Question

A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the administrator of the​ exam, scores are normally distributed with mu equals 517. The teacher obtains a random sample of 2200 ​students, puts them through the review​ class, and finds that the mean math score of the 2200 students is 524 with a standard deviation of 118 . Complete parts​ (a) through​ (d) below. ​(a) State the null and alternative hypotheses. Let mu be the mean score. Choose the correct answer below. A. Upper H 0 : mu less than 517 ​, Upper H 1 : mu greater than 517 B. Upper H 0 : mu equals 517 ​, Upper H 1 : mu not equals 517 C. Upper H 0 : mu greater than 517 ​, Upper H 1 : mu not equals 517 D. Upper H 0 : mu equals 517 ​, Upper H 1 : mu greater than 517 Your answer is correct. ​(b) Test the hypothesis at the alpha equals 0.10 level of significance. Is a mean math score of 524 statistically significantly higher than 517 ​? Conduct a hypothesis test using the​ P-value approach. Find the test statistic. t 0 equals2.78 ​(Round to two decimal places as​ needed.) Find the​ P-value. The​ P-value is 0.003 . ​(Round to three decimal places as​ needed.) Is the sample mean statistically significantly​ higher? No Yes Your answer is correct. ​(c) Do you think that a mean math score of 524 versus 517 will affect the decision of a school admissions​ administrator? In other​ words, does the increase in the score have any practical​ significance? ​No, because the score became only 1.35 ​% greater. Your answer is correct. ​Yes, because every increase in score is practically significant. ​(d) Test the hypothesis at the alpha equals0.10 level of significance with nequals 350 students. Assume that the sample mean is still 524 and the sample standard deviation is still 118 . Is a sample mean of 524 significantly more than 517 ​? Conduct a hypothesis test using the​ P-value approach. Find the test statistic. t 0 equals1.11 ​(Round to two decimal places as​ needed.)

Answer 1

Let X be the working hours of employees        
         
n = 2200  Sample Size     
x̅ = 524  Sample Mean     
s = 118  Sample Standard Deviation     
μ = 517  Population Mean     
Let α = 0.10   Level of significance = 5%     
         
Since the population standard deviation is unknown, we use t-test        
a) The null and alternative hypotheses are        
Answer : Option D        
Ho : μ = 517        
H1 : μ > 517        
         
b) Test statistic t is given by         

Test statistic = t = 2.78        
         
This is a one tailed right side test        
         
df = Degrees of freedom = n - 1 = 2200 - 1 = 2199        
We find the p-value using Excel function T.DIST.RT        
p-value = T.DIST.RT(2.7825, 2199)        
                 = 1 - 0.9839 = 0.0027       
p-value = 0.003   Rounded to 3 decimals     
         
Since 0.0027 < 0.10        
that is p-value < α        
we Reject Ho        
         
Is the sample mean statistically significant        
Answer : YES        
         
c) The increase in score does not have a practical significance , because it does not show         
a considerably higher increase        
Answer :        
No, because the score became only 1.35 ​% greater.        
         
d) n = 350        
Test statistic t is given by         
    
Test statistic = t = 1.11        
         
This is a one tailed right side test        
         
df = Degrees of freedom = n - 1 = 350 - 1 = 349        
We find the p-value using Excel function T.DIST.RT        
p-value = T.DIST.RT(1.11, 349)        
                 = 1 - 0.9839 = 0.134       
p-value = 0.134   Rounded to 3 decimals     
         
Since 0.134 > 0.10        
that is p-value > α        
we DO NOT reject Ho        
         
Conclusion :        
There does not exist sufficient statistically significant evidence at 10% level of significance        
to conclude that the sample mean of 524 is significantly more than 517 ​