Download Trying to find critical value for test statistics of 2

Trying to find critical value for test statistics of 2.36 with 10 subjects, 7 had higher scores, 2 had
lower scores, and 1 score did not change. Using the traditional method of hypothesis testing. Use an
alpha level of .01.
First step in finding the critical value is to determine if the test is one tailed test or two tailed test.
Assume the above test is one tailed test.
Level of significance is 1% ( alpha level)
Find the size of sample (n). Here it is 10.
Since the sample size is less 30 it’s a small sample test.
Next we determine the test being used. Here it is t test.
Open your table and go to the column t .01 . Since test statistic follow t distribution with (n-1)
degrees of freedom. Here degrees of freedom is 9.
In your chart find the row with 9 degrees of freedom.
Your critical value is cell intersecting t .01 column and 9 degrees of freedom row.
Here the value is 2.821.
In case of two tailed test, we divide .01/2 = .005. Then we go to
t 0.005 . Follow the same steps as
mentioned above. The critical value for two tailed test cell intersecting t 0.005 column and 9 degrees
of freedom row which is 3.250.
The test statistic is 2.36. Since the calculated value is less than critical value (both two
tailed and one tailed) we accept H0 (null hypothesis)
Trying to find the critical value of scores of girls are smarter than boys on tests. Assume the
degrees of freedom for this problem is 27. The test statistic was 2.69 (alpha equals .05). If the
mean of the boys score was lower than the mean of the girls score can she reject her null
hypothesis?
Here Ho : girls are same as boys on tests(Mean of boys score = Mean of girls score)
H1: Girls are smarter than boys (Mean of girls score > Mean of boys score)
So its right tailed test.
Here degrees of freedom is 27
Test statistic: 2.69
Critical value is 1.703
Conclusion: Since calculated value is more than critical value (2.69 > 1.703) we reject H0 and
conclude that girls are smarter than boys.
No. We cannot reject null hypothesis just if mean of boys are lower than the mean of girls.
We should also consider variation in score. We have to calculate the test statistic.
In the problem given below I suppose it is not education. It should be cars. If education
is correct then the problem is meaningless. Please check
Finding critical value, t score of randomly selected 217 students and found that they had education
with a mean of 7.89 years and a standard deviation of 3.67 years. They also randomly selected 152
faculty and found that they had education with a mean of 5.99 years and a standard deviation of
3.65 years. Use a 0.01 significance level to test the claim that student cars are older than faculty
cars.
Correct problem solution
Finding critical value, t score of randomly selected 217 students and found that they had cars with a
mean of 7.89 years and a standard deviation of 3.67 years. They also randomly selected 152 faculty
and found that they had cars with a mean of 5.99 years and a standard deviation of 3.65 years. Use
a 0.01 significance level to test the claim that student cars are older than faculty cars.
H0: student cars are as old as faculty cars
H1: student cars are older than faculty cars.(right tailed test)
Level of significance 1%
Test statistic:
7.89  5.99
= 4.91
3.67 * 3.67 3.65 * 3.65

217
152
Critical value: Here the sample size is very large therefore t tends to normal as n becomes larger.
So we take Z value = 2.33
Conclusion: Since calculated value is more than critical value we reject H0 and conclude that
student cars are older than faculty cars.