Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Answer:
A = pi 9/16 cm^2
Step-by-step explanation:
The circumference is 3/2 pi
The circumference is given by
C = 2 * pi *r
3/2 pi = 2 * pi *r
Divide each side by 2 *pi
3/2 pi / 2 pi = 2 pi r /2pi
3/4 cm = r
Now we can find the area
A = pi r^2
A = pi ( 3/4) ^2
A = pi 9/16 cm^2
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
&
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
=
=
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4

= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 -
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
Answer:
The 99% confidence interval for the population mean is 22.96 to 26.64
Step-by-step explanation:
Consider the provided information,
A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,
The confidence interval if 99%.
Thus, 1-α=0.99
α=0.01
Now we need to determine 
Now by using z score table we find that 
The boundaries of the confidence interval are:

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64
Why....are you asking..$50 is least expensive. Because you use about 36 gigabytes at least so 5•x=
Or $50+12