“APA” is an abbreviation for The American Psychological Association. Regardless of the type of research that is being conducted, the formatting standards maintained by the APA as it applies to statistical research, should always be utilized when presenting data in a professional manner.

__Details__All figures which contain decimal values should be rounded to the nearest hundredth. Ex. .105 = .11. Reporting p-values being the exception to this rule. P-values should, in most cases, be reported in a format which contains two decimals. The exception occurring when a greater amount of specificity is required to illustrate the details of the findings.

Another rule to keep in mind pertains to leading zeroes. A leading zero prior to a decimal place is only required if the represented figure has the potential to exceed “1”. If the value cannot exceed “1”, then a leading zero is un-necessary.

Below are examples which demonstrate the most common application of the APA format.

**Chi-Square**

**Template:**

A chi-square test of independence was performed to examine the relation between

**CATEGORY**and

**OUTCOME**. The relation between these variables was found to be significant at the p < .05 level, χ2 (

**DEGREES OF FREEDOM**, N =

**SAMPLE SIZE**) =

**X-Squared Value**, p =

**p - value**.

**- OR -**

**A chi-square test of independence was performed to examine the relation between**

**CATEGORY**and

**OUTCOME**. The relation between these variables was not found to be significant at the p < .05 level, χ2 (

**DEGREES OF FREEDOM**, N =

**SAMPLE SIZE**) =

**X-Squared Value**, p =

**p - value**.

**Example:**

While working as a statistician at a local university, you are tasked to evaluate, based on survey data, the level of job satisfaction that each member of the staff currently has for their occupational role (Assume a 95% Confidence Interval).

The data that you gather from the surveys is as follows:

130 Satisfied 20 Unsatisfied

30 Satisfied 20 Unsatisfied

80 Satisfied 20 Unsatisfied

20 Satisfied 10 Unsatisfied

A chi-square test of independence was performed to examine the relation between occupational role and job satisfaction. The relation between these variables was found to be significant at the p < .05 level, χ2 (3, N = 330) = 18.56, p < .001.

Post hoc comparisons using the Tukey HSD test indicated that the mean score for the

__General Faculty__130 Satisfied 20 Unsatisfied

__Professors__30 Satisfied 20 Unsatisfied

__Adjunct Professors__80 Satisfied 20 Unsatisfied

__Custodians__20 Satisfied 10 Unsatisfied

**# Code #**

Model <- matrix(c(130, 30, 80, 20, 20, 20, 20, 10), nrow = 4, ncol=2)

N <- sum(130, 30, 80, 20, 20, 20, 20, 10)

chisq.test(Model)

N

# Console Output #Model <- matrix(c(130, 30, 80, 20, 20, 20, 20, 10), nrow = 4, ncol=2)

N <- sum(130, 30, 80, 20, 20, 20, 20, 10)

chisq.test(Model)

N

# Console Output #

*Pearson's Chi-squared test*

data: Model

X-squared = 18.857, df = 3, p-value = 0.0002926

> N

[1] 330data: Model

X-squared = 18.857, df = 3, p-value = 0.0002926

> N

[1] 330

**APA Format:**A chi-square test of independence was performed to examine the relation between occupational role and job satisfaction. The relation between these variables was found to be significant at the p < .05 level, χ2 (3, N = 330) = 18.56, p < .001.

__Tukey HSD__**Template:**Post hoc comparisons using the Tukey HSD test indicated that the mean score for the

**CONDITION A**(M =**Mean1**, SD =**Standard Deviation1**) was significantly different than**CONDITION B**(M =**Mean2,**SD = S**tandard Deviation2**), p =**p-value**.

