In: Statistics and Probability
13. You wish to test the following claim (Ha) at a significance
level of α=0.002.
Ho:p1 = p2
Ha:p1 > p2
You obtain 17 successes in a sample of size n1=387 from the first
population. You obtain 15 successes in a sample of size n2=476 from
the second population. For this test, you should NOT use the
continuity correction, and you should use the normal distribution
as an approximation for the binomial distribution. What is the
p-value for this sample? (Report answer accurate to four decimal
places.)
p-value =
14. You wish to test the following claim (Ha) at a significance
level of α=0.001.
Ho:μ1 = μ2
Ha:μ1 > μ2
You believe both populations are normally distributed, but you do
not know the standard deviations for either. However, you also have
no reason to believe the variances of the two populations are not
equal. You obtain a sample of size n1=23 with a mean of M1=85.7 and
a standard deviation of SD1=19.6 from the first population. You
obtain a sample of size n2=19 with a mean of M2=71.7 and a standard
deviation of SD2=15.4 from the second population.
What is the p-value for this test? For this calculation, use the
conservative under-estimate for the degrees of freedom as mentioned
in the textbook. (Report answer accurate to four decimal
places.)
p-value =
13)
Ho: p1 - p2
= 0
Ha: p1 - p2
> 0
first sample size,
n1= 387
number of successes, sample 1 =
x1= 17
proportion success of sample 1 ,
p̂1= x1/n1= 0.044
second sample size,
n2
= 476
number of successes, sample 2 =
x2 =
15
proportion success of sample 1 , p̂ 2= x2/n2
= 0.031512605
difference in sample proportions, p̂1 - p̂2
= 0.012415044
pooled proportion , p
= (x1+x2)/(n1+n2)= 0.0371
std error ,SE = =SQRT(p*(1-p)*(1/n1+
1/n2)= 0.0129
Z-statistic = (p̂1 - p̂2)/SE =
0.9599
z-critical value , Z* =
1.6449
p-value =
0.1685 [excel
functon: =normsdist(-0.9599)
P-value >α, fail to reject Ho
-------------------------------
14)
Ho : µ1 - µ2
= 0
Ha : µ1-µ2
> 0
Level of Significance , α =
0.001
mean of sample 1,
x̅1= 85.7
standard deviation of sample 1, s1 =
19.6
size of sample 1,
n1= 23
mean of sample 2,
x̅2= 71.7
standard deviation of sample 2, s2 =
15.4
size of sample 2,
n2= 19
difference in sample means = x̅1-x̅2
= 14.0000
std error , SE = √(s1²/n1+s2²/n2) =
5.4023
t-statistic = ((x̅1-x̅2)-µd)/SE
= 2.5915
DF = min(n1-1 , n2-1 )= 18
p-value=0.0092
p-value>α , Do not reject null hypothesis