Question 751436: I am on crunch time, done 90% of my homework for this and other classes, now, I am stuck. Please save me....
PART ONE--> Imagine that you have population data with an average height of 5 feet 10 inches. Conduct a one-sample t-test to determine whether your sample population is significantly different from the general population.
MY DATA::::
(64, 65, 61, 63, 59, 64, 61, 63, 63)
DATA FROM MINI TAB:::
One-Sample T: Satisfaction
Test of mu = 70 vs not = 70
Variable N Mean StDev SE Mean 95% CI T P
Satisfaction 9 7.556 0.882 0.294 (6.878, 8.233) -212.42 0.000
PART TWO-->Imagine that you have population data with an average satisfaction with a job score of 5. Conduct a one-sample t-test to determine whether your sample population is significantly more or less satisfied than the general population.
MY DATA:::
(8, 8, 7, 9, 8, 7, 9, 6, 7)
DATA FROM MINI TAB:::
One-Sample T: Height
Test of mu = 5 vs not = 5
Variable N Mean StDev SE Mean 95% CI T P
Height 9 62.556 1.878 0.626 (61.112, 63.999) 91.93 0.000
HELP?? Be sure to use the proper df and provide all the steps of calculations.
HELP??Include an explanation of how you arrived at the critical value. Interpret the results using statistical analysis.
HELP??With the analysis and results, explain whether you can conclude that your sample is taller or shorter than the general population.
HELP??Describe how the results would be altered if you had twice as many subjects.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Note: It looks like you mixed up your tables for part 1 and 2.
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Part 1
Note: I'm assuming you're using a significance level of alpha = 0.05
Hypothesis
Ho: mu = 70
Ha: mu =/= 70
This is a two tailed test
Critical Values are: -2.306 and 2.306
Note: I used the calculator at this link to get the critical values. Use df = 8
To compute the test statistic, we'll need these values
mu = 70 (given)
xbar = 62.5555555555556 (add up all the data values and divide by 9)
s = 1.87823794493077 (to find this by hand is a pain, so use a calculator)
Test Statistic
t = (xbar-mu)/(s/sqrt(n))
t = (62.5555555555556-70)/(1.87823794493077/sqrt(9))
t = (62.5555555555556-70)/(1.87823794493077/(3))
t = (62.5555555555556-70)/(0.626079314976925)
t = (-7.44444444444444)/(0.626079314976925)
t = -11.8905772261759
This test statistic -11.8905772261759 is definitely NOT between the critical values of -2.306 and 2.306
So the test statistic is in the rejection region which means we reject the null hypothesis. So the only thing left to do is accept the alternative hypothesis which means that mu =/= 70 (mu doesn't equal 70)
So the sample population is significantly different from the general population.
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Part 2
This is the same as part 1, just with different numbers. Let me know if you still need help with this part. Thank you.
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