Question

Using a = 0.05, test for a difference in the
means.

group
n
mean sample std dev

1
20
8.3
3.0

2
20
7.2
3.0

ttest u1 = u2 ( VS NE) t= 1.16

Answer #1

ttest age == 56
One-sample t test
------------------------------------------------------------------------------
Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
age | 2,857 49.15576 .3310101 17.69279 48.50671 49.8048
------------------------------------------------------------------------------
mean = mean(age) t = -20.6768
Ho: mean = 56 degrees of freedom = 2856
Ha: mean < 56 Ha: mean != 56 Ha: mean > 56
Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) =
1.0000
State the null hypothesis, would you reject or accept...

Question 3: Independent-Samples t-Test
Group
Statistics
type of school
N
Mean
Std. Deviation
Std. Error Mean
reading score
public
168
51.8452
10.42279
.80414
private
32
54.2500
9.19677
1.62578
Independent
Samples Test
Levene's Test for
Equality of Variances
t-test for Equality of
Means
F
Sig.
t
df
Sig. (2-tailed)
Mean Difference
Std. Error
Difference
95% Confidence
Interval of the Difference
Lower
Upper
reading score
Equal variances
assumed
.564
.453
-1.217
198
.225
-2.40476
1.97519
-6.29986
1.49034
Equal variances not
assumed
-1.326...

4.
Group
Statistics
protestant
N
Mean
Std. Deviation
Std. Error Mean
HIGHEST YEAR OF SCHOOL
COMPLETED
.00
1260
13.32
3.219
.091
1.00
745
13.62
2.810
.103
Using this information
a) Conduct an independent samples t-test for these means
b) Find the effect size

3.
Group
Statistics
newengland
N
Mean
Std. Deviation
Std. Error Mean
HIGHEST YEAR OF SCHOOL
COMPLETED
.00
1937
13.38
3.073
.070
1.00
81
14.72
2.955
.328
a) Conduct an independent samples t-test for these means

Null hypothesis (H0): The true difference in mean
hemoglobin change for pregnant women not exposed to herbicides
versus those who are somewhat exposed is 0 g/dL.
(µbottle – µboth = 0 g/dL)
Research hypothesis (H1): The true difference in mean
hemoglobin change for pregnant women not exposed to herbicides
versus those who are somewhat exposed is not 0 g/dL.
(µbottle – µboth ≠ 0 g/dL)
SAS Output:
The TTEST Procedure
Variable: change
group
Method
Mean
95% CL Mean
Std Dev...

Group Statistics
Group
N
Mean
Std. Deviation
Std. Error Mean
DRP Score
1
21
51.48
11.007
2.402
2
23
41.52
17.149
3.576
An educator conducted an experiment to test whether new directed
reading activities in the classroom will help elementary school
pupils improve some aspects of their reading ability. She arranged
for a third grade class of 21 students to follow these activities
for an 8-week period. A control classroom of 23 third graders
followed the same curriculum without the...

One-sample t test
------------------------------------------------------------------------------
Variable |
Obs
Mean Std. Err. Std. Dev.
[95% Conf. Interval]
---------+--------------------------------------------------------------------
alctry | 16,149 991.0647
.0066429 .8441679
991.0517 991.0777
------------------------------------------------------------------------------
mean =
mean(alctry)
t = 1.1e+05
Ho: mean =
292.0672
degrees of freedom = 16148
Ha: mean <
292.0672 Ha: mean
!= 292.0672
Ha: mean > 292.0672
Pr(T < t) =
1.0000 Pr(|T| >
|t|) = 0.0000
Pr(T > t) = 0.0000
Write out the research question and write a short paragraph 3-4
sentences...

For each of the following data sets, carry out a one-tailed
t-test, with α = 0.05, and the alternative hypothesis ????????: U1
> U2
In each case, give the p-value (if applicable) or explain why
the data are inconsistent with the alternative hypothesis. State
whether or not the null hypothesis is rejected.
a) Sample 1: n=10 , mean= 10.8
Sample 2: n= 10, mean= 10.5 df= 18 SE= 0.23
b) Sample 1: n= 100, mean 750
Sample 2: n=100, mean=...

one group of mathematics students took a distance learning class
while another group took the same course in a regular classroom.
the sample data is available in the table below
statistic distance . regular
Number of students . 40 50
mean scores 30.2 . 42.5
sample std dev 2.4 . 2.5
a) is there a significance difference in the mean scores? test
the claim using a= 0.05 significance level
b.) construct a 95% confidence interval for the difference
between the...

The MINITAB printout shows a test for the difference in
two population means.
Two-Sample
T-Test and CI: Sample 1, Sample 2
Two-sample T for Sample
1 vs Sample 2
N
Mean
StDev
SE Mean
Sample 1
5
29.00
3.00
1.3
Sample 2
7
28.89
3.63
1.4
Difference = mu (Sample
1) - mu (Sample 2)
Estimate for
difference: 0.11
95% CI for difference:
(-4.3, 4.5)
T-Test of difference =
0 (vs not =):
T-Value = 0.06 P-Value
= 0.96...

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