Question 1 (a)

  • Estimation for median

    The median is the value with half of the P-T ratios at or below it
and half of the values at or above it. to find the position of the
median in the ordered list of For n observations in a group, use 2
observations. For states west of the Mississippi (n = 24) the median
falls between the 12th and 13th value in the ordered list, and both
the 12th and 13th values fall in the interval 15—16. For states east
of the Mississippi (n = 26) the median falls between the 13th and 14th
value in the ordered list, and both of these values also fall in the
interval 15—16. From the histogram, cumulative frequencies for the two
groups are shown in the table below. Interval 12-13 13-14 14-15 15-16
West 1 1+4=5 1 4 + 6 = 11 1 6 = 14 East 2+4=6 10 2 11 = 21 Thus, the
median P-T ratio for both groups is at least 15 students per teacher
and at most 16 students per teacher.

Question 1 (c)

The medians of the two distributions are about the same, as
 determined in part (a). The distribution of P-T ratios for states that
 are west of the Mississippi River is skewed to the riqht, indicating
 that the mean will probably be higher than the median. The rough for
 the east group indicates that the mean will be close to the median.
 Thus, the mean for the west group will probably be greater than the
 mean for the east group.

  • Mean, Median, and Skew


Question 1 (a)

The study was an experiment because treatments (D-cycloserine or
 placebo) were imposed by the researchers on the people with

Question 3 (a)

  • Pay attention to the notation

    Let Y denote the number of flights Sam must make until he receives
his first upgrade. The random variable Y follows a geometric
distribution with p = 0.1. The probability that Sam's upgrade will
occur after his third flight is calculated below. P(Y24) < 3) =0.729

Question 3 (b)

Let p denote the probability that Sam will be upgraded to first
 class on a particular flight. Let X denote the number of upgrades Sam
 will receive in 20 flights. The random variable X follows a binomial
 distribution with n = 20 independent trials and p = 0 1. The
 probability that Sam will be upgraded exactly 2 times in his next 20
 flights is calculated as follows. 20 18) 2 = 0 2852

  • Binomial distribution

    s\! SNI r \_ubxd sapt.p s u ueo iXi(x - u) = (x)d

    P (X = c) = binompdf(n,p, c) n -\> number of trials p -\>
probability of success This finds the probability of exactly c
successes, for some number c.

    P (X c) = binomcdf(n, p, c) n -\> number of trials p -\> probability
of success This finds the probability of c or fewer successes.

Question 3 (c)

Let X denote the number of upgrades Sam will receive in 104 flights.
 The random variable X follows a binomial distribution with n = 104
 independent trials and p = 0.1. Thus, p (x \> 20) = I-P(X 20)
 -1-0.9986 0.0014. Because this probability is so small, it is very
 unlikely that Sam would receive more than 20 upgrades in 104 flights
 if the airline's claim is correct. This would be expected to happen
 less than 1 percent of the time, indicating that one should be
 surprised if Sam receives more than 20 upgrades during the next year.

Question 4 (b)

  • Conditions for a chi-square inference procedure

    The following conditions for inference are met: 1. The students were
randomly selected. 2. The expected cell counts should be at least 5.
The computer output indicates that all expected counts are greater
than 5. The smallest expected cell count is 6.825.

Question 4 (d)

Because the null hypothesis was rejected a Type I error may have
 been made. A Type I error is concluding that there is an association
 between the perceived effect of part-time work on academic achievement
 and the average time spent on part-time jobs when, in reality, there
 is no association between the two variables.

  • Type I error and Type II error

    HYPOTHESIS TESTING OUTCOMES Reality The Null Hypothesis Is True
Accurate Type I Error h The Null Hypthesis Is True The Alternative
Hypothesis is True The Alternative Hypothesis is True Type Il Error
Accurate O

Question 5 (a)

  • Conditions for one-proportion Z interval

    1. Random sample 2. Large sample (nÉ210 and

  • Calculation for one-proportion Z interval


    0.41617(1-0.41617) 978 ±2.57583 2,350 0.41617±0.02619 (0.38998,
0.44236) 2,350

  • Interpretation for one-proportion Z interval

    Based on the sample, we are 99 percent confident that the proportion
of the vaccine-eligible people in the United States who actually got
vaccinated is between 0.39 and 0.44. Because 0.45 is not in the 99
percent confidence interval, it is for the population proportion of
vaccine-eligible people who received the vaccine. In other words, the
confidence interval is inconsistent with the belief that 45 percent of
those eligible got vaccinated.

