It is known that in 2013 the population mean for the writing section of the SAT is 488, with a standard deviation of 114. A sample of 100 students taking the SAT whose family income was less than $20,000 had a verbal score of 429. a. Perform a two-tailed hypothesis test to determine whether the group whose family income was less than $20,000 had a significantly different score on average than the population. State the test statistic and your decision. (3pts) b. Assume that rather than a sample of 100, the sample size was only 10. Perform the same hypothesis test again with this new sample size. State the test statistic and your decision. (2pts)

We are given that the population mean for the writing section of the SAT is 488 i.e. [tex]\mu[/tex] = 488 and Population Standard deviation,[tex]\sigma[/tex] = 114 .

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 488

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 488

The test statistics we will use here is;

[tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1) where, [tex]Xbar[/tex] = Sample mean

n = sample size = 100

So, test statistics = [tex]\frac{429-488}{\frac{114}{\sqrt{100} } }[/tex] = -5.1754

Since, significance level is not given in question so we assume it to be 5% and at this level of significance the z table gives critical value of -1.96 to 1.96. Now our test statistics falls below the critical value of -1.96 which means it falls in the rejection region so we have sufficient evidence to reject null hypothesis.

So, we conclude that the group whose family income was less than $20,000 had a significantly different score on average than the population.

(b) Now only the sample size have been revised to 10 instead of 100.

So, new test statistics = [tex]\frac{429-488}{\frac{114}{\sqrt{10} } }[/tex] = -1.64

Since, significance level is not given in question so we assume it to be 5% and at this level of significance the z table gives critical value of -1.96 to 1.96. Now our test statistics falls inside the range of critical value of -1.96 and 1.96 as -1.96 < -1.64 which means it does not fall in the rejection region so we have sufficient evidence to accept null hypothesis.

So, we conclude that the group whose family income was less than $20,000 had a score of 488.