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March Statewide Invitational
Answers and Solutions
1.
2.
3.
4.
5.
A
D
D
B
E
6. E
7. E
8. E
9. C
10. C
11.
12.
13.
14.
15.
D
A
B
D
C
Statistics Individual
16.
17.
18.
19.
20.
B
A
B
A
B
21.
22.
23.
24.
25.
D
B
A
D
D
26.
27.
28.
29.
30.
D
B
E
B
B
1. A – The missing mean of the girls’ scores is (5+5+5+5+5+4+4+4+4+1) / 10 = 4.2 and the median is (5+4) / 2 = 4.5.
The missing Q3 for the boys’ scores is 2 and entering the data into a list in the calculator and running the 1-Var Stats
function yields a sample standard deviation of 1.251666. The mean difference is (4+4+4+4+3+2+2+2-4) / 10 = 2.5 and
the median difference is (4+3) / 2 = 3.5. The requested sum is (4.2 + 4.5 + 2 + 1.251666 + 2.5 + 3.5) = 17.951666.
2. D – The four assumptions and conditions listed all require verification prior to performing a two-sample t-test.
3. D – Since Mr. Callow simply used his 20 readily available students, it is a convenience sample.
4. B – The table below summarizes the results (Note: the three outliers are somewhat obvious):
Lower Fence
Upper Fence
Outlier
Outlier Z-Score
Girl
4 – 1.5(5 – 4) = 2.5
5 + 1.5(5 – 4) = 6.5
1
(1 – 4.2) / 1.229273 = -2.603165
Boy
1 – 1.5(2 – 1) = -0.5
2 + 1.5(5 – 4) = 3.5
5
(5 – 1.7) / 1.251666 = 2.636486
Difference
2 – 1.5(4 – 2) = -1
4 + 1.5(4 – 2) = 7
-4
(-4 – 2.5) / 2.460804 = -2.641413
The mean of the Z-Scores is (-2.603165 + 2.636486 – 2.641413) / 3 = -0.8694 when rounded.
5. E – Since all four assumptions and conditions listed in question #2 are violated, the independent two-sample t-test is
clearly eliminated and since at least one of those conditions is also required in each of the other inference procedure
options, they are all eliminated from contention. This also forces E as the correct answer for the next two questions.
6. E – Since multiple inference assumptions and conditions are violated, the requested computations are meaningless.
7. E – Since multiple inference assumptions and conditions are violated, no valid conclusions can be drawn.
8. E – Since Matt is the only boy who passed the exam, the probability is 1.
9. C – Since there are 160 total students at Mr. Callow’s school with presumably 160 / 4 = 40 seniors of which 20 are
girls and 20 are boys, the probability of randomly selecting 10 girls and 10 boys in a SRS of 20 out of the 40 seniors is
(20C10)(20C10) / 40C20 = 0.248 when rounded.
10. C – The chi-square tests have no Normality assumption nor do they make any assumptions about any population
parameters. The other choices all do with respect to either the data coming from a Normally distributed population or
they require sufficiently large samples so that their sampling distributions are approximately Normal (according to the
CLT) with particular assumptions made about the means and standard deviations, whether known or unknown.
46
23
11. D – There are 46 total 4-game sweeps out of 234 total championship series, so the probability is: 234 = 117.
12. A – There are 18 + 20 = 38 MLB World Series which lasted 4 or 6 gamers out of 92 total series. Therefore, the
38
19
probability is: 92 = 46.
13. B – There are 66 total NBA Championship series and a total of 53 + 66 = 119 series that lasted 5 or 6 games, of
66+119−40
145
which 16 + 24 = 40 qualify as both. Therefore, the probability is:
= 234.
234
16
4
14. D – There are 16 out of 76 NHL Stanley Cup series which lasted 7 games, so the probability is: 76 = 19.
1
March Statewide Invitational
Statistics Individual
15. C – The 99% confidence interval for the proportion of 7-game Stanley Cup Championship Series is given by:
16 60
×
± 2.576√767676 → (0.09006, 0.33099) → (9%, 33%) when rounded to the nearest whole percent. Alternatively,
entering x = 16, n = 76, and C-Level = .99 into the 1-PropZInt in the calculator produces the same confidence interval.
