PYB110 Exam Notes PDF

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2019 PYB110 Exam Notes...

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Hypothesis Testing •

Used when the sample is N = 1

1.0 Steps 1) Create research hypothesis and null hypothesis RH (H1 or H^) à µ1 > µ2 NH (H0) à µ1 = µ2 2) Determine comparison distribution •

Assume the NH is true and assume the population parameters Ø Therefore, assume µ1 = µ2

3) Determine cut-off scores on the comparison distribution where the NH would be rejected •

Most cut-off values are 0.05 (5%), unless the question states otherwise

4) Calculate the Z score for your sample •

Substitute values into its respective positions in the Z formula Z=

("#$%) '(

5) Decide whether the NH should be rejected •

Does the calculated Z score exceed the cut-off value? If so, reject the H0

2.0 One-Tailed vs Two-Tailed Tests • •

Cut-off at 0.05 à Z = 1.645 Cut-off at 0.025 à Z = ±1.96 Ø If the question poses a directional hypothesis, use Z = 1.645 Ø If the question poses a non-directional hypothesis, use Z = ±1.96

3.0 Type I and Type II Error • • • •

Type I error à rejecting the H0 and accepting H1 when the H0 is actually true Ø Probability of making a Type I error is 0.05 (alpha) Type II error à accepting the H0 and rejecting H1 when H0 is actually false Ø Probability of making a Type II error is beta The ability to correctly reject the truly false hypothesis increases power Researchers like to have 0.8 (80%) of power

4.0 Example A randomly selected individual, after going through an experimental treatment, has a score of 27 on a particular measure. The scores of people in general on this measure are normally distributed with a mean of 19 and a standard deviation of 4. The researcher predicts an effect, but does not predict a particular direction of effect. Using the 5% significance level, what should you conclude? 1) Create research hypothesis and null hypothesis H1 à People who undergo the experimental procedure H0 à People in general 2) Identify comparison distribution •

Assuming H0 = µ1 = µ2, the mean of the comparative distribution is 19 whereas the standard deviation is 4

3) Determine cut-off •

The question states that a significance level of 5% is to be used. The hypothesis is nondirectional, so the cut-off value is equivalent to a Z score of ±1.96 (with 2.5% on either end of the distribution)

4) Calculate the Z score of the sample Z= =

("#$%) '( ()*#+,) -

=2 5) Consider whether the H0 should be rejected •

The patient received a Z score of +2, which exceeds the cut-off value of Z = +1.96. Therefore, the null hypothesis should be rejected and the patient’s result is significant.

Distribution of Means •

The comparison distribution consists of means, not single scores

1.0. Steps 1) Create research hypothesis and null hypothesis RH (H1 or H^) à µ1 > µ2 NH (H0) à µ1 = µ2 2) Identify comparison distribution 2a) Assume that the mean of a distribution of means is the same as the mean of the population µM = µ 2b) The variance of the population of means is the variance of the population divided by the number of individuals 𝜎2M =

/) 0

2c) The standard deviation of a distribution of means is the square root of the variance of the distribution of means 𝜎M =

/ Ö0

3) Determine cut-off scores on the comparison distribution where the NH would be rejected • •

Cut-off at 0.05 à Z = 1.645 Cut-off at 0.025 à Z = ±1.96 Ø If the question poses a directional hypothesis, use Z = 1.645 Ø If the question poses a non-directional hypothesis, use Z = ±1.96

•

If the question poses another cut-off percentage, use the Z-table to determine what the cut-off Z value is Ø If it is a directional hypothesis à Determine the Z value Ø If it is a non-directional hypothesis à Divide the cut-off value by 2 to get a cut-off on either side

4) Calculate the Z score of the sample (use the 𝜎 formula prior to calculate SD) Z=

