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Applicable Fall 09 Exam 1 Solutions Note: In the Fall 2009 semester, Dr. Thompson used a t–table that differs slightly from the one given at the review. These solutions reflect the usage of the t–table that accompanies this particular exam. 1. Answer: (D) The question reports that Q1 = 66 and Q3 = 74 Therefore, to conduct an outlier check, we must compute the fences i.e.
• lower fence = Q1 – 1.5(IQR) = 66 – 1.5(74 – 66) = 54
• upper fence = Q3 + 1.5(IQR) = 74+ 1.5(74 – 66) = 86 Of course, this question just wants the lower fence i.e. 86 2. Answer: (C) As per the previous question, we have an upper fence of 86 and a lower fence of 54. Therefore, we have two low outliers i.e.
52 53 62 63 63 64 64 66 66 67 67 68 69 69 70 70 71 71 72 72 72 73 73 73 73 73 74 74 75 75 75 76 77 78 81 82
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3. Answer: (A) The formula for the appropriate number of intervals is k = 1 + 3.3log(n) = 1 + 3.3log(36) = 6.13, thus we should use 6 or 7 intervals. However, ALL answer choices show 7 intervals, the width should be… Max Min# intervals
−= 82 52
7−
= 4.3 → 5
A – Good – The width is consistent with the formula and all intervals contain data. Perfect. B – No Good – The width is NOT consistent with the formula i.e. we are supposed to round UP. C – No Good – Even though the width seems a bit much, it’s ‘passable’. However, if we go up to the last bin i.e. 48 + (7)(6) = 90, we know that the last bin goes from 84 to 90. Since there are no data in this range, this bin is empty. D – No Good – Much as a width of 5 is good, the first bin of 55.1 to 60 misses the first few data values i.e. 52 & 53. 4. Answer: (C) As we can see, there are 7 players below 66 inches i.e.
52 53 62 63 63 64 64 66 66 67 67 68 69 69 70 70 71 71 72 72 72 73 73 73 73 73 74 74 75 75 75 76 77 78 81 82
= = =7
.19436
xp
n
SE =−(1 )p pn
=.194(1 .194)
36−
= .066
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5. Answer: (D) As per the formula sheet, when making a CI for the median, we use the .4n – 2 rule i.e.
.4(36) – 2 = 12.4 → 12 Therefore, to make the interval, we take the 12th smallest and 12th largest values. However, since the question only wants the lower value, we will take the 12th from the bottom…
52 53 62 63 63 64 64 66 66 67 67 68 69 69 70 70 71 71 72 72 72 73 73 73 73 73 74 74 75 75 75 76 77 78 81 82
Therefore, the lower endpoint of the interval is 68. 6. Answer: (E) As per the solution to #5, we will take the 12th value from the top i.e.
52 53 62 63 63 64 64 66 66 67 67 68 69 69 70 70 71 71 72 72 72 73 73 73 73 73 74 74 75 75 75 76 77 78 81 82
Therefore, the upper endpoint of the interval is 73. 7. Answer: (E) Since we are given both ticket prices in each city, the data values must be paired. Therefore, we only care about the mean and standard deviation of the differences i.e.
SE = 15.99
d
d
sn
= = 5.3
4
8. Answer: (C) Since we are told to disregard the answer in question #7, let’s use the new SE along with df = nd – 1 = 9 – 1 = 8 using the truncated t-‐‑table…
80% 90% 95% 98% 99% Clevel df ↓ : : : : : :
6 to 10 1.397 1.860 2.306 2.896 3.355 So, the resulting confidence interval is
* d
d
sd t
n± × → *d t SE± ×
4.56 + (1.860)(4.00) → –2.88 to +12.00
Since the interval does include 0, we conclude that there is NO significant difference in the prices of the tickets i.e. they are roughly the same. 9. Answer: (E) As per the formula sheet, we know the formula for a confidence interval for one mean is given as…
* sx tn
± → x ME±
So, let’s solve for the ME using the critical t value with df = n – 1 = 40 – 1 = 39
80% 90% 95% 98% 99% Clevel df ↓ : : : : : :
31 to 40 1.309 1.696 2.041 2.454 2.749
Therefore, ME = * st
n= 1.1012.041
40× = .355
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10. Answer: (C) As per the formula sheet, we know that the pooled variance is…
− + −=
+ −
2 22 1 1 2 2
1 2
( 1) ( 1)2p
n s n ss
n n=
2 2(40 1)1.101 (60 1)1.30240 60 2
− + −+ −
= 1.503
Therefore, the pooled standard deviation = 1.503 = 1.226 11. Answer: (B) We are told to assume that sp = 1.800 Therefore, we will compute our confidence interval with df = n1 + n2 – 2 = 60 + 40 – 2 = 98
80% 90% 95% 98% 99% Clevel df ↓ : : : : : :
61 to 100 1.295 1.670 2.000 2.389 2.659 Therefore, the confidence interval is given as…
− ± ×⎛ ⎞
+⎜ ⎟⎝ ⎠
* 21 2
1 2
( )1 1
px x t sn n
→ − ± × +⎛ ⎞⎜ ⎟⎝ ⎠
2 1 1(5.373 6.320) 2.000 1.8
40 60→ (–1.68, –.212)
So, this confidence interval indicates that we are 95% confident that the mean usage in Gainesville is between 212 kilowatt hours less and 1680 kilowatt hours less than Orlando. In other words, we believe that Gainesville usage is less. Note: In this question, since we were NOT told what level of confidence to use, we will use our ‘default’ level of 95%. 12. Answer: (A) As per the formula sheet, we know that the SE (incorporating the population size) is…
SE =1
s N nNn
−−
= 1.101 2500 402500 140
−−
= .173
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16. Answer: (B) Since the question states that the population SD is unknown we know that we are going to be using the tcalc. Since the test is a one-‐‑tailed upper tail test (i.e. H1: µμ > 20) we will draw our rejection region with the upper tail shaded. So, we will match our one-‐‑tailed α = .05 and df = n – 1 = 12 – 1 = 11…
0.1 0.05 0.025 0.01 0.005 One Tail df ↓ : : : : : :
11 to 15 1.363 1.796 2.201 2.718 3.106 Therefore, we will reject H0 if our tcalc > 1.796 17. Answer: (D) Note: For this question, the sample size changes from n = 12 to n = 21 From the formula sheet, we know that the value of a test statistic for one mean is…
0calc
xt
sn
µ−= =
20.8 201.4
21
− = 2.62
18. Answer: (D) We are told to assume that our tcalc = 2.43, so we will look up this value on the t-‐‑table with df = n – 1 = 21 – 1 = 20…
0.1 0.05 0.025 0.01 0.005 One Tail df : : : ↑ ↑ :
16 to 20 1.337 1.746 2.120 2.583 2.921
↑
tcalc = 2.43
Therefore, our p-‐‑value is between 0.01 and 0.025