16
76
16. B – Since either of the two teams can win the series and the 4th win must occur in the last game of the series, the three
missing probabilities (expected proportions) for a championship series lasting 5, 6, or 7 games are the following binomial
probabilities: P(5 Games) = 2[4C3(.5)3(.5)1(.5)] = 4(.5)4 = 0.2500, P(6 Games) = 2[5C3(.5)3(.5)2(.5)] = 10(.5)5 = 0.3125,
and P(7 Games) = 2[6C3(.5)3(.5)3(.5)] = 20(.5)6 = 0.3125. Therefore, the long-run average number of games a best-ofseven championship series is expected to last is: µx = E(X) = 4(0.1250) + 5(0.2500) + 6(0.3125) + 7(0.3125) = 5.8125.
17. A – The long-run average amount of deviation from the expected value for the length of any championship series is
the standard deviation of its discrete probability distribution. First, the variance of the distribution is:
Var(X) = (4 – 5.8125)2(0.1250) + (5 – 5.8125)2(0.2500) + (6 – 5.8125)2(0.3125) + (7 – 5.8125)2(0.3125) = 1.02734375.
Taking the square root yields SD(X) = 1.0136 when rounded. Alternatively: entering 4, 5, 6, and 7 into List 1 and
0.1250, 0.2500, 0.3125, and 0.3125 into List 2 of the calculator and running the 1-Var Stats function yields 𝜎𝑥 = 1.0136.
18. B – Multiplying the expected proportions by the total observed count of each league produces the following
completed table (with the filled in missing values in bold type):
NBA Championship
NHL Stanley Cup
MLB World Series
Series
Expected
Observed
Expected Observed Expected Observed Expected
Total
Length Proportion
Count
Count
Count
Count
Count
Count
Observed
4 Games
0.1250
8
8.25
20
9.5
18
11.5
46
5 Games
16
18
19
53
0.2500
16.5
19
23
6 Games
24
22
20
66
0.3125
20.625
23.75
28.75
7 Games
18
16
35
69
0.3125
20.625
23.75
28.75
Total
1.0000
66
66
76
76
92
92
234
The appropriate inference procedure is a Chi-Square goodness-of-fit test comparing the observed counts to expected
counts for each sports league separately. Entering each observed count list and each expected count list into a pair of lists
in the calculator and running the X2 GOF-Test with df = 3 on the appropriate pair of lists produces the following table of
Chi-Square test statistics and P-values (rounded to four decimal places):
NBA Championship
NHL Stanley Cup
MLB World Series
𝜒 2 = 0.0901; df = 3
𝜒 2 = 14.3158; df = 3
𝜒 2 = 8.3913; df = 3
2
2
P(𝜒 > 0.0901) = 0.8232
P(𝜒 > 14.3158) = 0.0025
P(𝜒 2 > 8.3913) = 0.0386
The only test that produces a statistically significant result at the 2.5% level is for the NHL since it is the only P-value less
than 0.025. Alternatively, the Chi-Square test statistic can be computed using the formula 𝜒 2 = ∑
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑−𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑)2
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
for each league and compared to the 𝜒 2 critical value with df = 3 and upper-tail probability of 0.025 from the appropriate
table, which is 9.35. Since only the NHL produces a test statistic greater than 9.35, it is the only result that is statistically
significant. In either case, there is exactly one of the three sports leagues (namely, the NHL) that show statistically
significant evidence at the 2.5% level that the two teams competing in the finals are not always equally matched.
19. A – This requires a Chi-Square test of homogeneity with the following hypotheses: Ho: Series length is distributed
homogeneously for each sports league and Ha: Series length is not distributed homogeneously for each sports league. The
table below shows the observed and expected counts (to the nearest thousandth) for each cell where each expected count
is computed by (Row Total)(Column Total) / (Table Total). For example: (46)(66) / 234 = 12.974.
