividing by N (actually √N),

with the same data (same sample & population mean and σ), but larger sample size, our p-value will be smaller (i.e. more likely to be significant) All statistical tests that produce p-values will be

sensitive to sample size – i.e. with enough people anything is significant at p < .05


Two implications of this formula: 2. If you recall, this formula was derived from the

formula for the normal distribution This means that your data must be normally

distributed to use this test validly However, this test is robust to violations of this

assumption – i.e. you can violate it, if you have (a) a large enough sample or (b) your population data is normally distributed

Why?


The Central Limit Theorem:

Given a population with mean μ and variance σ2, the sampling distribution of the mean (the distribution of sample means) will have a mean equal to μ (i.e., μ = μ) and a variance (σ2) equal to σ2/N (and standard deviation, σ = σ/√N). The distribution will approach the normal distribution as N, the sample size, increases.

X

XX


Example #1: You want to test the hypothesis that the current

crop of Kent State freshman are more depressed than Kent State undergraduates in general. What is your sample and what is your population? What is your Ho and your H1? Are you using a one- or two-tailed test? Assuming that for current Kent State freshman, their

mean depression score is 15, while the mean for all previous Kent State undergrads (N = 100,000) is 10, and their standard deviation is 5


= 5/.0158 = 316.46

Look up value in Table E10 with value in “Smaller Portion”

p < 0.0000 Since this is less than .05 (or .025 if we were using

a two-tailed test), we could conclude that the current batch of freshman is significantly more depressed than previous undergrads Also notice the effect that our large N had on our p-value

000,100

51015

z


Example #2: You want to test the hypothesis that a treatment conducted

at the ward you are working on in a hospital is more beneficial than the average treatment used in other wards of the hospital Get into groups of 2 or more State Ho and H1

State if you’re using a one- or two-tailed test and why. Given that the mean wellness score of the people on your

ward is 87, and that the mean for the entire hospital (N = 20) is 76 and the standard deviation is 15, is your ward significantly better? State the p-value that supports your claim.


Example #2:

= 11/3.36 = 3.27

Smaller Portion = p < .0006

20

157687

z


Most often, however, we don’t know the μ and σ, because this is what we’re trying to estimate with our sample in the first place

The formula for the t statistic accomplishes this by substituting s2 for σ2 in the formula for the z statistic

Because of this substitution, we have a different statistic, which requires that we use a different table than E.10

Don’t worry too much about why it’s different (you can skip pg. 274 in your book if you want, you won’t be tested on it)


Testing mean versus standard deviation for distribution of sample means I.e. standard error

Testing mean versus standard deviation for sample

N

sX

t

N

Xz


After computing our t statistic, we need to compare it with the t-table (called the Student’s T-Table; E.6, pg. 519)

First, we will need to become familiar with the concept of degrees of freedom or df df = N – 1 This represents the number of individual subjects

data points that are free to vary, if you know the mean or s already


For example: If we already know that a particular set of data has a mean

of 5, and 10 scores in total (n = 10) Once we have nine of those scores, we can calculate the

tenth, however, if we have eight scores we do not know what the other two scores could be We can solve x + 5 = 10, but not x + y = 10, because

in the latter we have more than one unknown (x and y) x and y could be 5 and 5, 8 and 2, 4 and 6, 7 and 3, etc.

Therefore, nine scores are free to vary, then the tenth is fixed


Two-Tailed Significance Level

df .10 .05 .02 .0110 1.812 2.228 2.764 3.16915 1.753 2.131 2.602 2.94720 1.725 2.086 2.528 2.84525 1.708 2.060 2.485 2.78730 1.697 2.042 2.457 2.750

100 1.660 1.984 2.364 2.626


df: we know = N – 1 Two-Tailed vs. One-Tailed Tests and Significance

Level (α): we determine in advance @ p = .05 for a Two-Tailed Test (with 10 df), the

