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

    Kouetbeæ

Question 1 (a)

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

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

    C:\\6432CA65\\FE01530B-89BD-4F8B-A3E1-55F12080AD12\_files\\image354.png

    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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