In: Statistics and Probability
6.251 Exercise 6.250 describes pulse rates collected from 10 students, once during a quiz and once during a lecture. the data are given in table 6.23 and stored in quiz pulse 10. we might expect mean pulse rates to increase under the stress of a quiz. use the information in exercise 6.250 and the data in table 6.23 to test whether the data provide sufficient evidence to support this claim. conduct a March paired test and explain why a matched paired test is appropriate.
matched pairs not march
We have to test mean pulse rates to increase under the stress of a quiz.
We have to use matched paired t-test because here we are taking observation on same 10 student for two different situations.In such situationwe shoulduse matched paired t-test .
Here we have
mean of pulse rate of student in quiz is 68.8
mean of pulse rate of student in lecture is 66.1
sample SD of pulse rate of student in quiz is 12.5
sample SD of pulse rate of student in lecture is 12.8
Thus
Now we have,
Total | mean | s | |||||||||||
Quiz | 75 | 52 | 52 | 80 | 56 | 90 | 76 | 71 | 70 | 66 | 688 | 68.8 | 12.506 |
Lecture | 73 | 53 | 47 | 88 | 55 | 70 | 61 | 75 | 61 | 78 | 661 | 66.1 | 12.80148 |
Difference | 2 | -1 | 5 | -8 | 1 | 20 | 15 | -4 | 9 | -12 | 27 | 2.7 | |
(diff -mean diff)2 | 0.49 | 13.69 | 5.29 | 114.49 | 2.89 | 299.29 | 151.29 | 44.89 | 39.69 | 216.09 | 888.1 |
Standard deviation of difference is
We test below hypothesis
H0: mean of pulse rate of student in quiz = mean of pulse rate of student in lecture
Ha: mean of pulse rate of student in quiz > mean of pulse rate of student in lecture
t= (Mean_Q - Mean_L)/ (Sd/sqrt(n))
=(68.8-66.1)/(9.93/sqrt(10))
=0.859834
P-value
p(T(df=9)> 0.8598)= 0.2614
P-value=0.2614
Critical Value= t(0.025,9)=2.26
t < critical Value We do not reject H0
Conclusion
Hence there is no sufficient evidence to support this claim of mean pulse rates to increase under the stress of a quiz