Question 5 (b)

The sample-size calculation uses 0.5 as the value of the proportion
 in order to provide the minimum required sample size to guarantee that
 the resulting interval will have a margin of error no larger than
 0.02. (0.02)2 2 2 576 2(0.02) = 4,147.36 Thus, a sample of at least 4,
 148 vaccine-eligible people should be taken in Canada.

  • Partially correct (P) if supporting work is shown, BUT the
response includes one or both of the following errors: 1. 0.41617
(the sample proportion) or 0.45 is used instead of 0 5. 2. An
incorrect critical z-value is used — unless the same incorrect
value was used in part (a).

  • Calculating Required Sample Size to Estimate Population Mean

    If prior estimate of population proportion exists: 1•96 - 778.564
1'96 = 1067.111 n = 13(1 — n = 0.76(1 - 0.030 If prior estimate of
population proportion does not exist: n = 0.25( n = 0.25(— 0.030

    Sample Size Formulas Margin of Error (ME) = z ME = 196 ( p(l — p)
p(l — p)z2 Formula based on t n- ({1tE)2 — score

Question 6 (b)

No. This is extrapolation bevond the ranqe of data from the
 experiment. Buffer strips narrower than 5 feet or wider than 15 feet
 were not investigated.

Extrapolated. Measured. Interpolated. C Measured. A B

Question 6 (c)

  • Describe distribution: Mean & Standard deviation

    Because the distribution of nitrogen removed for any particular
buffer strip width is normally distributed with a standard deviation
of 5 parts per hundred, the sampling distribution of the mean of four
observations when the buffer strips are 6 feet wide will be normal
with mean 33.8+3.6 X 6 = 55.4 = = 2.5 parts per hundred. parts per
hundred and a standard deviation of

Question 6 (d)

The distribution of the sample mean is normal, so the interval that
 has probability 0.95 of containing the mean nitrogen content removed
 from four buffer strips of width 6 feet extends from 55.4—1.96 x 2.5 =
 50.5 parts per hundred to 55.4 + 1.96x2.5 = 60.3 parts per hundred.

  • Confidence interval

    Mean (o known) Mean (o unknown) Standard Deviation Proportion
(exact) Proportion (estimate) Distribution Z-distribution
T-distribution Chi-squared distribution F-distribution Z-distribution
Minitab Path Stat \> Basic Stat \> I-sample Z \> Options Stat \> Basic
Statistics \> 1- Sample t Stat \> Basic Statistics \> Display
Descriptive Statistics Stat \> Basic Statistics \> 1- Proportion g x x
Formula z 2 n-l rt-1J—a/2 Rower p n S n n-l 2 n

Question 6 (e)

85 80 75 70 65 60 55 50 85 v? 80 65 60 55 50 6 7 8 9 10 11 12 13
 Width of Buffer Strip (feet) 6 7 8 9 10 11 12 13 Width of Buffer Strip
 (feet) If we think that the sample mean nitrogen removed at a
 particular buffer width might reasonably be a sample regression line
 will result from connecting any point in the interval above 6 to any
 point in the interval above 13. With this in mind, the dashed lines in
 the plots above represent extreme cases for possible sample reqression
 lines. From these plots, we can see that there is a wider range of
 possible slopes in the second plot (on the right) than in the first
 plot (on the left). Because of this, in the sampling distribution of
 b, the estimator for the slope of the regression line, will be smaller
 for the first studv Dlan (with four observations at 6 feet and four
 observations at 13 feet) than it would be for the second study plan
 (with four observations at 8 feet and four observations at 10 feet).
 Therefore, the first study plan (on the left) would provide a better
 estimator of the slope of the regression line than the second study
 plan (on the right).

Question 6 (f)

To assess the linear relationship between width of the buffer strip
 and the amount of nitrogen removed from runoff water, more widths
 should be used. To detect a nonlinear relationship it would be best to
 use buffer widths that were spaced out over the entire range of
 interest. For example, if the range of interest is 6 to 13 feet, eight
 buffers with widths 6, 7, 8, 9, 10, 11, 12 and 13 feet could be used.

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