How do you work out 23percent of 600
1 answer:
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Answer:
There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).
P-value=0.042.
Step-by-step explanation:
The question is incomplete:
The data of the scores for each student is:
Before After
430 465
485 475
520 535
360 410
440 425
500 505
425 450
470 480
515 520
430 430
450 460
495 500
540 530
We will generate a sample for the difference of scores (before - after) and test that sample.
The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]
This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.
The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).
Then, the null and alternative hypothesis are:
The significance level is 0.01.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.042) is bigger than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).