__Analysis of Variance (ANOVA)__

(One Way)

Template:

(One Way)

Template:

There was a significant effect of the

**CATEGORY**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(1)**,

**Degrees of Freedom(2)**) =

**F Value**, p =

**p - value**).

**- OR -**

There was not a significant effect of the

**CATEGORY**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(1)**,

**Degrees of Freedom(2)**) =

**F Value**, p =

**p - value**).

**Example:**

A chef wants to test if patrons prefer a soup which he prepares based on salt content. He prepares a limited experiment in which he creates three types of soup: soup with a low amount of salt, soup with a high amount of salt, and soup with a medium amount of salt. He then servers this soup to his customers and asks them to rate their satisfaction on a scale from 1-8.

Low Salt Soup it rated: 4, 1, 8

Medium Salt Soup is rated: 4, 5, 3, 5

High Salt Soup is rated: 3, 2, 5

(Assume a 95% Confidence Interval)

# Code #

satisfaction <- c(4, 1, 8, 4, 5, 3, 5, 3, 2, 5)

salt <- c(rep("low",3), rep("med",4), rep("high",3))

salttest <- data.frame(satisfaction, salt)

results <- aov(satisfaction~salt, data=salttest)

summary(results)

# Console Output #

# Code #

satisfaction <- c(4, 1, 8, 4, 5, 3, 5, 3, 2, 5)

salt <- c(rep("low",3), rep("med",4), rep("high",3))

salttest <- data.frame(satisfaction, salt)

results <- aov(satisfaction~salt, data=salttest)

summary(results)

# Console Output #

Df Sum Sq Mean Sq F value Pr(>F)

salt

**2**1.92 0.958

**0.209 0.816**

Residuals

**7**32.08 4.583

**APA Format:**

**There not was a significant effect of the level of salt content on patron satisfaction at the p<.05 level for the three conditions (F(2, 7) = 0.21, p = 0.82).**

**(Two Way)**

Template:

Hypothesis 1:

Template:

Hypothesis 1:

There was a significant effect of the

**CATEGORY**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(1)**,

**Degrees of Freedom(2)**) =

**F Value**, p =

**p - value**).

**- OR -**

There was not a significant effect of the

**CATEGORY**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(1)**,

**Degrees of Freedom(2)**) =

**F Value**, p =

**p - value**).

**Hypothesis 2:**

There was a significant effect of the

**CATEGORY2**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(2)**,

**Degrees of Freedom(4)**) =

**F Value**, p =

**p - value**).

**- OR -**

There was not a significant effect of the

**CATEGORY2**on the

**OUTCOME**for

**SCENARIO**at the p <. 05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(2)**,

**Degrees of Freedom(4)**) =

**F Value**, p =

**p - value**).

**Hypothesis 3:**

There was a statistically significant interaction effect of the

**CATEGORY1**on the

**CATEGORY2**at the p < .05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(3)**,

**Degrees of Freedom(4)**) =

**F Value**, p =

**p - value**).

**- OR -**

There was not a statistically significant interaction effect of the

**CATEGORY1**on the

**CATEGORY2**at the p < .05 level for the

**NUMBER OF CONDITIONS**(F(

**Degrees of Freedom(3)**,

**Degrees of Freedom(4)**) =

**F Value**, p =

**p - value**).

**Example:**

Researchers want to test study habits within two schools as they pertain to student life satisfaction. The researchers also believe that the school that each group of students is attending may also have an impact on study habits. Students from each school are assigned study material which in sum, totals to 1 hour, 2 hours, and 3 hours on a daily basis. Measured is the satisfaction of each student group on a scale from 1-10 after a 1 month duration.

