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KEY_Statistics_Test_4.pdf.docx-Stascs Test 4 11/16/17 Mulple Choice:
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KEY Statistics Test 4.pdf.docx-Sta...
KEY_Statistics_Test_4.pdf.docx-Stascs Test 4 11/16/17 Mulple Choice:
##### Page 1
Stascs Test 4
11/16/17
Mulple Choice:
1. A test to screen for a minor ailment is similar to hypothesis tesng, with a null hypothesis of no sickness, and an
alternave 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 paent is not sick, it is beer 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 chi-square goodness-of-ﬁt test?
A.
The probabilies that a family with two children will have 2 boys, 1 boy and 1 girl, and 2 girls are ¼, ½ and
¼, respecvely.
B.
There is no relaonship between frequent cell phone use (yes, no) and ADHD diagnosis (yes, no).
C.
There is no relaonship 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.
4.

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5.
6. Which of the following is FALSE?
A.
small values of alpha minimize the risk of type I errors.
B.
the statement of the alternave 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 alternave hypothesis
,
clearly state the
decision
, and write a
concluding
statement
.
7. The mean lasng me of 2 compeng 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 eﬀecve 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 eﬀecve 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 populaon would suﬀer from this side eﬀect. Out of 650 people parcipang 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 populaon would suﬀer cysts as a
side eﬀect. The FDA will reject the drug.

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