A university wants to know more about the knowledge of students regarding international events. They are concerned that their students are uninformed in regards to news from other countries. A standardized test is used to assess students’ knowledge of world events (national reported mean=65, S=5). A sample of 30 students are tested (sample mean=58, Standard Error=3.2). Compute a 99 percent confidence interval based on this sample's data. How do these students compare to the national sample?
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A 99% confidence interval can be calculated using the formula: sample mean ± (critical value * standard error). The critical value for a 99% confidence interval is approximately 2.576.
So, the confidence interval would be: 58 ± (2.576 * 3.2) = 58 ± 8.24
This gives a confidence interval of 49.76 to 66.24.
This means that we are 99% confident that the true mean score of the university students on the test is between 49.76 and 66.24.
Comparing this to the national mean score of 65, it appears that the university students may be slightly less knowledgeable about international events than the average student nationally, as their mean score is lower and the confidence interval barely includes the national mean. However, since the national mean is within the confidence interval, it's also possible that there's no significant difference. Further testing would be needed to confirm.
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