(Assume a 95% Confidence Interval)

School A:

1 Hour of Study Time: 7, 2, 10, 2, 2

2 Hours of Study Time: 9, 10, 3, 10, 8

3 Hours of Study Time: 3, 6, 4, 7, 1

School B:

1 Hour of Study Time: 8, 5, 1, 3, 10

2 Hours of Study Time: 7, 5, 6, 4, 10

3 Hours of Study Time: 5, 5, 2, 2, 2

satisfaction <- c(7, 2, 10, 2, 2, 8, 5, 1, 3, 10, 9, 10, 3, 10, 8, 7, 5, 6, 4, 10, 3, 6, 4, 7, 1, 5, 5, 2, 2, 2)

studytime <- c(rep("One Hour",10), rep("Two Hours",10), rep("Three Hours",10))

school = c(rep("SchoolA",5), rep("SchoolB",5), rep("SchoolA",5), rep("SchoolB",5), rep("SchoolA",5), rep("SchoolB",5))

schooltest <- data.frame(satisfaction, studytime, school)

results <- aov(lm(satisfaction ~ studytime * school, data=schooltest))

summary(results)

satisfaction <- c(7, 2, 10, 2, 2, 8, 5, 1, 3, 10, 9, 10, 3, 10, 8, 7, 5, 6, 4, 10, 3, 6, 4, 7, 1, 5, 5, 2, 2, 2)

studytime <- c(rep("One Hour",10), rep("Two Hours",10), rep("Three Hours",10))

school = c(rep("SchoolA",5), rep("SchoolB",5), rep("SchoolA",5), rep("SchoolB",5), rep("SchoolA",5), rep("SchoolB",5))

schooltest <- data.frame(satisfaction, studytime, school)

results <- aov(lm(satisfaction ~ studytime * school, data=schooltest))

summary(results)

Which produces the output:

*Df Sum Sq Mean Sq F value Pr(>F)*

studytime

school

studytime:school

Residuals

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

studytime

**2**62.6 31.300**3.809 0.0366 ***school

**1**2.7 2.700**0.329 0.5718**studytime:school

**2**7.8 3.900**0.475 0.6278**Residuals

**24**197.2 8.217---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

**APA Format:**

There was a significant effect as it pertains to study time impacting student stress levels at the p < .05 level for the three conditions (F(2, 24) = 3.81, p = .04).

There was not a significant effect as it relates to the school attended impacting student stress levels at the p < .05 level for the two conditions (F(1, 24) = 0.329, p > .05).

There was not a statistically significant interaction effect of the school variable on the study time variable at the p < .05 level (F(2, 24) = 0.475, p > .05).

**TukeyHSD(results)**

*> TukeyHSD(results)*

Tukey multiple comparisons of means

95% family-wise confidence level

Fit: aov(formula = lm(satisfaction ~ studytime * school, data = schooltest))