NBA Championship
NHL Stanley Cup
MLB World Series
Series
Observed
Expected Observed Expected Observed Expected
Length
Count
Count
Count
Count
Count
Count
4 Games
8
12.974
20
14.940
18
18.085
5 Games
16
14.949
18
17.214
19
20.838
6 Games
24
18.615
22
21.436
20
25.949
7 Games
18
19.462
16
22.410
35
27.128
2
March Statewide Invitational
2
The test statistic is: 𝜒 = ∑
Statistics Individual
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑−𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑)2
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
=
(8−12.974)2
12.974
+
(16−14.949)2
14.949
+
(24−18.615)2
18.615
+⋯+
(35−27.128)2
27.128
≈ 11.057
Using the appropriate table, the Chi-Square critical values with df = (4 – 1)(3 – 1) = 6 for the four given significance
levels are summarized as:
Letter Choice:
A) 10%
B) 5%
C) 2%
D) 1%
Critical Value:
𝜒 2 = 10.64
𝜒 2 = 12.59
𝜒 2 = 15.03
𝜒 2 = 16.81
Since the test statistic of 11.057 only exceeds the critical value of 10.64 and not any of the others, the results are only
statistically significant at the 10% level, at best. Alternatively, entering the 12 observed frequencies into a 4 x 3 matrix in
the calculator and running the 𝜒 2 − 𝑇𝑒𝑠𝑡 produces the same approximate test statistic and an approximate P-value of
0.0866, which is less than 10% but not any of the other choices.
20. B – This is the very definition of Simpson’s Paradox.
46
21. D – This is a one-proportion Z-test of: Ho: P = 0.125 vs. Ha: P > 0.125. The sample proportion is 𝑝̂ = 234 ≈ 0.1966
and the test statistic is 𝑍 =
0.1966−0.125
(0.125)(0.875)
234
≈ 3.31. This yields a right tail P-value of 0.0005 when using the chart.
√
Otherwise, entering the values of Po = 0.125, x = 46, and n = 234 into the 1-PropZTest in the calculator produces
essentially the same test statistic and a P-value of 0.000465, when rounded. In either case, the null hypothesis is rejected
at even the 1% level.
22. B – The correct set of ordered pairs (x, y) is: {(1, 6), (2, 5), (3, 3), (4, 2), (5, 1), (6, 4)}. Entering the x-values into
List 1 and the y-values into List 2 of the calculator and running one of the Linear Regression functions yields a Pearson
correlation of -0.657, when rounded to the nearest thousandth.
23. A – It is helpful to notice that the correlation matrix is symmetric with 1s down the main diagonal since the
correlation between any pair of variables is the same regardless of order and any variable correlates perfectly with itself.
Therefore, only the correlation between Item 2 and Total Score requires actual computation. Entering the student Item 2
results into List 1 and their corresponding Total Score results into List 2 of the calculator and running one of the Linear
Regression functions produces the missing point-biserial correlation of -0.2582, when rounded. The completed matrix is:
Item 1
Item 2
Item 3
Item 4
Total Score
Correlation Matrix
Item 1
1
-0.7746
0.0000
0.5000
0.2582
Item 2
1
0.0000
-0.4667
-0.7746
-0.2582
Item 3
0.0000
0.0000
0.0000
0.0000
1
Item 4
0.2582
0.0000
1
-0.4667
0.7746
Total Score
0.5000
0.7746
1
-0.2582
0.0000
By the definitions given, there are 6 non-redundant item pair phi correlations and 4 non-redundant point-biserial
correlations. The strongest phi correlation is P = -0.7746 and the strongest point-biserial correlation is B = 0.7746. The
trace of the matrix is the sum of the entries along the main diagonal, which is 5. Therefore, P + B + T = 5.
24. D – By the description given on how to correctly interpret a point-biserial correlation, the correct order is 4, 1, 3, 2
since they progress from the strongest positive to the strongest negative as 0.7746, 0.5000, 0.0000, -0.2582, respectively.
