Question 632837: Hello there, I am asking if you can please explain to me how to do this?
Perform a two-tailed hypothesis test on both the legal services satisfaction and the sentence satisfaction variable’s data using a .05 significance level
Begin by creating a null and an alternate statement.
Use Microsoft Excel to process your data.
Copy and paste the results of the output to your report in Microsoft Word.
Identify the significance level, the test statistic, and the critical value.
State whether you are rejecting or failing to reject the null hypothesis statement.
The legal services
6.2
4.2
6.2
4.3
3.2
4.5
4.7
5.3
4.9
3.4
5.5
5.8
6.2
4.2
5.7
6.4
3.7
6.2
6.5
5.1
6.2
6.3
6.3
5.2
4.6
Sentence sat.
5.5
3.6
3.8
5.6
5.5
4.4
3.1
6.9
4.6
6.8
5.5
3.8
3.2
4.6
6.5
6.1
3.9
4.6
5.5
4.4
4.7
6.2
4.7
5.5
4.8
So far I have come up with this:
t-Test: Two-Sample Assuming Unequal Variances
6.2 5.5
Mean 5.191666667 4.929166667
Variance 1.032971014 1.187373188
Observations 24 24
Hypothesized Mean Difference 0
df 46
t Stat 0.863027781
P(T<=t) one-tail 0.196299822
t Critical one-tail 1.678660414
P(T<=t) two-tail 0.392599645
t Critical two-tail 2.012895599
Thank you so much for your time.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Perform a two-tailed hypothesis test on both the legal services satisfaction and the sentence satisfaction variable’s data using a .05 significance level
Begin by creating a null and an alternate statement.
Use Microsoft Excel to process your data.
Copy and paste the results of the output to your report in Microsoft Word.
Identify the significance level, the test statistic, and the critical value.
State whether you are rejecting or failing to reject the null hypothesis statement.
The legal services(x1 DATA)
6.2
4.2
6.2
4.3
3.2
-----
4.5
4.7
5.3
4.9
3.4
-----
5.5
5.8
6.2
4.2
5.7
-----
6.4
3.7
6.2
6.5
5.1
-----
6.2
6.3
6.3
5.2
4.6
=======================================
Sentence sat: x2 DATA
5.5
3.6
3.8
5.6
5.5
-----
4.4
3.1
6.9
4.6
6.8
-----
5.5
3.8
3.2
4.6
6.5
----
6.1
3.9
4.6
5.5
4.4
-----
4.7
6.2
4.7
5.5
4.8
----
I used a TI-84 calculator and got the following:
-----
2-Sample Ttest
--------------------
6.2 5.5
Mean 5.23 ; 4.95
standard deviations:: 1.2 ; 1.07
Observations 25 ; 25
df = 47.85
========================
Hypothesized Mean Difference
Ho: u1-u2 = 0
Ha: u1-u2 # 0
=======================
t Stat :: t(x1bar - x2bar) = 0.948
p-value = 2*P(t > 0.948, , with df 47.85) = 0.35
==========================================
Ho:
Ha:
Identify:::::
the significance level:: 5% or 0.05
------
the test statistic::: 0.848
-------
the critical value::: +-invT(0.025,47.85) = +-2.01
State whether you are rejecting or failing to reject the null hypothesis statement.
Since the p-value is greater than 5%, fail to reject Ho.
================
Cheers,
Stan H.
|
|
|