Critical Value of T, or the value of t equal to a “Smaller Value” of 2.5% (if we were using the z-table), is 2.228 Once we calculate our t-score, we compare it with our

rejection region and our critical t to determine if we reject H0 or fail to reject it


p = .025; z = 1.96; t = 2.228 Our rejection region is the area in red – it is all in one tail because we’re using a

one-tailed test (in this example), and is in the lower half of the distribution because we’re interested in scores below the population mean (again, in this example)

If we were interested in scores above the population mean, we would move our rejection region to the other tail


Factors that influence the z and t statistics: The difference between the sample mean and population

mean – greater differences = greater t and z values The magnitude of s (or s2) – since we’re dividing by s,

smaller values of s result in larger values of t or z [i.e. we want to decrease variability in our sample (error)]

The sample size – the bigger the bigger t and z The significance level (α) – the smaller the α, the higher the

critical t to reject Ho – although raising α also raises our Type I Error, so we probably won’t want to do this without good reason

Whether the test is one- or two-tailed – two-tailed tests split α into two tails of p< .025, instead of one tail at p < .05


Example #1: You’ve administered a therapy for people with anorexia that

will supposedly assist them in gaining weight. The following data are amount of weight gained in pounds over your 16 session therapy for 29 participants. Does this represent a significantly increased degree of weight gain compared to the average weight gained without treatment (-.45 lbs.)?

What are Ho and H1? Will you be using a one- or two-tailed test? Why? Based on this, what is your df? What is your

critical t?


Example #1:

1.7 -9.1

.7 2.1

-.1 -1.4

-.7 1.4

-3.5 -.3

14.9 -3.7

3.5 -.8

17.1 2.4

-7.6 12.6

1.6 1.9

11.7 3.9

6.1 .1

1.1 15.4

-4.0 -.7

20.9


Example #1: Critical t (df = 28; two-tailed test at .05) = 2.048

Sample Mean = 87.2/29 = 3.0069

s2 = (1757.8 – [(87.2)2/29])/ 28 = 53.41

s = 7.3085

t = (3.0069 - -.45)/(7.3085/√29) = 2.5472, p < .05 t > Critical t and in our rejection region, which is above the

population mean (since we’re only interested in people gaining weight), therefore we reject Ho and conclude that our treatment is more effective than no treatment at all


Example #2: A family therapist states that parents talk to their

teenagers an average of 27 minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time parents spend in conversation with their teenage children. For the n = 12 parents, the study revealed that following times (in minutes) devoted to conversation in a week:


Example #2: Do the psychologists findings differ significantly from the

therapist’s claim? If so, is the family expert’s claim an overestimate or underestimate of actual time spent talking to children? Use the .05 level of significance with two tails.

29 22 19 25 27 28

21 22 24 26 30 22


Example #2: H0 = μ = 27 Critical t (df = 11) = ±2.201 Mean = 24.58, s2 = 12.24, (s/√n) = 1.00, and t(11)

= -2.41 Reject H0, and conclude that the data are

significantly different (less time) from the therapist’s claim


Often, if we’re reporting the results of our experiments to the public, or the results of an assessment (psychological or otherwise) to a client, we want to emphasize to them that our measurements are made with error, or that our samples include sampling error We can do this by including intervals around the

scores we report, indicating that the “true” score measured without error lies in this interval


This is what is known as a Confidence Interval In keeping with the p < .05 tradition, we are often

looking for the 95% confidence interval, or the scores that 95% of our distribution lie, but we can do this for any interval

They are calculated just like for z-scores, where we plug the t values into the formula and work backwards


For a 95% CI: Given your df (we’ll assume df = 9 for this

example) and type of test (assume a two-tailed test for now), look up your critical values of t from Table E.6 (t = ± 2.262)

Plug into your formula with your Sample Mean and s (which we’ll assume are 1.463 and .341, respectively), and solve for μ


For a 95% CI:± 2.262 = (1.463 – μ)/(.341/√10)

μ = ±2.262(.108) + 1.463 = ±.244 + 1.462

.244 + 1.463 = 1.707

-.244 + 1.463 = 1.219 CI.95 = 1.219 ≤ μ ≤ 1.707

hypothesis testing comparing one sample to its population

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