## Finding the t-test and deciding on the hypotheses

Here, the means, standard deviations, and t-tests for the two different sets of data are calculated. Conclusions on hypotheses on each are given, and the results are all explained. The pooled variance is also calculated.

**Finding the mean and standard deviation for the first set of participants
**

1a) what is the Mean for the Before Therapy group?

Answer: 10.875

1b)what is the Standard Deviation for the Before Therapy group?

Answer: 2.390955

1c) what is the Mean for the After Therapy group?

Answer: 10.5625

1d) what is the Standard Deviation for the After Therapy group?

Answer: 2.393568

**Deciding on the appropriate test
**

1e) what type of t-test is required here?

Answer: paired sample t-test

**Deciding on the hypothesis
**

1f) based on your findings; can we reject the null hypothesis? Report your findings using the format below, as demonstrated in the lab. Be sure to clearly state the variables. If there is a significant difference, clarify which direction it is going.

Answer: We cannot reject the null hypothesis. There is not a significant difference in mean between before therapy and after therapy.

t(15)= 0.628 , p= 0.540

**Finding the T-Test
**

1g) Calculate this same t-test by hand. Round to two decimals. You do not have to show your calculations. Just state what you calculated the Standard Error (s_D┴¯ ) to be. (Remember, the Standard Error is the denominator of the t-ratio)

Standard Error (s_D┴¯ ) =1.99

**Problem 2
**

Researchers were interested in which of two methods were best for retention. They had all subjects read a chapter in a textbook. Then, one group of subjects ('Study') was encouraged to study the material for 60 minutes. The other group of subjects (‘Quiz’) was not allowed to study but was actually given a 60-minute quiz on the material. Then, all subjects were required to complete a final test on the material. Perform an appropriate t-test on the final test scores with R to see if there is a difference in performance between the Study and Quiz groups

**Finding the mean and standard deviation for the second set of participants
**

2a) what is the Mean for the Study group?

Answer: 18.7857

2b) What is the Standard Deviation for the Study group?

Answer: 5.950806

2c) What is the Mean for the Quiz group?

Answer: 26.5

2d) what is the Standard Deviation for the Quiz group?

Answer: 6.449806

**Deciding on the appropriate test
**

2e) what type of t-test is required here?

Answer: Independent two-sample t-test

**Deciding on the hypothesis
**

2f) based on your findings; can we reject the null hypothesis? Report your findings using the format below. Be sure to clearly state the variables. If there is a significant difference, clarify which direction it is going.

**Answer:
**

We can reject the null hypothesis. There is a significant difference in the mean between before and after.

T ( 28) = -3.3873, p= 0.002

**Finding the Pooled Variance
**

2g) Calculate the Pooled Variance (s_p^2) for this same t-test by hand. Hint: To save much time, use the Standard Deviations from the R output. Round to two decimals places. You do not have to show your calculations.

Pooled Variance (s_p^2) =6.22

Q3

> #Q1

> Assign3 <- read.delim("C:/Users/CSSC/Downloads/Assign3.txt")

> ETURN<-data.frame(Assign3)

>mean(ETURN$BeforeTherapy,na.rm=TRUE )

[1] 10.875

>sd(ETURN$BeforeTherapy, na.rm = TRUE)

[1] 2.390955

>mean(ETURN$AfterTherapy, na.rm= TRUE)

[1] 10.5625

>sd(ETURN$AfterTherapy)

[1] 2.393568

>t.test(ETURN$BeforeTherapy,ETURN$AfterTherapy,paired = TRUE, na.rm= TRUE)

Paired t-test

data: ETURN$BeforeTherapy and ETURN$AfterTherapy

t = 0.62795, df = 15, p-value = 0.5395

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

95 percent confidence interval:

-0.7482174 1.3732174

sample estimates:

mean of the differences

0.3125

> #Q2

>mean(ETURN$Study,na.rm= TRUE)

[1] 18.78571

>sd(ETURN$Study, na.rm=TRUE)

[1] 5.950806

>mean(ETURN$Quiz, na.rm=TRUE)

[1] 26.5

>sd(ETURN$Quiz, na.rm=TRUE)

[1] 6.449806

>t.test(ETURN$Study,ETURN$Quiz, na.rm=TRUE)

Welch Two Sample t-test

data: ETURN$Study and ETURN$Quiz

t = -3.4061, df = 27.907, p-value =

0.002016

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

95 percent confidence interval:

-12.354269 -3.074302

sample estimates:

mean of x mean of y

18.78571 26.50000