Question #6467

A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).
What is the Z critical value (Z cv ) using a 0.05 alpha level?
Should H0 be rejected? What should the researcher conclude?
Determine the 95 % confidence interval for the population mean, based on the sample mean.

Expert's answer

Let Ho: average private high scools score (m)<=aver. general score (M).

So

we can use one-tailed z-test.

z-score equals

Z=(m-M)*sqrt(n)/σ=(1030-1000)*sqrt(90)/200=1.42

Z cv=1.645 (we take it from

normal distribution table for alpha=0.05)

Z<Z cv, so we can't reject Ho

and conclude that the observed data set provides no strong evidence against the

null hypothesis.

The 95% confidence interval based on the sample mean (now we

suppose that population mean is uknown with known standard deviation) is

m±Z*σ/sqrt(n), where Z*=Z cv for level alpha/2 (! because to get an interval we

need to use two-sided estimation). Z*=1.96 and 95% confidence interval is

(1030-41.32;1030+41.32)=(988.68,1071.32)

So

we can use one-tailed z-test.

z-score equals

Z=(m-M)*sqrt(n)/σ=(1030-1000)*sqrt(90)/200=1.42

Z cv=1.645 (we take it from

normal distribution table for alpha=0.05)

Z<Z cv, so we can't reject Ho

and conclude that the observed data set provides no strong evidence against the

null hypothesis.

The 95% confidence interval based on the sample mean (now we

suppose that population mean is uknown with known standard deviation) is

m±Z*σ/sqrt(n), where Z*=Z cv for level alpha/2 (! because to get an interval we

need to use two-sided estimation). Z*=1.96 and 95% confidence interval is

(1030-41.32;1030+41.32)=(988.68,1071.32)

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