In: Statistics and Probability
A simple random sample of 100 adults is obtained, and each person’s red blood cell count (in cells per microliter) is measured. The sample mean is 5.22 and the sample standard deviation is 0.53. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 5.3, the calculated value of t-test statistic is (Given that H0: µ = 5.3, Ha:µ < 5.3)
choose
-15.09
15.09
-1.509
1.509
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Three students took a statistics test before and after coaching, but coaching did not effect the scores of students i.e mean change in scores is zero. Their scores are as follows:
Students |
A |
B |
C |
Before |
71 |
88 |
63 |
After |
70 |
89 |
60 |
The value of t-test statistic for matched pairs is
choose
8.66
0.866
-0.866
-8.66
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The basic procedure of hypothesis testing is to make an initial assumption about the population parameter, collect evidence and decide whether to "reject" or "not reject" our initial assumption.
choose
True
False
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1)
population mean μ= | 5.3 |
sample mean 'x̄= | 5.220 |
sample size n= | 100.00 |
sample std deviation s= | 0.530 |
std error 'sx=s/√n= | 0.0530 |
test stat t ='(x-μ)*√n/sx= | -1.509 |
2)
S. No | before | after | diff:(d)=x1-x2 | d2 |
1 | 71 | 70 | 1 | 1.00 |
2 | 88 | 89 | -1 | 1.00 |
3 | 63 | 60 | 3 | 9.00 |
total | = | Σd=3 | Σd2=11 | |
mean dbar= | d̅ = | 1.0000 | ||
degree of freedom =n-1 = | 2.000 | |||
Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) = | 2.0000 | |||
std error=Se=SD/√n= | 1.1547 | |||
test statistic = | (d̅-μd)/Se = | 0.8660 |
3)
true