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Chapter 9: Q54 E (page 398)

Teen Court is a juvenile diversion program designed to circumvent the formal processing of first-time juvenile offenders within the juvenile justice system. The article "An Experimental Evaluation of Teen Courts" (J. of Experimental Criminology, 2008: 137-163) reported on a study in which offenders were randomly assigned either to Teen Court or to the traditional Department of Juvenile Services method of processing. Of the \(56TC\) individuals, 18 subsequently recidivated (look it up!) during the 18 -month follow-up period, whereas 12 of the 51 DJS individuals did so. Does the data suggest that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so? State and test the relevant hypotheses using a significance level of 0.10.

Short Answer

Expert verified

There is not sufficient evidence to support the claim that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so.

Step by step solution

01

Step 1: Given information

\(\begin{array}{l}{\rm{Sample size of first sample: }}{n_1} = 56{\rm{ and }}{x_1} = 18\\{\rm{Sample size of second sample:}}{n_2} = 51{\rm{ and }}{x_2} = 12\\{\rm{Level of significance }}\alpha = 0.10\end{array}\)

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\{H_a}:{p_1} \ne {p_2}\end{array}\)

02

Test statistic

\({\rm{ The sample proportion is the number of successes divided by the sample size: }}\)

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}} = \frac{{18}}{{56}} \approx 0.3214\\{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}} = \frac{{12}}{{51}} \approx 0.2353\\{{\hat p}_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{18 + 12}}{{56 + 51}}\\ = \frac{{30}}{{107}}\\ \approx 0.2804\end{array}\)

Determine the value of the test statistic:

\(\begin{array}{c}z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }}\\ = \frac{{0.3214 - 0.2353}}{{\sqrt {0.2804(1 - 0.2804)} \sqrt {\frac{1}{{56}} + \frac{1}{{51}}} }}\\ \approx 0.99\end{array}\)

03

Finding P-value

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. Determine the P-value using table the normal probability table in the appendix:

\(\begin{array}{c}P = P(Z < - 0.99{\rm{ or }}Z > 0.99)\\ = 2P(Z < - 0.99)\\ = 2(0.1611)\\ = 0.3222\end{array}\)

If the P-value is smaller than the significance level, then reject the null hypothesis:

\(P > 0.10 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. Determine the P-value using table the normal probability table in the appendix:

\(\begin{array}{c}P = P(Z < - 0.99{\rm{ or }}Z > 0.99)\\ = 2P(Z < - 0.99)\\ = 2(0.1611)\\ = 0.3222\end{array}\)

If the P-value is smaller than the significance level, then reject the null hypothesis:

\(P > 0.10 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

04

Final conclusion

There is not sufficient evidence to support the claim that the true proportion of TC individuals who recidivate during the specified follow-un period differs from the proportion of DJS individuals who do so.

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Most popular questions from this chapter

Scientists and engineers frequently wish to compare two different techniques for measuring or determining the value of a variable. In such situations, interest centers on testing whether the mean difference in measurements is zero. The article "Evaluation of the Deuterium Dilution Technique Against the Test Weighing Procedure for the Determination of Breast Milk Intake" (Amer: J. of Clinical Nutr., \(1983: 996 - 1003\)) reports the accompanying data on amount of milk ingested by each of\(14\)randomly selected infants.

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;\;\;Infant\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;\;\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;5}\\{D method\;\;\;\;\;\;\;\;\;\;\;1509\;\;\;\;\;1418\;\;\;\;\;1561\;\;\;\;\;1556\;\;\;\;\;2169}\\{W method\;\;\;\;\;\;\;\;\;\;1498\;\;\;\;\;1254\;\;\;\;\;1336\;\;\;\;\;1565\;\;\;\;\;2000}\\{jifference\;\;\;\;\;\;\;\;\;\;\;\;11\;\;\;\;\;\;\;\;\;\;164\;\;\;\;\;\;\;225\;\;\;\;\;\;\; - 9\;\;\;\;\;\;\;\;\;\;169}\end{array}\)

