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Unformatted text preview: Test 11A AP Statistics Name: Directions: Work on these sheets. Tables and formulas appear on a separate sheet. Part 1: Multiple Choice. Circle the letter corresponding to the best answer. 1. In preparing to use a t procedure, suppose we were not sure if the population was normal. In which of the following circumstances would we not be safe using a t
procedure?
(a) A stemplot of the data is roughly bell shaped.
(b) A histogram of the data shows moderate skewness.
A stemplot of the data has a large outlier.
d) The sample standard deviation is large.
(e) The t procedures are robust, so it is always safe. . The weight of 9 men have mean 92 = 175 lbs. and standard deviation 5 = 15 lbs. What is the standard error of the mean?
(a) 58.3
(b) 19.4
5
d) 1.7
(e) None of the above. The answer is 3. What is the critical value t* which satisfies the condition that the t distribution with 8 degrees of freedom has probability 0.10 to the right of t*?
@1397 (b) 1.282 ’ (c) 2.89 (d) 0.90 (e) None of the above. The answer is . Suppose we have two SRSs from two distinct populations and the samples are
independent. We measure the same variable for both samples. Suppose both
populations of the values of these variables are normally distributed but the means
and standard deviations are unknown. For purposes of comparing the two means,
we use
Twosample t procedures )Matched pairs t procedures
(c) z procedures
(d)The leastsquares regression line
(e) None of the above. The answer is . In preparing to perform a test to compare the means of two independent
populations, which of the following would lead us to believe that the t procedures
are not safe to use?
(a) The sample medians and means for the two groups were slightly different.
(b) The distributions of the data were moderately skewed.
c The data are integers between 1 and 10 and so cannot be normal.
Only the most severe departures from normality.
(e) None of the above. The answer is Part2: Free Response
Answer completely, but be concise. Write sequentially and Show all steps. '~\ An SRS 'of 16 Henrico County Schools’ seniors had a mean SATMath score of 32 = 500 and a standard deviation of s = 100. We know that the population of SATMath scores
for seniors in the district is approximately normally distributed. We wish to determine
a 90% confidence interval .for the mean SATMath score it for the population of all
seniors in the district. 6. Using the above data, the critical value for the
confidence interval has how many degrees of freedom? bbpk : L{ 7. What is the critical value for the 90% confidence
interval? Identify the curve and the critical value on the sketch provided. \,’l§3 8. Find the 90% confidence interval. 5'00 5: msg (if—[’3 : (4(9tﬁl§%3.<23>
:39 9. Suppose the population of all high school seniors has a mean SATMath score of
450. We wish to see if the data provide evidence that the mean SAT Math score it for the population of seniors in the district is larger than 450. Write the hypotheses
for such a test. Mai/Qéqg'o QIN>LE30 10. Perform the appropriate test of significance to test these hypotheses and report the
test statistic and the Pvalue. 11. Write your conclusions in simple English. 5._ *‘S New $AT Kb ”“36? M’— +\—\.n Wp%,¢.ao MM I Chapter 11 2 Test 11A XFor the SAT Math data above, find the power of the test against the alternative
LL: 500 at the 5% significance level. Assume that o—  100. x You regard your calculated power to be too low. How could you change the
problem to increase the power? Verify your answer. Nitrites are often added to meat products as preservatives. In a study of the effect of
these chemicals on bacteria, the rate of uptake of a radiolabeled amino acid was
measured for a number of cultures of bacteria, some growing in a medium to which
nitrites had been added. Here are the summary statistics from this study. Group H x s
Nitrite 30 7880 1115
Control 30 8112 1250 14. Carry out a test of the research hypothesis that nitrites decrease amino acid uptake
and report your results ‘ 5 1 1,, ... n a Test 11A art“ Chapter 11 r p L’XTTG’AYC5 S\ kr~ The table below gives the pretest and posttest scores on the MLA listening test in .7
Spanish for 20 high school Spanish teachers who attended an intensive summer course in Spanish. _ 3
Subject Pretest Posttest Subject Pretest Posttest L
1 3o 29 W
2 28 3o .. 12 29 28 /
3 31 32 — 13 31 34 
4 26 3O — 14 29 32 "
5 20 16 15 34 32
6 3O 25 16 2O 27 "
7 34 31 17 26 28 '
8 15 18 ~ 18 25 29 "’
9 28  33  19 31 32 ’
10 2O 25 ' 20 29 32 "' listening skills. State an appropriate H0 and Ha. Be sure 15. We hope to show that attending the institute improves ,1) v a: + PF‘Q
: a P’
