Question

You want to construct a 92% confidence interval. The correct z* to use is

A) 1.645

B) 1.41

C) 1.75

2) Suppose the average Math SAT score for all students taking the exam this year is 480 with standard deviation 100. Assume the distribution of scores is normal. The senator of a particular state notices that the mean score for students in his state who took the Math SAT is 500. His state recently adopted a new mathematics curriculum, and he wonders if the improved scores are evidence that the new curriculum has been successful. Since over 10,000 students in his state took the Math SAT, he can show that the P-value for testing whether the mean score in his state is more than the national average of 480 is less than 0.0001. We may correctly conclude that

A) these results are not good evidence that the new curriculum has improved Math SAT scores.

B) there is strong statistical evidence that the new curriculum has improved Math SAT scores in his state.

C) although the results are statistically significant, they are not practically significant, since an increase of 20 points is fairly small.

3) Suppose the average Math SAT score for all students taking the exam this year is 480 with standard deviation 100. Assume the distribution of scores is normal. A SRS of four students is selected and given special training to prepare for the Math SAT. The mean Math SAT score of these students is found to be 560, 80 points higher than the national average. We may correctly conclude

A) the results are statistically significant at level α = 0.05, but they are not practically significant.

B) the results are not statistically significant at level α = 0.05, but they are practically significant.

C) the results are neither statistically significant at level α = 0.05 nor practically significant.

Answer 1

1) The correct z* to to construct a 92% confidence interval is 1.75

2) these results are not good evidence that the new curriculum has improved Math SAT scores.

3) the results are not statistically significant at level α = 0.05, but they are practically significant.

Explanation:

1) z-score for 92% confidence level is ≈ 1.75

2) P-value for testing whether the mean score in senator's state is more than the national average of 480 is less than 0.0001 means that

the probability that the sample is drawn from the population where senator's state is more than the national average of 480 is <0.0001.

Thus we have to reject this hypothesis since the probability is too small.

3) if we calculate the statistic of the sample we get (560-480)/100=0.8 where

• 560 is the mean score of trained 4 students
• 480 is the mean math sat score of this year
• 100 is the standard deviation of the test.

Since t-critical at 0.05 significance for 3 degrees of freedom ≈ 2.35 is bigger than 0.8, the result is not significant statistically.

But 80 points higher than the national average is a practically significant result since its effect size is large.