$studytime

diff lwr upr p adj

Three Hours-One Hour -1.3 -4.5013364 1.901336 0.5753377

Two Hours-One Hour 2.2 -1.0013364 5.401336 0.2198626

$school

diff lwr upr p adj

SchoolB-SchoolA -0.6 -2.760257 1.560257 0.571817

$`studytime:school`

diff lwr upr p adj

Three Hours:SchoolA-One Hour:SchoolA -0.4 -6.005413 5.2054132 0.9999178

Two Hours:SchoolA-One Hour:SchoolA 3.4 -2.205413 9.0054132 0.4401459

One Hour:SchoolB-One Hour:SchoolA 0.8 -4.805413 6.4054132 0.9976117

Three Hours:SchoolB-One Hour:SchoolA -1.4 -7.005413 4.2054132 0.9696463

Two Hours:SchoolB-One Hour:SchoolA 1.8 -3.805413 7.4054132 0.9157375

Two Hours:SchoolA-Three Hours:SchoolA 3.8 -1.805413 9.4054132 0.3223867

One Hour:SchoolB-Three Hours:SchoolA 1.2 -4.405413 6.8054132 0.9844928

Three Hours:SchoolB-Three Hours:SchoolA -1.0 -6.605413 4.6054132 0.9932117

Two Hours:SchoolB-Three Hours:SchoolA 2.2 -3.405413 7.8054132 0.8260605

One Hour:SchoolB-Two Hours:SchoolA -2.6 -8.205413 3.0054132 0.7067715

Three Hours:SchoolB-Two Hours:SchoolA -4.8 -10.405413 0.8054132 0.1240592

Two Hours:SchoolB-Two Hours:SchoolA -1.6 -7.205413 4.0054132 0.9470847

Three Hours:SchoolB-One Hour:SchoolB -2.2 -7.805413 3.4054132 0.8260605

Two Hours:SchoolB-One Hour:SchoolB 1.0 -4.605413 6.6054132 0.9932117

Two Hours:SchoolB-Three Hours:SchoolB 3.2 -2.405413 8.8054132 0.5052080

Tukey multiple comparisons of means

95% family-wise confidence level

Fit: aov(formula = lm(satisfaction ~ studytime * school, data = schooltest))

$studytime

diff lwr upr p adj

Three Hours-One Hour -1.3 -4.5013364 1.901336 0.5753377

Two Hours-One Hour 2.2 -1.0013364 5.401336 0.2198626

**Two Hours-Three Hours 3.5 0.2986636 6.701336 0.0302463**$school

diff lwr upr p adj

SchoolB-SchoolA -0.6 -2.760257 1.560257 0.571817

$`studytime:school`

diff lwr upr p adj

Three Hours:SchoolA-One Hour:SchoolA -0.4 -6.005413 5.2054132 0.9999178

Two Hours:SchoolA-One Hour:SchoolA 3.4 -2.205413 9.0054132 0.4401459

One Hour:SchoolB-One Hour:SchoolA 0.8 -4.805413 6.4054132 0.9976117

Three Hours:SchoolB-One Hour:SchoolA -1.4 -7.005413 4.2054132 0.9696463

Two Hours:SchoolB-One Hour:SchoolA 1.8 -3.805413 7.4054132 0.9157375

Two Hours:SchoolA-Three Hours:SchoolA 3.8 -1.805413 9.4054132 0.3223867

One Hour:SchoolB-Three Hours:SchoolA 1.2 -4.405413 6.8054132 0.9844928

Three Hours:SchoolB-Three Hours:SchoolA -1.0 -6.605413 4.6054132 0.9932117

Two Hours:SchoolB-Three Hours:SchoolA 2.2 -3.405413 7.8054132 0.8260605

One Hour:SchoolB-Two Hours:SchoolA -2.6 -8.205413 3.0054132 0.7067715

Three Hours:SchoolB-Two Hours:SchoolA -4.8 -10.405413 0.8054132 0.1240592

Two Hours:SchoolB-Two Hours:SchoolA -1.6 -7.205413 4.0054132 0.9470847

Three Hours:SchoolB-One Hour:SchoolB -2.2 -7.805413 3.4054132 0.8260605

Two Hours:SchoolB-One Hour:SchoolB 1.0 -4.605413 6.6054132 0.9932117

Two Hours:SchoolB-Three Hours:SchoolB 3.2 -2.405413 8.8054132 0.5052080

**twohours <- c(9, 10, 3, 10, 8, 7, 5, 6, 4, 10)**

threehours <- c(3, 6, 4, 7, 1, 5, 5, 2, 2, 2)

mean(twohours)

sd(twohours)

mean(threehours)

sd(threehours)

threehours <- c(3, 6, 4, 7, 1, 5, 5, 2, 2, 2)

mean(twohours)

sd(twohours)

mean(threehours)

sd(threehours)

*> mean(twohours)*

[1] 7.2

> sd(twohours)

[1] 2.616189

>

> mean(threehours)

[1] 3.7

> sd(threehours)

[1] 2.002776

[1] 7.2

> sd(twohours)

[1] 2.616189

>

> mean(threehours)

[1] 3.7

> sd(threehours)

[1] 2.002776

**APA Format:**

Post hoc comparisons using the Tukey HSD test indicated that at the p < .05 level, the mean score for the level of stress exhibited by students who studied for Two Hours (M = 7.20, SD = 2.62), was significantly different as compared to the scores of the students who studied for Three Hours (M = 3.70, SD = 2.00), p = .03.

