What is the least common denominator (LCD)
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Answer:
The null and alternative hypothesis are:
P-value=0.1996
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .
Step-by-step explanation:
This is a hypothesis test for the difference between populations means.
The claim is that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample X (one-tailed), of size nX=80 has a mean of 7.8 and a standard deviation of 1.06.
The sample Y (two-tailed), of size nY=80 has a mean of 7.64 and a standard deviation of 1.32.
The difference between sample means is Md=0.16.
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
This test is a right-tailed test, with 158 degrees of freedom and t=0.8453, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.1996) is greater than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .