In: Statistics and Probability
QUESTION 37
Match the appropriate term to the definition or technique. Each technique can be used only once. As such, choice the best definition
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QUESTION 30
A bank auditor claims that credit card balances are normally distributed with a mean of $2,870 and a standard deviation of $1000. You randomly select 25 credit card holders. What is the probability that their mean credit card balance is less than $2,500
a. |
0.0 |
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b. |
0.0197 |
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c. |
.0322 |
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d. |
.0505 |
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e. |
1.0 |
What is the Z value for variable x equal to 15 and the mean is 20 with a standard deviation of 5
a. |
-2.0 |
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b. |
2.0 |
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c. |
-1.0 |
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d. |
1.0 |
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e. |
Cannot answer as there is not enough information provided |
question 37
A. hypothesis test
the most suitable definition is : A test to compare means of sample or proportion data to a claims.
because here one makes a claim and then creates the null and alternative hypothesis and then carry out testing based on the given data at hand. hypothesis test is appropriate for one population problem or two population problems
B. ANOVA
but when the testing extends to three or more populations and to compare the behaviour of the populations based on 1 characteristics then general hypothesis test fails here. What we need is Analysis of variance that is ANOVA
the most suitable defintion is: A statistical hypothesis test for comparison of three or more sample means.
C. Two way Anova
when one wants to carry out anova but based on two factors then we need two way anova
the most suitable defintion is: Extends to situations where two factors serve to explain variation in response
D. Chi square/ goodness of fit testing
All the above mentioned tests are basically parametric tests where we have a underlying population structure. But there are instances where we can not assume any parametric form. This test comes handy that time.
hence the most suitable definition is : A non parametric test to compare the distribution of observed data to expected data.
question 30.
let X be the random variable denoting the credit card balance in $
then X~N(2870,10002)
now we have a random sample of (X1,X2,...,X25) 25 credit card balances.
then mean be Xbar=(X1+X2+...+X25)/25
since Xbar is a linear combination of independent normal variates, it is also a normal variate with mean
E[Xbar]=E[X1+X2+...+X25]/25=25*E[X]/25=2870
and variance= V[Xbar]=V[X1+X2+...+X25]/252=25*V[Xbar]/252=V[Xbar]/25
hence standard deviation=sqrt(V[Xbar])=1000/5=200
so Xbar~N(2870,2002)
so P[Xbar<2500]=P[(Xbar-2870)/200<(2500-2870)/200]=P[Z<-1.85] where Z~N(0,1)
=0.03215677 [using pnorm() function in R]
=0.0322 [option C]
Z value= [x-mean(X)]/SD(X)
now x=15 mean(X)=20 SD(X)=5
so Z value=(15-20)/5=-1.0 [option C]