Question 1 (a)

The distribution of Robin's tip amounts is skewed to the right.
 There is Q\_ggp between the largest tip amount (in the $20 to $22.50
 interval) and the second largest tip amount (in the $12.50 to $15
 interval), and the largest tip amount appears to be an outlier. The
 median tip amount is between $2.50 and $5.00. Robin's tip amounts
 vaufrom a mmimum of between $0 and $2.50 to a maximum of between
 $20.00 and $22.50. About 78 Dercent of the tin amounts are between $0
 and $5.

Essentially correct (E) if the response includes reasonable comments
 on the following five components: 1. 2. 3. 4. 5. Shape (skewed right)
 Outlier (at least one) OR gap (one tip amount greater than $20, next
 highest at most $15) Center between $2.50 and $5.00 (median) or
 between $2.62 and $5.13 (mean) Variability, by noting that the tip
 amounts vary from about $0 to at most $22.50, or that a majority of
 tip amounts are between $0 and a value greater than or equal to $5, or
 by providing a correct numerical approximation of a measure of
 variability Context (tip amounts)

Question 2 (a)

Step 1: States a correct pair of hypotheses. Ho The proportion of
 children who would choose each snack is the same regardless of which
 type of ad is viewed. • The proportion of children who would choose
 each snack differs based on which type of ad is Ha viewed. Step 2:
 Identifies a correct test procedure (by name or formula) and checks
 appropriate conditions. The appropriate procedure is a chi-square test
 of homogeneity. The conditions for this test are satisfied because (1)
 the question states that the children were randomly assigned to
 groups, and for the six cells of the table are allEEäE3 as seen in the
 following table that lists expected counts beside observed counts.
 Group Total Choco-Zuties 21 (18.67) 13 (18.67) 22 (18.67) 56
 Apple-Zuties 4 (6.33) 12 (6.33) 3 (6.33) 19 Total 25 25 25 75

Step 3: Calculates the appropriate test statistic and p-value. The
 test statistic is calculated as X = , which is 0.292 + 0.860 + + 5.07
 C) + 0.595 + 1.754 10.291. The p-valueis Rx \> 10291) 0.006. Step 4:
 States a correct conclusion in the context of the study, using the
 result of the statistical test. Because the p-value is very small (for
 instance, much smaller than a = 0.05), we reject the null hypothesis
 at the 0.05 level (and at the 0.01 level). The data provide convincing
 statistical evidence that the proportions who would choose each snack
 differ based on which ad is viewed.

Question 3 (a)

The explanatory variable is the smoking. The response variable iscE
 the person develops Alzheimer's disease during the course of the

Essentially correct (E) if both variables are described correctly. A
 correct description includes some degree of status of the variables,
 such as smoking versus not smoking and developing Alzeimer's versus
 not developing Alzheimer's.

Question 3 (b)

This is an observational study because the people in the study were
 not. assigned tn a certain clearee of cigaLe.LLe-sDQEing. Rather, the
 degree of cigarette smoking for each person was
 .n.assi.u.elu.hssz.u.ed and recorded, by the researchers.

Question 5 (a)

  • Denote the procedure used

    The appropriate procedure is a one-sample z-interval for a
population proportion. The problem stated the conditions for inference
have been met, so they do not need to be checked. A 95 percent
confidence interval for the population proportion is given as b ± z
which is n 0.37 0.63 0.37 ± 0.03 = (0.34,0.40). 0.37 ± 1.96 1,048 We
are 95 percent confident that the population proportion of all adults
in the U -S. who would have chosen the economy statement is between
0.34 and 0.40.

Question 5 (b)

One of the conditions for inference that was met is that the number
 who chose the economy statement and the number who did not choose the
 economy statement are both greater than 10. Explain why it is
 necessary to satisfy that condition.

The condition is necessary because the formula for the confidence
 interval relies on the fact that the binomial distribution can be
 approximated by a normal distribution which then results in the
 sampling distribution of 13 being approximately normal. The
 approximation does not work well unless both ni and n(l — b) are at
 least 10.

Normal approximation of binomial distribution Binom(10, 0.3)
 Binom(100, 0.3) When n is large, so that (np\>10) and (nq\>10), the
 binomial distribution Binom(n, p) can be approximated by

What if p is too small that np<10 even when n is large • Poisson
 distribution can be used to approximate the binomial distribution when
 n is large np is small • where X=np.

Question 5 (c)

The suggested procedure is not appropriate because one of the
 requirements for using a two-sample z-interval for a difference
 between proportions is that the two proportions are based on two In
 the situation described the two proportions come from a single sample
 and thus are not independent.

results matching ""

    No results matching ""