Chapter 10 Extra practice

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated the persistence of this pesticide on various types of surfaces. Researchers applied a 0.5% emulsion of diazinon to glass and plasterboard. After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On glass, 18 cockroaches died, while on plasterboard, 25 died. Construct and interpret a 90% confidence interval for the difference in the proportion of cockroaches that die on each surface.

State: Our parameters of interest are p1 = the proportion of all cockroaches similar to those in the study that die on glass and p2,= the proportion of all cockroaches similar to those in the study that die on plasterboard. We want to estimate the difference p1 – p2, at a 90% confidence level. Plan: We should use a 2-sample z-interval for p1 – p2. Random: Cockroaches were randomly assigned to one of the two treatment groups, plasterboard or glass. Large Counts/Normal: Number of successes and failures in the two groups are 25, 11, 18, and 18, all of which are at least 10, it is safe to do Normal calculations. 10%: Since no sampling took place, the 10% condition does not apply.

Do: Conclude: We are 90% confident that the interval from –0 Do: Conclude: We are 90% confident that the interval from –0.3808 to –0.0081 captures the true difference in the proportion of cockroaches that die on glass versus the proportion that die on plasterboard.

2. A study of iron deficiency among infants compared blood hemoglobin levels of a random sample of one-year-old infants who had been breast-fed to a random sample of one-year old infants who had been fed with standard infant formula. Here are the results. We wish to test the hypothesis H0: μB – μF = 0 against Ha: μB – μF ≠ 0 where μB and μF are the population mean blood hemoglobin levels for breast-fed and formula-fed infants, respectively. a) Assuming the conditions have been met, calculate the test statistic and P-value for this test.

a) Using Table B and conservative df = 18, the two tailed P-value is between 0.1 and 0.2. 2 *tcdf(lower: 1.645, upper: 999999, df = 18) = 0.1173 Using the calculator and df = 37.60, P-value = 0.1065.

b) Draw the appropriate conclusion at the α = 0. 05 level b) Draw the appropriate conclusion at the α = 0.05 level. b) Fail to reject H0. Since our p-value, 0.1173, is greater than  = 0.05, there is not enough evidence to suggest that there is a difference in the mean blood hemoglobin levels of the infants in the two groups.