Question

2. The Graduate Record Examination (GRE) is a test required for admission to many U.S. graduate schools. Students' scores on the quantitative portion of the GRE follow a normal distribution with standard deviation 8.8. (Source: http://www.ets.org/.) Suppose a random sample of 10 students took the test, and their scores are given below: 152, 126, 146, 149, 152, 164, 139, 134, 145, 136. Test the claim that the mean verbal reasoning score is different from 150 using the P-value approach at the 0.05 level of significance. Han P-value (three decimal places) Ha: Conclusion (Reject or Fail to Reject): Interpretation: 3. The National Center for education Statistics monitors many aspects of elementary and secondary education nationwide. Their 1996 numbers are often used as a baseline to assess changes. In 1996, 31% of students reported that their mothers had graduated from college. In 2000, responses from 8368 students found that that figure had grown to 34%. Is this evidence of an increase in education level among mothers? Use a = 0.05 level of significance. Hai Ha: P-value (three decimal places) Conclusion (Reject or Fail to Reject): Interpretation:

Answer 1

2. Given that the population sd of the GRE scores is $\sigma=8.8$ . The given data is

	GRE
	152
	126
	146
	149
	152
	164
	139
	134
	145
	136
Total	1443

The mean of the given data is $\bar{x}=\frac{\sum_{i=1} ^{n}x_{i}}{n}$

$\bar{x}=\frac{1443}{10}=144.3$

The Null and alternate hypothesis are:

$H_{0}:\mu=150$ and    p-value = 0.0406

OST +11:"H

Conclusion: Reject .

он

Interpretation: Since the Null hypothesis is rejected, we conclude that there are enough evidence to claim that the mean verbal reasoning score is different from 150.

3. This is a case of Z-test for one proportion problem.

It is given that in 1996 the 31% of students' mothers were graduates.

In 2000, the proportion is 34% from a survey of 8368 students.

$H_{0}:p=0.31$

$H_{1}:p>0.31$

p-value=0.000

Conclusion: Reject the Null hypothesis

Interpretation: Since p-value<0.05, we reject the Null hypothesis. We conclude that there is an evidence of increase in mother's education level.

---------------------------------------------------------------------Work outs-----------------------------------------------------------------------------------

The Z-test statistic:

$Z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$

$Z=\frac{144.3-150}{8.8/\sqrt{10}}$

$Z=\frac{-5.7}{8.8/3.1623}$

$Z=\frac{-5.7}{2.7828}$

$Z=-2.0483$

The p-value =0.0406(By using the 2*NORM.S.DIST(-2.0483,TRUE) function in EXCEL).

3). The test statistic:$Z=\frac{p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$

$Z=\frac{0.34-0.31}{\sqrt{\frac{0.31(1-0.31)}{8368}}}$

$Z=\frac{0.03}{\sqrt{\frac{0.2139}{8368}}}$

  $Z=\frac{0.03}{\sqrt{0.00002556}}$

  $Z=\frac{0.03}{0.005056}$

  $Z=5.9337$