Question

The Survey of study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher who suspects that older students have better attitudes toward school gives the SSHA to 25 students who are at least 30 years of age. Their mean score is = 133.2. Assuming that sigma = 30 for the population of older students, carry out a test of H0 : mu = 115 Ha : mu>115 What is the P-value of your test? 0.9976 0.0024 0.0012 0.9988

Answer 1

Answer:

$p_v =P(z>3.033)=0.001209$  

Best option: Pv=0.0012

Step-by-step explanation:

1) Data given and notation      

$\bar X=133.2$ represent the mean score for the sample  

$\sigma=30$ represent the standard deviation for the population      

$n=25$ sample size      

$\mu_o =115$ represent the value that we want to test    

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

z would represent the statistic (variable of interest)      

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

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean score is at least 15 for people older than 30, the system of hypothesis would be:    

Null hypothesis:$\mu \geq 115$      

Alternative hypothesis:$\mu > 115$      

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:      

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)      

z-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:      

$z=\frac{133.2-115}{\frac{30}{\sqrt{25}}}=3.033$      

Calculate the P-value      

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

$p_v =P(z>3.033)=0.001209$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the the actual mean score for people older than 30 seems to be higher than 115