**(Repeated Measures)**

Template:

Template:

**Example:**

Researchers want to test the impact of reading existential philosophy on a group of 8 individuals. They measure the happiness of the participants three times, once prior to reading, once after reading the materials for one week, and once after reading the materials for two weeks. We will assume an alpha of .05.

Before Reading = 1, 8, 2, 4, 4, 10, 2, 9

After Reading = 4, 2, 5, 4, 3, 4, 2, 1

After Reading (wk. 2) = 5, 10, 1, 1, 4, 6, 1, 8

library(lme4) # You will need to install and enable this package #

library(nlme) # You will also need to install and enable this package #

happiness <- c(1, 8, 2, 4, 4, 10, 2, 9, 4, 2, 5, 4, 3, 4, 2, 1, 5, 10, 1, 1, 4, 6, 1, 8 )

week <- c(rep("Before", 8), rep("Week1", 8), rep("Week2", 8))

id <- c(1,2,3,4,5,6,7, 8)

survey <- data.frame(id, happiness, week)

model <- lme(happiness ~ week, random=~1|id, data=survey)

anova(model)

library(lme4) # You will need to install and enable this package #

library(nlme) # You will also need to install and enable this package #

happiness <- c(1, 8, 2, 4, 4, 10, 2, 9, 4, 2, 5, 4, 3, 4, 2, 1, 5, 10, 1, 1, 4, 6, 1, 8 )

week <- c(rep("Before", 8), rep("Week1", 8), rep("Week2", 8))

id <- c(1,2,3,4,5,6,7, 8)

survey <- data.frame(id, happiness, week)

model <- lme(happiness ~ week, random=~1|id, data=survey)

anova(model)

**This method saves some time by producing the output:**

*numDF denDF F-value p-value*

(Intercept)

week 2 14

(Intercept)

**1****14**37.21053 <.0001week 2 14

**1.04624****0.3772**There was not a significant effect of the health assessment on the survey questions related to stroke concern at the p < .05 level for the five conditions (F(1, 14) = 1.05, p > .05).

__Student’s T-Test__**(One Sample T-Test)**

**Template:**

(Right Tailed)

(Right Tailed)

There was a significant increase in the

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the historically assumed mean (M =

**Historic Mean Value**); t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value**.

**- OR -**

There was not a significant increase in the

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the historically assumed mean (M =

**Historic Mean Value**); t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value**.

Example:

Example:

A factory employee believes that the cakes produced within his factory are being manufactured with excess amounts of corn syrup, thus altering the taste. 10 cakes were sampled from the most recent batch and tested for corn syrup composition. Typically, each cake should comprise of 20% corn syrup. Utilizing a 95 % confidence interval, can we assume that the new batch of cakes contain more than a 20% proportion of corn syrup?

The levels of the samples were:

.27, .31, .27, .34, .40, .29, .37, .14, .30, .20

**N <- c(.27, .31, .27, .34, .40, .29, .37, .14, .30, .20)**

**t.test(N, alternative = "greater", mu = .2, conf.level = 0.95)**

**# " alternative = " Specifies the type of test that R will perform. "greater" indicates a right tailed test. "left" indicates a left tailed test."two.sided" indicates a two tailed test. #**

*One Sample t-test*

data: N

t = 3.6713, df = 9, p-value = 0.002572

alternative hypothesis: true mean is greater than 0.2

95 percent confidence interval:

0.244562 Inf

sample estimates:

mean of x

0.289

data: N

t = 3.6713, df = 9, p-value = 0.002572

alternative hypothesis: true mean is greater than 0.2

95 percent confidence interval:

0.244562 Inf

sample estimates:

mean of x

0.289

**mean(N)**

sd(N)

sd(N)

> mean(N)

[1] 0.289

>

> sd(N)

[1] 0.07665942

> mean(N)

[1] 0.289

>

> sd(N)

[1] 0.07665942

**APA Format:**

A one sample t-test was conducted to compare the level of corn syrup in the current sample batch of cakes, to the assumed historical level of corn syrup contained within previously manufactured cakes.