("#$µ) 1

5) Decide whether the NH should be rejected •

Does the calculated Z score exceed the cut-off value? If so, reject the H0

2.0 Example A researcher wants to test the effect of a communication skills workshop on students’ use of verbal fillers during a presentation. Verbal fillers are words such as “um,” “uh,” and “you know” that people commonly use in conversations and when giving presentations. The researcher conducts a study in which 25 students attend communication skills workshops and then give a half-hour presentation on a topic of their choice. The presentations are recorded and the researcher later counts the number of verbal fillers each student used during his or her presentation. It is known from previous studies that students typically use a mean of 53 verbal fillers during a half-hour presentation of this kind, with a standard deviation of 7, and the distribution of verbal fillers follows a normal curve. The 25 students who took a communication skills workshop used a mean of 48 verbal fillers. The researcher wants to carry out the Z test using the 1% significance level, and an effect in either direction would be important (that is, the researcher is interested in whether communications workshops could increase or decrease the use of verbal fillers). 1) Create research hypothesis and null hypothesis H1 à People in the workshop H0 à People in general 2) Identify comparison distribution 2a) Assume that the mean of a distribution of means is the same as the mean of the population 53 (M) = 53 𝜎M 2b) The variance of the population of means is the variance of the population divided by the number of individuals 𝜎 2M = 𝜎

2

M=

/)

0 *$2$* )3

= 1.96 2c) The standard deviation of a distribution of means is the square root of the variance of the distribution of means 𝜎M= 𝜎M=

/ Ö0 * Ö)3

= 1.4

3) Determine cut-off scores on the comparison distribution where the NH would be rejected • • •

The researcher wants the cut-off value to be at 1% It is a non-directional hypothesis so the cut-off is divided into 0.5% The Z value for 0.5% equates to a cut-off value of ±2.58

4) Calculate the Z score of the sample Z=

(-4#$35) +.-

= -3.57 5) Decide whether the NH should be rejected •

The Z score of -3.57 exceeds the cut-off value of -2.58, therefore the null hypothesis is testing

T-Tests for Single Samples •

A t-test is used instead of A Z-test when the population SD is unknown Ø A question that gives the SD à Use a Z-test Ø A question that does not give you the SD but gives you the SS à Use a T-test

1.0 Steps 1) Determine the populations and the H1 and H0 2) Determine the characteristics of the comparative distribution 2a) Estimate the variance by calculating the sum of squares S2 = =

∑("#%)8 0#+$ 99 :;

2b) Estimate the variance of the comparative distribution using the answer from above 𝑆)M =

S) N

2c) Estimate the standard deviation with the answer from above $$$$$$$$$$$SA = BS ) 𝑀 3) Determine cut-off scores on the comparison distribution where the NH would be rejected • •

If population 1 > population 2 = +ve cut-off If population 1 < population 2 = -ve cut-off

4) Substitute values into t-score formula t=

%#$D 'E

5) Decide whether the NH should be rejected •

Does the calculated t-score exceed the cut-off value? If so, reject the H0

2.0 Example Suppose a researcher was studying the psychological effects of a devastating flood in a small rural community. Specifically, the researcher was interested in how hopeful (versus unhopeful) people felt after the flood. The researcher randomly selected 10 people from this community to complete a short questionnaire. The key item on the questionnaire asked how hopeful they felt, using a 7-point scale from extremely unhopeful (1) to neutral (4) to extremely hopeful (7). The researcher wanted to know

whether the ratings of hopefulness for people who had been through the flood would be consistently above or below the neutral point on the scale (4). The sum of squares is 32.10 and the mean is 4.7. 1) Create research hypothesis and null hypothesis Population 1 à People in the flood Population 2 à People who were neither hopeful nor unhopeful 2) Identify comparison distribution •

The mean for both populations is 4

2a) Estimate the variance by calculating the sum of squares S2 =

∑("#%)8 0#+$

$$$$= $$

32.10 10 − 1$

= 3.57 2b) Estimate the variance of the comparative distribution using the answer from above 𝑆)M =

=

S) N

5.3* +K

= 0.36 2c) Estimate the standard deviation with the answer from above

$$$$$$$$$$$SA = B S ) 𝑀 = √0.36 = 0.6

3) Determine cut-off scores on the comparison distribution where the NH would be rejected • • • Ø

The researcher wants to do a 0.1 significance level The hypothesis is non-directional so a two-tailed t-test will be used The df is 9 Therefore, the cut-off value is 3.250

4) Substitute values into t-score formula t= =

%#$D 'E -.*K#$K.N

= 1.17 5) Decide whether the NH should be rejected •

t = 1.17 which does not exceed the cut-off value of ±3.250, therefore the null hypothesis shall be retained

T-Test for Dependent Means • •

Also known as a repeated measures test Used when comparing the results from before and after of a sample

