In: Statistics and Probability

College students were randomly sorted into one of two groups.
Members of each group performed a series of mental tasks while
music was playing in the background. One group listened to pop
music and the other to country music. It is hypothesized that music
will influence the number of tasks completed. What can be concluded
with an *α* of 0.10? The results are below:

pop |
country |

_{1} = 34.34
n _{1} = 16 |
_{2} = 37.05
n _{2} = 9 |

**a)** What is the appropriate test statistic?

---Select---naz-testOne-Sample t-testIndependent-Samples
t-testRelated-Samples t-test

**b)**

Condition 1:

---Select---country musicmental taskpop musicmusicschool

Condition 2:

---Select---country musicmental taskpop musicmusicschool

**c)** Obtain/compute the appropriate values to make a
decision about *H*_{0}.

(Hint: Make sure to write down the null and alternative hypotheses
to help solve the problem.)

critical value =________ ; test statistic =________

Decision: ---Select---Reject H0Fail to reject H0

**d)** If appropriate, compute the CI. If not
appropriate, input "na" for both spaces below.

[_____ ,______ ]

**e)** Compute the corresponding effect size(s) and
indicate magnitude(s).

If not appropriate, input and/or select "na" below.

*d* =______ ; ---Select---natrivial effectsmall effectmedium
effectlarge effect

*r*^{2} =______ ; ---Select---natrivial effectsmall
effectmedium effectlarge effect

**f)** Make an interpretation based on the
results.

Students that listened to pop music significantly completed more tasks.

Students that listened to country music significantly completed more tasks.

Music had no significant effect on the number of tasks completed.

a) Independent-Samples t-test

b)

condition 1: pop music

condition 2: country music

c)

Ho : µ1 - µ2 = 0

Ha : µ1-µ2 ╪ 0

Level of Significance , α =
0.05

Sample #1 ----> 1

mean of sample 1, x̅1= 34.340

standard deviation of sample 1, s1 =
5.330

size of sample 1, n1= 16

Sample #2 ----> 2

mean of sample 2, x̅2= 37.050

standard deviation of sample 2, s2 =
4.370

size of sample 2, n2= 9

difference in sample means = x̅1-x̅2 =
34.3400 - 37.1 =
-2.71

pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 5.0170

std error , SE = Sp*√(1/n1+1/n2) =
2.0904

**t-statistic = ((x̅1-x̅2)-µd)/SE = (
-2.7100 - 0 ) /
2.09 = -1.2964**

Degree of freedom, DF= n1+n2-2 =
23

**t-critical value , t* = ± 1.714 ** (excel
formula =t.inv(α/2,df)

Decision: | t-stat | < | critical value |, so,
**Fail to Reject Ho
**

d)

Degree of freedom, DF= n1+n2-2 =
23

t-critical value = t α/2 =
1.7139 (excel formula =t.inv(α/2,df)

pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 5.0170

std error , SE = Sp*√(1/n1+1/n2) =
2.0904

margin of error, E = t*SE = 1.7139
* 2.09 = 3.58

difference of means = x̅1-x̅2 =
34.3400 - 37.050 =
-2.7100

**confidence interval is
Interval Lower Limit= (x̅1-x̅2) - E =
-2.7100 - 3.5827 =
-6.293
Interval Upper Limit= (x̅1-x̅2) + E =
-2.7100 + 3.5827 =
0.873**

e)

d, r² = Na

f)

Music had no significant effect on the number of tasks completed.

College students were randomly sorted into one of two groups.
Members of each group performed a series of mental tasks while
music was playing in the background. One group listened to pop
music and the other to country music. It is hypothesized that music
will effect the number of tasks completed. What can be concluded
with an α of 0.10? The results are below:
pop
country
1 = 39.31
1 =
5.43
n1 = 10
2 = 44.15
2 =...

College students were randomly sorted into one of two groups.
Members of each group performed a series of mental tasks while
music was playing in the background. One group listened to pop
music and the other to country music. It is hypothesized that more
tasks will be completed with pop than with country music. What can
be concluded with α = 0.05? The results are below:
pop
country
mean 1 = 47.14
sigma hat 1 =3.77
n1 = 15
mean...

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