There was a significant increase in the amount of corn syrup in the recent batch of cakes (M = .29, SD = .08), as compared to the historically assumed mean (M =.20); t(9) = 3.67, p = .003.

**(Two Sample T-Test)**

**Template:**

**(Two Tailed)**

**There was a significant difference in the**

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the

**GROUP B**(M =

**Mean of GROUP B**, SD =

**Standard Deviation of GROUP B**), t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value**.

**-OR-**

There was not a significant difference in the

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the

**GROUP B**(M =

**Mean of GROUP B**, SD =

**Standard Deviation of GROUP B**), t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value**.

A scientist creates a chemical which he believes changes the temperature of water. He applies this chemical to water and takes the following measurements:

70, 74, 76, 72, 75, 74, 71, 71

He then measures temperature in samples which the chemical was not applied.

74, 75, 73, 76, 74, 77, 78, 75

Can the scientist conclude, with a 95% confidence interval, that his chemical is in some way altering the temperature of the water?

**N1 <- c(70, 74, 76, 72, 75, 74, 71, 71)**

N2 <- c(74, 75, 73, 76, 74, 77, 78, 75)

t.test(N2, N1, alternative = "two.sided", var.equal = TRUE, conf.level = 0.95)

N2 <- c(74, 75, 73, 76, 74, 77, 78, 75)

t.test(N2, N1, alternative = "two.sided", var.equal = TRUE, conf.level = 0.95)

*Two Sample t-test*

data: N2 and N1

t = 2.4558, df = 14, p-value = 0.02773

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

0.3007929 4.4492071

sample estimates:

mean of x mean of y

75.250 72.875

data: N2 and N1

t = 2.4558, df = 14, p-value = 0.02773

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

0.3007929 4.4492071

sample estimates:

mean of x mean of y

75.250 72.875

**mean(N1)**

sd(N1)

mean(N2)

sd(N2)

sd(N1)

mean(N2)

sd(N2)

> mean(N1)

[1] 72.875

>

> sd(N1)

[1] 2.167124

>

> mean(N2)

[1] 75.25

>

> sd(N2)

[1] 1.669046

> mean(N1)

[1] 72.875

>

> sd(N1)

[1] 2.167124

>

> mean(N2)

[1] 75.25

>

> sd(N2)

[1] 1.669046

**APA Format:**

A two sample t-test was conducted to compare the temperature of water prior to the application of the chemical, to the temperature of water subsequent to the application of the chemical

There was a significant difference in the temperature of water prior to the application of the chemical (M = 72.88, SD = 2.17), as compared to the temperature of the water subsequent to the application of the chemical (M = 75.25, SD = 1.67); t(14) = 2.46, p = .03.

**(Paired T-Test)**

Template:

(Right Tailed)

Template:

(Right Tailed)

There was a significant increase in the

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the

**GROUP B**(M =

**Mean of GROUP B**, SD =

**Standard Deviation of GROUP B**), t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value.**

**- OR -**

There was not a significant increase in the

**GROUP A**(M =

**Mean of GROUP A**, SD =

**Standard Deviation of GROUP A**), as compared to the

**GROUP B**(M =

**Mean of GROUP B**, SD =

**Standard Deviation of GROUP B**), t(

**Degrees of Freedom**) =

**t-value**, p =

**p-value.**

**Example:**

A watch manufacturer believes that by changing to a new battery supplier, that the watches that are shipped which include an initial battery, will maintain longer lifespan. To test this theory, twelve watches are tested for duration of lifespan with the original battery.

The same twelve watches are then re-rested for duration with the new battery.

Can the watch manufacturer conclude, that the new battery increases the duration of lifespan for the manufactured watches? (We will assume an alpha value of .05).