1.0 Steps 1) Determine the populations and the H1 and H0 2) Determine the characteristics of the comparative distribution •

If the null hypothesis is true, the mean of the population of difference score is 0

2a) Calculate the difference score After – before = difference score 2b) Calculate the mean of the difference scores M=

∑" 0

2c) Calculate the deviation score Difference score – mean = deviation 2d) Calculate the squared deviation and add all values to calculate the SS Deviation2 2e) Estimate the variance by calculating the sum of squares S2 = =

∑("#%)8 0#+$ 99 :;

2f) Estimate the variance of the comparative distribution using the answer from above 𝑆)M =

S) N

2g) Estimate the standard deviation with the answer from above $$$$$$$$$$$SA = B S ) 𝑀

3) Determine cut-off scores on the comparison distribution where the NH would be rejected • •

Directional à one-tailed test Non-direction à two-tailed test

4) Substitute values into t-score formula t=

%#$D 'E

5) Decide whether the NH should be rejected

2.0 Example A team of researchers examined the brain systems involved in human romantic love. One issue was whether romantic love engages a part of the brain called the ventral tegmental areas (VTA for short; a brain area that is engaged when people win money, are given cocaine, and other such ‘rewards’). Thus, the researchers recruited 10 participants who had very recently fallen ‘madly in love’. (For example, to be in the study participants had to think about their partner at least 80% of their waking hours.) Participants brought a picture of their beloved with them, plus a picture of a familiar, neutral person of the same age and sex as their beloved. Participants then went in to the functional magnetic resonance imaging (fMRI) machine and their brain was scanned while they looked at the two pictures—30 seconds at the neutral person’s picture, 30 seconds at their beloved, 30 seconds at the neutral person, and so forth.

1) Determine the populations and the H1 and H0 Population 1 à Individuals like those in the study Population 2 à Individuals whose brain activation in the VTA area of interest is the same when looking at a picture of their beloved and the neutral person

2) Determine the characteristics of the comparative distribution •

If the null hypothesis is true, 𝜇 = 0

2a) Calculate the difference score *Shown in table above 2b) Calculate the mean of the difference scores *Shown in table above 2c) Calculate the deviation score *Shown in table above 2d) Calculate the squared deviation and add all values to calculate the SS *Shown in table above SS = 5.660 2e) Estimate the variance by calculating the sum of squares S2 = =

3.NNK +K#+$ 3.NNK ,

= 0.629

2f) Estimate the variance of the comparative distribution using the answer from above $$$𝑆 ) M =

0.629 10

= 0.063 2g) Estimate the standard deviation with the answer from above $$$$$$$$$SA = √0.063 = 0.251

3) Determine cut-off scores on the comparison distribution where the NH would be rejected • • •

The hypothesis is directional so a one-tailed test will be used A significance level of 0.05 will be used Cut-off value = 1.833

4) Substitute values into t-score formula t= = t=

%#$D 'E %QRS$TU$VWUUQXQSYQ$ZYTXQ#$D 'E +.-#$K K.)3+

= 5.58 5) Decide whether the NH should be rejected •

t = 5.58 which exceeds the cut-off value of ±0.251, therefore the null hypothesis shall be rejected

T-Test for Independent Means •

Used when comparing two means from two different samples

1.0 Steps 1) Determine the populations and the H1 and H0 2) Determine the characteristics of the comparative distribution •

If the null hypothesis is true, the mean of the population of difference score is 0

2a) Estimate the variance by calculating the sum of squares First sample: 2

S = =

Second sample:

∑("#%)8 0#+$ 99

S2 = =

:;

∑("#%)8 0#+$ 99 :;

2b) Calculate the pooled estimate of the population variance • • •

In theory, researchers assume the homogeneity of variance In practice, two separate estimates of the variance will be calculated (as per the formulas above) Therefore, the two estimates need to be averaged to get a single overall estimate – this is called the pooled estimate (S2Pooled) S2Pooled =

:;$[; $\]^_`a$+ :;$b[b]`$

(s21) +

:;[;$\]^_`a$) :;$b[b]`$

(s22)

NOTE: dftotal = (df1) + (df2)

2c) Calculate the variance of each for two distributions of means using the pooled estimate from above

First sample: S2M1 =

Second sample:

9 8c[[`a: d$ [;$\]^_`a $+

S2M2 =

9 8c[[`a: d$ ;e[^$\]^_`a $)

