Download Answers to Multiple Choice Questions

Answers to Multiple Choice Questions
1.
E
a is false because we either reject or fail to reject H0. We cannot say if H0 is true or false.
b is wrong because the Ha is stated as a mean or a proportion, not a z-value or t-value
c is wrong. SRSs are needed for both tests of significance.
d It may or may not be true. We just haven’t rejected it.
e YES-> the larger the p-value the more likely one is to accept (fail to reject) H0
2.
E
a is false because we assume H0 is true.
b- no, the p-value is the probability that we would get an answer as extreme as the one we got, given that the H0
is true. It is NOT the probability that H0 is true.
c- no, the opposite- a larger p-value means you are more likely to accept the H0. That means you are less likely
to reject the H0.
d- no, you should make these decisions before you collect data
e- YES- one-sided means we only care if data is less than or greater than a cutoff
3.
A parameter is the value of some variable in a whole population (not in sample) that we are trying to
measure. So the answer is E)- we are measuring the amount of time that middle schoolers spend in arcades.
4.
We know that the PTA are the ones conducting the study and they think that middle schoolers spend more
than 2 hours at arcades. So this is our Ha because this is what the researchers believe to be true:
Ha : 𝜇 > 2
5.
6.
None of these- the p-value means that there is 3.2% chance of getting an answer as extreme as the one the
PTA did if the H0 were true. So if middle schoolers only spent 2 hours at the arcade, then there is only a
3.2% chance that we would get the answers we did. (That’s a pretty small percentage).
Type 1 errors are false positives- saying that something is true when it isn’t. In this case, it’s saying that the
strength has dropped too low when it hasn’t. That means that the company will stop making the bungee
cords and will lose money. The answer is C.
7. Type II errors are false negatives- for example, saying someone doesn’t have ESP when they actually do.
The answer is C.
8. The question asks us to find the probability that the consumer will reject the claim even though the claim is
true. So let’s assume that it’s true, and that the coffee machine really does put out 12 ounces each time.
 Our H0: 𝜇= 12 and Ha: 𝜇 < 12.

Calculate the standard error:
SE =


𝑠
√𝑛
where s is the standard deviation of the sample (which the problem tells us is 0.9) and n is
the number of samples (which the problem tells us is 5).
Calculate the degrees of freedom:
df = n- 1= 5-1 = 4
Calculate the t-statistic:
𝑥̅ − 𝜇 11.5 − 12
𝑡=
=
= −1.24
.9
𝑆𝐸
√5

Use normalcdf (-1000,-1.24) to find the probability that the consumer thinks the coffee machine is
dispensing less coffee than it says it should.
=.107
The answer is B.
9.
The answer is C.
𝑠

Calculate SE = =

8.3) and n is the number of samples (which the problem tells us is 40).
Calculate the degrees of freedom:
√𝑛
where s is the standard deviation of the sample (which the problem tells us is
df = n – 1 = 39

Calculate the t-statistic:
𝑡=
𝑥̅ − 𝜇 59.5 − 58.4
=
8.3
𝑆𝐸
√40

Because the pharmaceutical company claims that their medicine lasts 58.4 minutes, we’re interested
in whether the effect lasts more or less than that time, so we’ll need to multiply our P-value by 2.
 Rule out A and B because they don’t multiply times 2. Rule out D because it has the incorrect
degrees of freedom.
10. The answer is B. You would change the significance level (𝛼), not the sample size, to change the probability
that you commit a Type I error.
11. The answer is D.
 H0: 𝜇 = 150 𝑔
 Ha: 𝜇 < 150 𝑔
𝑠

SE:

df = 14

𝑡=

√
=
𝑛
𝑥̅ −𝜇
𝑆𝐸
12.81
√15
=
145.8−150
12.81
√15
= −1.27
P-value: normalcdf(-100000000, -1.27) = .10 . This means that, given that the null hypothesis is
true, at a 𝛼 = .05 significance level , the odds of getting a value as extreme as 145.8 is actually pretty
likely. So, we cannot reject the null hypothesis.
12. Omit- not in this playlist
13. Type I is a false positive, so that would mean saying that the storm is hitting but it really doesn’t. Type II is
a false negative, so that would mean saying that the storm won’t hit but it really does. The answer is B.
14. The answer is_______. We use a z-test for one sample because this is a proportion problem.
 H0: 𝑝 = .08
 Ha: 𝑝 ≠ .08
𝑝 (1−𝑝)
(.086)(1−.086)

𝜎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 ∶ √

(I got the .086 because 198 out of 2300 patients had heart attacks and 198/2300 = .086.