For this, we will utilize the code:

**N1 <- c(376, 293, 210, 264, 297, 380, 398, 303, 324, 368, 382, 309)**

N2 <- c(337, 341, 316, 351, 371, 440, 312, 416, 445, 354, 444, 326)

t.test(N2, N1, alternative = "greater", paired=TRUE, conf.level = 0.95 )

N2 <- c(337, 341, 316, 351, 371, 440, 312, 416, 445, 354, 444, 326)

t.test(N2, N1, alternative = "greater", paired=TRUE, conf.level = 0.95 )

Paired t-test

data: N2 and N1

t = 2.4581, df = 11, p-value = 0.01589

alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval:

12.32551 Inf

sample estimates:

mean of the differences

45.75

Paired t-test

data: N2 and N1

t = 2.4581, df = 11, p-value = 0.01589

alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval:

12.32551 Inf

sample estimates:

mean of the differences

45.75

mean(N1)

sd(N1)

mean(N2)

sd(N2)

mean(N1)

sd(N1)

mean(N2)

sd(N2)

> mean(N1)

[1] 325.3333

>

> sd(N1)

[1] 56.84642

>

> mean(N2)

[1] 371.0833

>

> sd(N2)

[1] 51.22758

> mean(N1)

[1] 325.3333

>

> sd(N1)

[1] 56.84642

>

> mean(N2)

[1] 371.0833

>

> sd(N2)

[1] 51.22758

**APA Format:**

A paired t-test was conducted to the lifespan duration of watches which contained the new battery, to the lifespan of watches which contained the initial battery.

There was a significant increase in the lifespan duration of watches which contained the new battery (M = 325.33, SD =56.85), as compared to the lifespan of watches which contained the initial battery (M = 371.08, SD = 51.23); t(11) = 2.46, p = .02.

__Regression Models__**Example:**

**(Standard Regression Model)**

x <- c(27, 34, 22, 30, 17, 32, 25, 34, 46, 37)

y <- c(70, 80, 73, 77, 60, 93, 85, 72, 90, 85)

z <- c(13, 22, 18, 30, 15, 17, 20, 11, 20, 25)

multiregress <- (lm(y ~ x + z))

x <- c(27, 34, 22, 30, 17, 32, 25, 34, 46, 37)

y <- c(70, 80, 73, 77, 60, 93, 85, 72, 90, 85)

z <- c(13, 22, 18, 30, 15, 17, 20, 11, 20, 25)

multiregress <- (lm(y ~ x + z))

*Call:*

lm(formula = y ~ x + z)

Residuals:

Min 1Q Median 3Q Max

-6.4016 -5.0054 -1.7536 0.8713 14.0886

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 47.1434 12.0381 3.916 0.00578 **

x 0.7808 0.3316 2.355 0.05073 .

z 0.3990 0.4804 0.831 0.43363

---

Residual standard error: 7.896 on 7 degrees of freedom

Multiple R-squared: 0.5249, Adjusted R-squared: 0.3891

F-statistic: 3.866 on 2 and 7 DF, p-value: 0.07394

lm(formula = y ~ x + z)

Residuals:

Min 1Q Median 3Q Max

-6.4016 -5.0054 -1.7536 0.8713 14.0886

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 47.1434 12.0381 3.916 0.00578 **

x 0.7808 0.3316 2.355 0.05073 .

z 0.3990 0.4804 0.831 0.43363

---

Residual standard error: 7.896 on 7 degrees of freedom

Multiple R-squared: 0.5249, Adjusted R-squared: 0.3891

F-statistic: 3.866 on 2 and 7 DF, p-value: 0.07394

**APA Format:**

A linear regression model was utilized to test if variables “x” and “z” significantly predicted outcomes within the observations of “y” included within the sample data set. The results indicated that while “x” (B = .781, p = .051) is a significant predictor variable, the overall model itself does not possess a worthwhile predictive capacity (r2 = .041).

