Question 1197674: Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores:
Grades 65 64.3 88 69 83.2 68.4 96 64.3 65 62.7
Test grades are believed to be normally distributed.
Use a significance level of 5%.
State the mean of the sample:
85.08
Incorrect
State the standard error of the sample means:
State the test statistic:
t
=
5.935830
Incorrect
State the p-value:
Answer by math_tutor2020(3816)   (Show Source):
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Answers:
Mean = 72.59
Standard Error = 3.769746
Test Statistic = 2.01
P-value = 0.0377
Conclusion = "Reject the null"
Interpretation = "The mean score is higher than 65"
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
mean = 72.59
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
SE = 3.769746
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
t = 2.01
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 0.0377
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 reject the null 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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