𝑧=
𝑝−𝑃
𝜎
=
.086− .08
.0058
𝑛
= √
2300
= .0058
= 1.03


P (1.03) = .1515 (I got this number by looking at a z-table)
Since the probability of getting a proportion as extreme as .086 is pretty likely (.1515 is pretty high
probability) given that the null hypothesis is true, we fail to reject the null hypothesis. The answer is
E.
15. Omit- not in this playlist
16. E. We can’t predict the probability of a Type II error. If it had asked about a Type I error, then we could
have answered, because 𝛼 is actually the probability of a Type I error.
17. H0: p = .5
Ha >.5

(I got the .54 because 305 out of 565 cancers occurred on the left side, and 305/565 = .54..

19.
20.
21.
22.
23.
(.54).46)
𝜎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 ∶ √

18.
𝑝 (1−𝑝)

𝑧=
𝑝−𝑃
𝜎
=
.54−.50
.02
𝑛
= √
565
= .02
=2
The answer is C. It is a one-sided test (the researcher wants to see if the cancers are more on only
one side), so you can eliminate B and E. I reject A because the .5 is not the mean of the sample.
This leaves C and D, and D has an equals sign after the z. The researcher is interested in anything
above 50%, not equal to 50%. The answer is C.
There is a 0.01 chance of getting a Type I error each time I do the test. The probability of getting no Type I
errors in all 5 trials is .99*.99*.99*.99*.99= .95. So the probability of getting at least one Type 1 error is
about 5% (1.00- .95 = .05). The answer is C.
A larger sample size with a mean problem means we reduce the probabilities of both Type I and Type II
errors, so eliminate B, D and E. Now we just have to figure out if the power has been increased or reduced.
Because we have reduced the chance of getting Type II errors, we have increased the power of the test. The
answer is C.
Omit – not in this playlist
When we reduce the 𝛼, we reduce the probability that we get a Type I error, but we increase the probability
that we get a Type II error. The more power, the less likely you are to get a Type II error. Since we just
increased the probability that we get a Type II error, we have reduced the power. The answer is C.
Okay, first off, if we increase the risk of Type II error, we decrease the power, so we can eliminate A and D
right off the bat. The further away from the true value of the population the sample is, the smaller our pvalue is going to be, and we are going to be more and more confident in our answers. We are reducing the
risk for any error, including Type II errors, which increases the power. The answer is B.
H0: 𝜇 = 6
 Ha: 𝜇 > 6

Next, I calculate the mean and standard deviation of the numbers in the table. Their mean is 6.48
and the standard deviation is 1.36.

SE:

df = 24

𝑡=
𝑠
√𝑛
=
𝑥̅ −𝜇
𝑆𝐸
1.36
√25
=
= .272
6.48−6
.272
= 1.76

The p-value (using a p-value calculator) is .04, which means that there is a 4% chance of getting a
mean as extreme as 6.48 given the null hypothesis is true. .04 is between .01 and .05 so the answer is
C.
24. A power of .75 is pretty high, so I’m not likely to commit a Type II error with a test this powerful. You can
eliminate A, because power is not related to Type I errors. B has a high probability of committing a Type II
error, and I just said that I’m not likely to commit a Type II error. Eliminate B. A Type II error is about
failing to reject a null hypothesis when it is incorrect. You can eliminate D, because that answer choice is
about the null hypothesis being correct. You can eliminate E, because again, it has a high probability of me
failing to reject a null hypothesis when it’s false, and I’m not likely to do that because the test has pretty high
power. The answer is C.
25. This question is asking about the probability that the students commit a Type II error, or fail to reject the
manager’s incorrect claim. So let’s do the test:
𝑠

SE:

df = 99

𝑡=
√
=
𝑛
𝑥̅ −𝜇
𝑆𝐸
150
√100
=
= 15
375−385
15
= −.66667
 P(-.66667) = .2525
So there is a 25% chance of getting an answer this extreme given that the null hypothesis is true. The
answer is B.