to identify the parameters appearing in the hypotheses.  3461;“ {:0 11111p>© 16. Make a graphical check for outliers or strong skewness in
the data that you will use in your statistical test, and
report your conclusions on the validity of the test. You
can use the space to the right. rejec '0 a e % significanc 17. Carry out E'test. Can you 3}” —
l t:' K I ”O
5 ’ B 1 LEE/2 39/5 33655.3 18. Give a 90% confidence interval for the mean‘increase in listening score due to
attending the summer institute. .3555
LLES S (TIE—‘1 (1:: '1 (Au 2 LQ‘MB I pledge that I have neither given nor received aid on this test. . W Ky Chapter 11 4 Test 11A (3) Yes, the result is significant at the 5% significance level because .0086 < ..05 (4) Yes, the result IS significant at the 5% significance level because .0086 < 0.5 (5) (— —10. 64, 0.439) by TI 83. (6) I am 90% confident that this interval captured the true mean difference 1n pulse
rates. Test 11A (1) c (2) c (3) a (4) a (5) d (6) 15 (7) t* = 1.753 (8) The 90% confidence interval is (456.17, 543.83). (9) H0: 0:450 and Ha: ”>450 (10) The test statistic is t = 2, and the Pvalue is .032. (11) There is sufficient evidence to reject H0 and conclude that the Henrico seniors
mean SATMath score is greater than 450 (the national mean). (12) For df = 15, reject H0 if t > 1.753, i.e., if 32 > 493.825. Taking s = 100 and
standardizing using the alternative 11 = 500, P02 > 493.825) = normalcdf(493.825, 1E99,  500, 25) = .60. (13) Increase the power by increasing the sample size. If n = 100, for example, Power =
normalcdf(493.825, 1E99, 500, 10) = . . (14) Let population 1 be the (nitrite) treatment group and let population 2 be the
control group. H0: 111 = 112 (or 111 — 112 = 0). Ha: 111 < 112 (or 111 — 112 < 0). By TI83, t = —.7586, df = 57.25855, and the Pvalue = .2256. There is insufficient evidence to reject
Ho. I can’t conclude that nitrites decrease amino acid uptake. (15) Let X = posttest score — pretest score. This is a matched pairs experimental design.
Assume that the increase in scores 0. (pre to post) is normally distributed and that the 20 teachers can be considered a representative sample of teachers who would enroll in ‘ this summer course. Define H0: u = 0 and Ha: u > 0.
(16) Here is a stemplot of the data, where the stems are the values of X and the Os serve
as tally marks: ‘ —5 0
34 0 —3 0
—2' o —1 00 0 1 00 2 000
3 0000
4 00 5 00 6 7 0 There is a slight left skew and no outliers. ' We are not greatly concerned about a small
amount of skewness with a sample size of 20. Chapter 11 2 Solutions \«m/ l . /'
\x‘w/ Chapter 11 Solutions Quiz 11.1A (1) X = difference between the nicotine content for the current filter and the new filter.
(2) H0: u=0andHaz “>0. ‘ (3) The population is all differences between the nicotine content for the current filter
and the new filter for all nine cigarette brands. ' (4) t = (1.32  0) / (2.35 / \/9) = 1.69. The degrees of freedom are 8. From the ttable, .05 <'
PValue < .10. By the TI—83, tcdf(1.685, 1E99, 8) = .065.
(5) There is only marginal evidence that the new filter is significantly better at reducing nicotine levels.
(6) t* = t.05 = 1.860. The 90% confidence interval is (—.14, 2.78). Quiz 11.13 (1) X = the amount of nitrogen in the amber specimens. (2) The stemplot shows some left skewness; however, for such a small sample, the
data are not unreasonably skewed. There are no outliers. 4 9 5 1 5 5 6 1334
6 55 (3) The curve is the t3 curve. The t* value is the upper critical value with .0250 area
above it. t’* = 2.306. (4) The 95% confidence interval is (54.781, 64.397). (Note: The sample size (9) is
relatively small; a larger sample size would yield more satisfactory results.) (5) We are 95% confident that this interval captures the true percent of nitrogen in
ancient age. Alternatively, we arrived at these results by a procedure that is correct 95%
of the time. Quiz 1112A (1) H0: H1 = H2, or H1  H2 = 0. Ha: H1 ¢ H2, or H1 — H2 at 0. (2) t = —.5; df = 2. (3) P(t2 < 0.5) = P(t2 > 0.5) > .25, so PValue = P( l t > 0.5) > .50. (4) Fail to reject H0. (5) There is no evidence to conclude that the means are not the same.
(6) The sample sizes are extremely small. Quiz 11.23
(1) Let pl = treatment mean and H2 2 control mean. H0: 111 = M2, or [.11 — M2 = 0. Ha: Ml < H2, or Mi — 112 < 0
(2) t = —2.45. df = 29 by the rule, or df = 57.8 by TI—83. The P—Value = .0086 (by TI—83). (Note: If students work this by hand and use the table, they will obtain a Pvalue
between .01 and .02 because of the discrepancy in degrees of freedom.) Chapter 11 1 Solutions ...
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 Spring '11
 Smith
 Statistics

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