2d) Calculate the variance of the distribution of differences between means using the answers from above – S2Difference S2Difference = √S2M1 + S2M2

3) Substitute values into t-score formula t =9

A+#$A) fghhijikli

4) Determine cut-off scores on the comparison distribution where the NH would be rejected 5) Decide whether the NH should be rejected

2.0 Example Twenty students were recruited to take part in the study. The 10 students randomly assigned to the expressive writing group wrote about their thoughts and feelings associated with their most traumatic life events. The 10 students randomly assigned to the control group wrote about their plans for the day. One month later, all of the students rated their overall level of physical health on a scale from 0 = very poor health to 100 = perfect health. The mean from the expressive group was 79 and the squared deviation was 850 whereas the mean from the control group was 68 and the squared deviation was 1002

1) Determine the populations and the H1 and H0 Population 1 à People from the expressive group Population 2 à People from the neutral group 2) Determine the characteristics of the comparative distribution •

If the null hypothesis is true, the mean of the population of difference score is 0

2a) Estimate the variance by calculating the sum of squares Expressive group: 2

S = S2 = =

Neutral group:

∑("#%)8 0#+$ 43K

S2 = S2 =

+K#+$ 43K

=

,

= 94.44

∑("#%)8 0#+$ +KK) +K#+$ +KK) ,

= 111.33

2b) Calculate the pooled estimate of the population variance S2Pooled =

, +4$

(94.44) +

, +4

(111.33)

= 102.89 dftotal = (10-1) + (10-1) = 18

2c) Calculate the variance of each for two distributions of means using the pooled estimate from above

Expressive group: 2

S

M1 =

Neutral group:

+K).4,

S2M2 =

+K

= 10.29

+K).4, +K

= 10.29

2d) Calculate the variance of the distribution of differences between means using the answers from above – S2Difference S2Difference = √10.29 + 10.29 = 4.54 3) Substitute values into t-score formula t=9 t=

A+#$A) fghhijikli

*,#N4 -.3-

= 2.42

4) Determine cut-off scores on the comparison distribution where the NH would be rejected • • • •

The researcher is interested in an effect in either direction, so a two-tailed test will be used df = dftotal (18 in this instance) The usual cut-off value is 0.05 Therefore, the critical value is ±2.101

5) Decide whether the NH should be rejected • The value of 2.42 exceeds the value of +2.101, so the null hypothesis is rejected

Eta-Squared (Dependent and Independent Means)

• • • • Ø

Interpret n2 like r2 n2 cannot be >1 Greater n2 = greater power Ratio of variance in an outcome variable that is explained by a predictor variable How much that variable X explains variable Y

1.0 Steps 1) Calculate t-score and df • Note which test you are administering as the df will be different Ø Dependent means à df = N – 1 Ø Independent means à dftotal = (N1 – 1) + (N2 – 1)

2) Substitute the values into the eta formula b8

n2 = b 8n:;

Chi-Square Tests • • •

Used when the variables are nominal variables Basic idea is that chi-tests compare how well an observed breakdown of people over various categories fits some expected breakdown df = Ncategories – 1

1.0 Logic for Goodness of Fit Chi-Tests • •

One variable Examines how well an observed frequency distribution of a nominal variable fits some expected pattern of frequencies

2.0 Steps for Goodness of Fit 1) Determine the actual, observed frequency for each category 2) Determine the expected frequency for each category E=

∑" 0

3) Minus the expected frequency from the observed frequency O–E 4) Square each difference (O – E)2 5) Divide each squared difference by the expected frequency for each category (O − E)) E 6) Add up each squared difference ∑ 7) Determine the cut-off value

•

Usually a 0.05 significance level

8) Decide whether the NH should be rejected

(O − E)) E

4.0 Logic for Independent Chi-Tests • • •

Two variables Examines whether the distribution of frequencies over the categories of one nominal variable is unrelated to the distribution of frequencies over the categories of a second nominal variable df = (NCATEGORIES – 1) ´ (NROWS – 1)

5.0 Steps for Independence 1) Determine the actual, observed frequency for each category 2) Determine the expected frequency for each category E=

q$×$s 0

3) Minus the expected frequency from the observed frequency O–E 4) Square each difference (O – E)2 5) Divide each squared difference by the expected frequency for each category (O −...