Question

In 2009, 250 students from North Dakota took the verbal portion of the SAT (standardized achievement test) and scored a mean of 525. Scores on this portion of the SAT are normally distributed having a mean of 500 and a standard deviation of 100.

1. Based off of hypothesis testing, how did this sample of students perform relative to the population?
2. Construct and interpret the 95% confidence interval
3. Determine the power of this finding, clearly indicating what this value tells us for this case.

Answer 1

As we can see that in the confidence interval, all the values are greater than 500 which is the population mean of SAT scores this means that the sample has done better as compared to the population.

At the 5% significance level, the decision criterion for the test is to reject H0 if Z > 1.645, where

Z = = (-500)/100/ = 1.645

(-500)/100/ = 1.645

= 510.4

Calculate the Z-statistic assuming the alternative hypothesis is true, i.e., μ1 = 525:

Z = (510.4 - 525)/100/

= -2.308

Power = P(reject H0|H0 is false) = P(Z>2.308) + P(Z<2-.308)

= 2*0.0105

= 0.021

i.e. the power of the test is 2.1%. It is the probability of rejecting null hypothesis when it is false.

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