Question

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. (Source.


www.ets.org) Suppose a random sample of 10 economics students took the test, and their scores are given below.


158, 166, 158, 149, 154, 164, 169, 144, 155, 160


Test the claim that the mean quantitative score is greater than 152.1 at the 0.10 level of significance.


i. State your null and alternative hypothesis:

ii. Describe what you type into your calculator to have it run the numbers.

iii. State your P-value

iv. State your conclusion is a way a non-stat student would understand.

Answer 1

Answer:

i) Null hypothesis:$\mu \leq 152.1$    

Alternative hypothesis:$\mu > 152.1$  

ii) Assuming that we have a TI84 or TI83 calculator we need to follow these steps:

a) Press STAT

b) Press on Edit > 1: Edit..., Enter

c) On the column L1, put all the 10 values below

d) Press 2nd and then Mode

e) Press on STAT

f) Move to the right to the option TESTS

g) Select the second option T-Test..

h) On Inpt: select Data, the value for $\mu_o =152.1$, List: L1, Freq: 1, and on the $\mu : >\mu_o$

i) Finally press on Calculate

iii) $p_v =P(t_{(9)}>2.316)=0.0229$    

iv) If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 152.1 at 10% of significance.  

Step-by-step explanation:

Data given and notation    

Data: 158, 166, 158, 149, 154, 164, 169, 144, 155, 160

We can find the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^n}{n-1}}$

$\bar X=157.7$ represent the sample mean    

$s=7.646$ represent the sample standard deviation    

$n=25$ sample size    

$\mu_o =152.1$ represent the value that we want to test    

$\alpha=0.1$ represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

i. State your null and alternative hypothesis:

We need to conduct a hypothesis in order to check if the mean is higher than 152.1, the system of hypothesis are :    

Null hypothesis:$\mu \leq 152.1$    

Alternative hypothesis:$\mu > 152.1$    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$t=\frac{157.7-152.1}{\frac{7.646}{\sqrt{10}}}=2.316$    

ii. Describe what you type into your calculator to have it run the numbers.

Assuming that we have a TI84 or TI83 calculator we need to follow these steps:

a) Press STAT

b) Press on Edit > 1: Edit..., Enter

c) On the column L1, put all the 10 values below

d) Press 2nd and then Mode

e) Press on STAT

f) Move to the right to the option TESTS

g) Select the second option T-Test..

h) On Inpt: select Data, the value for $\mu_o =152.1$, List: L1, Freq: 1, and on the $\mu : >\mu_o$

i) Finally press on Calculate

iii. State your P-value

P-value    

First we need to calculate the degrees of freedom given by:  

$df=n-1=10-1=9$  

Since is a one-side upper test the p value would be:    

$p_v =P(t_{(9)}>2.316)=0.0229$    

iv. State your conclusion is a way a non-stat student would understand.

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 152.1 at 10% of significance.