(Non-Standard Regression Model)

(Non-Standard Regression Model)

**Example:**

**# Model Creation #**

Age <- c(55, 45, 33, 22, 34, 56, 78, 47, 38, 68, 49, 34, 28, 61, 26)

Obese <- c(1,0,0,0,1,1,0,1,1,0,1,1,0,1,0)

Smoking <- c(1,0,0,1,1,1,0,0,1,0,0,1,0,1,1)

Cancer <- c(1,0,0,1,0,1,0,0,1,1,0,1,1,1,0)

# Summary Creation and Output #

CancerModelLog <- glm(Cancer~ Age + Obese + Smoking, family=binomial)

summary(CancerModelLog)

# Output #

Age <- c(55, 45, 33, 22, 34, 56, 78, 47, 38, 68, 49, 34, 28, 61, 26)

Obese <- c(1,0,0,0,1,1,0,1,1,0,1,1,0,1,0)

Smoking <- c(1,0,0,1,1,1,0,0,1,0,0,1,0,1,1)

Cancer <- c(1,0,0,1,0,1,0,0,1,1,0,1,1,1,0)

# Summary Creation and Output #

CancerModelLog <- glm(Cancer~ Age + Obese + Smoking, family=binomial)

summary(CancerModelLog)

# Output #

Call:

glm(formula = Cancer ~ Age + Obese + Smoking, family = binomial)

Deviance Residuals:

Min 1Q Median 3Q Max

-1.6096 -0.7471 0.5980 0.8260 1.8485

Coefficients:

Estimate Std. Error z value Pr(>|z|)

(Intercept) -2.34431 2.25748 -1.038 0.2991

Age 0.02984 0.04055 0.736 0.4617

Obese -0.38924 1.39132 -0.280 0.7797

Smoking 2.54387 1.53564 1.657 0.0976 .

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 20.728 on 14 degrees of freedom

Residual deviance: 16.807 on 11 degrees of freedom

AIC: 24.807

Number of Fisher Scoring iterations: 4

Call:

glm(formula = Cancer ~ Age + Obese + Smoking, family = binomial)

Deviance Residuals:

Min 1Q Median 3Q Max

-1.6096 -0.7471 0.5980 0.8260 1.8485

Coefficients:

Estimate Std. Error z value Pr(>|z|)

(Intercept) -2.34431 2.25748 -1.038 0.2991

Age 0.02984 0.04055 0.736 0.4617

Obese -0.38924 1.39132 -0.280 0.7797

Smoking 2.54387 1.53564 1.657 0.0976 .

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 20.728 on 14 degrees of freedom

Residual deviance: 16.807 on 11 degrees of freedom

AIC: 24.807

Number of Fisher Scoring iterations: 4

**# Generate Nagelkerke R Squared #**

# Download and Enable Package: "BaylorEdPsych" #

PseudoR2(CancerModelLog)

# Console Output #

# Download and Enable Package: "BaylorEdPsych" #

PseudoR2(CancerModelLog)

# Console Output #

*McFadden Adj.McFadden Cox.Snell*

0.2328838 -0.2495624 0.2751639

Adj.Count AIC Corrected.AIC

0.5714286 23.9005542 27.9005542

**Nagelkerke**McKelvey.Zavoina Effron0.2328838 -0.2495624 0.2751639

**0.3674311**0.3477522 0.3042371 0.8000000Adj.Count AIC Corrected.AIC

0.5714286 23.9005542 27.9005542

**APA Format:**

A logistic regression model was utilized to test if a model containing the variables “Age”, “Smoking Status”, and “Obesity”, could predict Cancer outcomes as it pertains to the individuals included within the sample data set. The results indicated that the model does not possess a worthwhile predictive capacity (Nagelkerke R-Square = .37).

## No comments:

## Post a Comment

Note: Only a member of this blog may post a comment.