Question 1197674
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Answers: 
Mean = <font color=red>72.59</font>
Standard Error = <font color=red>3.769746</font>
Test Statistic = <font color=red>2.01</font>
P-value = <font color=red>0.0377</font>
Conclusion = "<font color=red>Reject the null</font>"
Interpretation = "<font color=red>The mean score is higher than 65</font>"


With exception of the mean, each decimal value is approximate. 
Round however your teacher instructs.


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Explanation:


Null Hypothesis: mu = 65
Alternative Hypothesis: mu > 65
The inequality sign in the alternative hypothesis tells us we have a right-tailed test.


The data set is
{65,64.3,88,69,83.2,68.4,96,64.3,65,62.7}
There are n = 10 items in this data set.


Add up all of the values in the data set to get 
65+64.3+88+69+83.2+68.4+96+64.3+65+62.7 = 725.9


Divide that sum by n = 10 to compute the mean
mean = (sum of values)/n
mean = (725.9)/10
<font color=red>mean = 72.59</font>
This value is exact without any rounding done to it.
This is the sample mean denoted as xbar. Its job is to estimate the population mean mu.


Use a graphing calculator, spreadsheet, or statistical software to compute the standard deviation of the data set. 
We want the sample standard deviation.
You should find the sample standard deviation is approximately s = 11.920985


Let's compute the standard error (SE)
SE = s/sqrt(n)
SE = 11.920985/sqrt(10)
SE = 3.76974645527023
<font color=red>SE = 3.769746</font>
Round this however your teacher instructs.


Now we can find the test statistic
t = (xbar - mu)/SE
t = (72.59 - 65)/3.769746
t = 2.01339825017389
<font color=red>t = 2.01</font>
Many stats textbooks I've come across will have the t statistic accurate to two decimal places. 
But if your teacher requires some other level of precision, then be sure to follow those instructions.


The sample size is n = 10
The degrees of freedom (df) is
df = n-1
df = 10-1
df = 9
Use a stats calculator, spreadsheet, or similar to find the p-value is roughly <font color=red>0.0377</font>
This tells the reader that P(T > 2.01) = 0.0377 approximately when df = 9.
The p-value may slightly vary depending if you use t = 2.01 or if you use more decimal digits in that test statistic such as t = 2.01339825017389
Either way, the p-value shouldn't differ too much.


The p-value being smaller than alpha = 0.05 tells us to <font color=red>reject the null</font> and conclude that mu > 65 appears to be the case. 


Note: we're doing a right tailed test, so be sure to find the area under the T distribution curve to the right of t = 2.01
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