25. D – Consider the following inequality for the ratio of the explained to unexplained variance for a correlation
𝑟2
coefficient: 1−𝑟2 > 1. One solution is to evaluate the expression for all six unique, non-zero / non-one correlation
coefficients in the matrix (one at a time) and notice that exactly two satisfy the inequality. Alternatively, one could solve
the inequality for the absolute value of r needed to satisfy the criterion for significance in this situation:
𝑟2
1−𝑟 2
> 1 → 𝑟 2 > 1 − 𝑟 2 → 2𝑟 2 > 1 → 𝑟 2 > 0.5 → |𝑟| > √. 5 ≈ 0.7071. The two correlation coefficients (in
absolute value) that exceed 0.7071 are the phi correlation between Item 1 and Item 2 and the point-biserial correlation
between Item 4 and Total Score.
26. D – Cronbach’s Alpha is: 𝛼 =
𝐾
𝐾−1
(1 −
∑ 𝑉𝑎𝑟(𝐼)
𝑉𝑎𝑟(𝑇)
4
) = 4−1 (1 −
0.250000+0.234375+0.000000+0.234375
This classifies this quiz as “Unacceptable” according to the given criterion.
3
0.250000
) = −2.5.
March Statewide Invitational
Statistics Individual
27. B – If symmetry was assumed, the mean (and median) should be located midway between the 10th and 90th percentiles
which is ($58,290 + $180,500) / 2 = $119,395. This is slightly larger than the given median of $97,070, indicating the
distribution is likely right skewed. Alternatively, the difference between the 90th percentile and the median is $180,500 –
$97,070 = $83,430 which is much greater than the difference between the median and the 10th percentile of $97,070 –
$58,290 = $38,780. This also indicates a likely skew to the right.
28. E – Assuming independence of at least 1,000 hypothesis tests at the 0.05 level, we have P(At least one Type-I Error)
= 1 – P(none) ≈ 1 − (0. 95)1000 ≈ 1. This also makes sense intuitively, since, as the number of independent hypothesis
tests increases without bound, the probability of committing at least one Type-I error would indeed approach 1, according
to the law of large numbers.
29. B – Let P0 = 0.5 represent the assumption that the coin is fair and P1 = 0.6 represent the assumption that the coin is
biased towards heads as indicated. The initial hypothesis test would use H0: P0 = 0.5 and Ha: P0 > 0.5 and a critical value
𝑝̂−0.5
of Z = 1.96 for ∝ = 0.025. Thus, the expression 1.96 ≤
represents the required sample proportion needed to
(0.5)(0.5)
𝑛
√
reject the null assuming the coin is fair. Likewise, if we assume the coin is biased at P1 = 0.6 and we a desire a power of
0.9, then the following expression gives the requirement to reject the null: −1.28 ≤
𝑝̂−0.6
(0.6)(0.4)
𝑛
. Solving each inequality for
√
(0.5)(0.5)
+
𝑛
𝑝̂ and setting them equal to each other produces 1.96√
simplify and solve for n: 1.96√
(0.5)(0.5)
+
𝑛
0.5 = −1.28√
(0.6)(0.4)
𝑛
0.5 = −1.28√
+ 0.6 → 1.96
(0.6)(0.4)
𝑛
0.5
+
√𝑛
+ 0.6 which we can then
0.24
𝑛
0.5 + 1.28√
= 0.1 →
2
0.98+1.28√0.24
)
0.1
1.96(0.5) + 1.28√0.24 = 0.1√𝑛 → 𝑛 ≥ (
≈ 258.2672 → 259 is the correct answer.
30. B – This is a binomial setting with n = 30 and P = 0.2. Thus, E(X) = 0.2(30) = 6 and SD(X) = √30(0.2)(0.8) ≈
2.19. This makes the lower bound 6 – 2(2.19) = 1.62 and the upper bound 6 + 2(2.19) = 10.38.
4