\(\begin{array}{*{20}{c}}{}&{Infant}&{}&{}&{}&{}\\{}&6&7&8&9&{10}\\{DD method}&{1760}&{1098}&{1198}&{1479}&{1281}\\{TW method}&{1318}&{1410}&{1129}&{1342}&{1124}\\{Difference}&{442}&{ - 312}&{69}&{137}&{157}\end{array}\)

\(\begin{array}{*{20}{c}}{}&{Infant}&{}&{}&{}\\{}&{11}&{12}&{13}&{14}\\{DD method}&{1414}&{1954}&{2174}&{2058}\\{TW method}&{1468}&{1604}&{1722}&{1518}\\{Difference}&{ - 54}&{350}&{452}&{540}\end{array}\)

a. Is it plausible that the population distribution of differences is normal?

b. Does it appear that the true average difference between intake values measured by the two methods is something other than zero? Determine the\(P\)-value of the test, and use it to reach a conclusion at significance level . \(05\).

The National Health Statistics Reports dated Oct. \(22,2008\), included the following information on the heights (in.) for non-Hispanic white females:

Sample sample Std. Error

Age Size Mean Mean

\(\begin{array}{*{20}{l}}{20 - 39}&{866}&{64.9}&{.09}\\{60 and older }&{934}&{63.1}&{.11}\\{}&{}&{}&{}\end{array}\)

  1. Calculate and interpret a confidence interval at confidence level approximately \(95\% \) for the difference between population mean height for the younger women and that for the older women.
  2. Let \({\mu _1}\) denote the population mean height for those aged \(20 - 39\) and \({\mu _2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \({H_0}:{\mu _1} - {\mu _2} = 1 and {H_a}:{\mu _1} - {\mu _2} > 1,\) and then carry out a test of these hypotheses at significance level \(.001\)
  3. Based on the \(p\)-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.
  4. What hypotheses would be appropriate if \({\mu _1}\) referred to the older age group, \({\mu _2}\) to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than \(1\)in.?

The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of \(43\) chronic schizophrenics, resulting in \(\bar x = 2.69\) and \({s_1} = 2.30,\), as well as for \(45\) normal subjects, resulting in \(\bar y = 6.35\) and \({s_2} = 4.03.\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha = .01\)

. (Hint: \({H_0}\) and \({H_a}\) here have a different form from the three standard cases. Let \({\mu _1}\) and \({\mu _2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta = 2{\mu _1} - {\mu _2}\). Write \({H_0}\) and \({H_a}\) in terms of \(\theta \), estimate \(\theta \), and derive \({\hat \sigma _{\tilde \theta }}\) (鈥淩educed Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients,鈥� Nature, July 28, 1972: 225鈥�226).) \(\alpha = .01\)

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar x = 18.12kgf/c{m^2}\)for the modified mortar \((m = 40)\) and \(\bar y = 16.87kgf/c{m^2}\) for the unmodified mortar \((n = 32)\). Let \({\mu _1}\) and \({\mu _2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.

a. Assuming that \({\sigma _1} = 1.6\) and \({\sigma _2} = 1.4\), test \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} > 0\) at level . \(01.\)

b. Compute the probability of a type II error for the test of part (a) when \({\mu _1} - {\mu _2} = 1\).

c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta = .10\) when \({\mu _1} - {\mu _2} = 1. \). If \(m = 40\), what value of n is necessary?

d. How would the analysis and conclusion of part (a) change if \({\sigma _1}\) and \({\sigma _2}\) were unknown but \({s_1} = 1.6\)and \({s_7} = 1.4?\)

An experiment was performed to compare the fracture toughness of high-purity \(18Ni\) maraging steel with commercial-purity steel of the same type (Corrosion Science, 1971: 723鈥�736). For \(m = 32\)specimens, the sample average toughness was \(\overline x = 65.6\) for the high purity steel, whereas for \(n = 38\)specimens of commercial steel \(\overline y = 59.8\). Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial purity steel by more than 5. Suppose that both toughness distributions are normal.

a. Assuming that \({\sigma _1} = 1.2\) and \({\sigma _2} = 1.1\), test the relevant hypotheses using \(\alpha = .001\).

b. Compute \(\beta \) for the test conducted in part (a) when \({\mu _1} - {\mu _2} = 6.\)
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