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KEY_Statistics_Test_4.pdf.docxStascs Test 4 11/16/17 Mulple Choice:
KEY Statistics Test 4.pdf.docxSta...
KEY_Statistics_Test_4.pdf.docxStascs Test 4 11/16/17 Mulple Choice:
Page 1
Stascs Test 4
11/16/17
Mulple Choice:
1. A test to screen for a minor ailment is similar to hypothesis tesng, with a null hypothesis of no sickness, and an
alternave hypothesis of sickness. If the null hypothesis is rejected treatment will be given. Otherwise, it will not.
Assuming the treatment can have very serious side eﬀects if the paent is not sick, it is beer to increase the probability
of:
A. making a Type 1 error, providing treatment when it is not needed.
B. making a Type 1 error, not providing treatment when it is needed.
C. making a Type 2 error, providing treatment when it is not needed.
D. making a Type 2 error, not providing treatment when it is needed.
2. Which of the following null hypotheses would be tested using a chisquare goodnessofﬁt test?
A.
The probabilies that a family with two children will have 2 boys, 1 boy and 1 girl, and 2 girls are ¼, ½ and
¼, respecvely.
B.
There is no relaonship between frequent cell phone use (yes, no) and ADHD diagnosis (yes, no).
C.
There is no relaonship between age and opinion on having children.
D.
The probability that a randomly selected person age 50 of older has moderate to severe hearing loss is 0.73.
3.
ANSWER: D
4.
ANSWER: B
Page 2
5.
ANSWER: C
6. Which of the following is FALSE?
A.
small values of alpha minimize the risk of type I errors.
B.
the statement of the alternave hypothesis never contains an equality
C. an increase in the risk of type I error also increases the risk of a type II error
D.
the signiﬁcance level of a test is equal to the probability of a type I error
Chi Square Goodness of Fit Formula:
Page 3
Free Response:
For every hypothesis test be sure to clearly indicate the
type of test
, the
null and alternave hypothesis
,
clearly state the
decision
, and write a
concluding
statement
.
7. The mean lasng me of 2 compeng ﬂoor waxes is to be compared. Thirty ﬂoors are randomly assigned to test
each wax. The following table is the result. Does the data indicate that wax 1 is more eﬀecve than wax 2? Test at a 5%
level of signiﬁcance. (Assume normality)
2 samp z Test
Ho: u1=u2
Ha: u1>u2
Pvalue=0.1310
Alpha = 0.05
Fail to Reject Ho. There is insuﬃcient evidence to show that wax 1 is more eﬀecve than wax 2
8. A new drug is being proposed for the treatment of hair loss. Unfortunately some users in early tests of the drug have
reported the growth of trichilemmal cysts as a side eﬀect. The FDA will reject the drug if it thinks that more than 5% of
the populaon would suﬀer from this side eﬀect. Out of 650 people parcipang in the drug trial, 80 of them report cyst
growth as a side eﬀect. At 1% signiﬁcance, do we have reason to believe that the FDA will reject the drug?
1 prop
Ho: p≤0.05
Ha: p>0.05
Pvalue=6.31x10
18
Alpha = 0.01
Reject Ho. There is suﬃcient evidence to show that more than 5% of the populaon would suﬀer cysts as a
side eﬀect. The FDA will reject the drug.
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