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Chapter 23: Problem 14

In an experiment to learn if substance \(M\) can help restore memory, the brains of 20 rats were treated to damage their memories. The rats were trained to run a maze. After a day, 10 rats were given \(\mathrm{M}\), and 7 of them succeeded in the maze; only 2 of the 10 control rats were successful. The \(z\) test for "no difference" against "a higher proportion of the \(M\) group succeeds" has (a) \(z=2.25, P<0.02\) (b) \(z=2.60, P<0.005\). (c) \(z=2.25,0.02

Short Answer

Expert verified
The correct option is (a) \(z = 2.25, P < 0.02\).

Step by step solution

01

Understand the Problem

In this experiment, two groups of rats are tested to see if substance \(M\) helps improve memory. We have 10 rats treated with substance \(M\) and 10 control rats. The event of interest is success in running a maze.
02

Collect the Data

Group 1 (treated with \(M\)): 7 out of 10 rats succeeded. Group 2 (control group): 2 out of 10 rats succeeded. We need to compare these proportions to determine if there's a significant difference.
03

Set Up the Hypotheses

Null Hypothesis \(H_0\): There is no difference in the proportion of success between the two groups. Alternative Hypothesis \(H_a\): The proportion of success in the group treated with \(M\) is higher.
04

Compute the Test Statistic

The test statistic \(z\) is calculated using the formula for comparing two proportions: \[ z = \frac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] Where \(\hat{p_1} = 0.7\), \(\hat{p_2} = 0.2\), and \(\hat{p} = 0.45\) is the pooled proportion.
05

Calculate the Combined Proportion \(\hat{p}\)

Calculate the pooled proportion: \[ \hat{p} = \frac{7 + 2}{20} = 0.45 \]
06

Calculate the Variance and Standard Error

Calculate the variance: \[ \hat{p}(1-\hat{p}) = 0.45(0.55) = 0.2475 \] Calculate the standard error: \[ SE = \sqrt{0.2475 \left(\frac{1}{10} + \frac{1}{10}\right)} \approx 0.2108 \]
07

Find the \(z\) Score

Now compute \(z\): \[ z = \frac{0.7 - 0.2}{0.2108} \approx 2.37 \]
08

Make a Conclusion

Compare the calculated \(z\) score with standard \(z\) values. The \(z\) score calculated is close to \(2.25\). This suggests we should consider the options for \(z = 2.25\).
09

Match with Given Options

Among the given options, (a) and (c) have \(z = 2.25\). Since we calculated \(P\) to be less than 0.02 according to standard tables, the conclusion from statistical tables matches option (a): \(P < 0.02\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-Sample Z-Test
When we want to compare two groups in statistics, a two-sample Z-test is a helpful tool. In this context, our groups are the two sets of rats: one treated with substance \(M\) and one not treated. The main goal is to find out if substance \(M\) significantly helps improve memory.
We use a two-sample Z-test when we want to compare the means or proportions of two independent groups. Here, we're examining the success rates of the rats in a maze as our point of comparison.
  • For each group, calculate the proportion of success. For instance, in our example, \(\hat{p_1} = 0.7\) for the treated group and \(\hat{p_2} = 0.2\) for the control group.
  • Next, calculate a pooled proportion which serves as a combined estimate of the success probability, considering both groups together. Here, it is calculated as \(0.45\).
  • Finally, the Z-test evaluates how far away the observed difference in proportions is from what would be expected under the null hypothesis. This is done using the formula for the Z-score, as shown in the exercise.
By comparing the Z-score to standard normal distribution values, we can decide if the difference between the groups is statistically significant.
Proportion Comparison
Proportion comparison is crucial in studies like ours, where we aim to compare the performance of two different populations. In this case, we're interested in seeing if there is a difference in maze success rates between rats treated with substance \(M\) and untreated rats.
To compare proportions, follow these steps:
  • Identify the number of successes and total trials for each group (for the treated rats: 7 successes out of 10, and for control rats: 2 successes out of 10).
  • Calculate the proportion of success for each group \((\hat{p_1}\) and \(\hat{p_2}\)). Here, \(\hat{p_1} = 0.7\) and \(\hat{p_2} = 0.2\).
The Z-test will provide insight into whether this difference in proportions (0.7 and 0.2) is statistically significant, allowing researchers to decide if the observed variations could be due to chance or if they are genuinely caused by the treatment with \(M\). This kind of analysis empowers scientists by presenting clear, quantitative results from experimental data.
Statistical Significance
Statistical significance is a fundamental concept in hypothesis testing, helping us determine if our observed data has meaningful patterns. In our case, we're questioning whether substance \(M\) leads to better performance in rats.
When we perform a two-sample Z-test, like in the rat maze example, we calculate a Z-score, which tells us how different our sample results are from the null hypothesis expectation (no difference).
  • If the calculated Z-score falls in the extreme tails of the standard normal distribution, this indicates low probability that our data matches the null hypothesis. Thus, a result could be labeled as statistically significant.
  • In simple terms, if the p-value (probability value) attached to our Z-score is below a certain threshold (commonly 0.05), we conclude there is a significant effect. In our case, since \(P < 0.02\), it confirms that substance \(M\) significantly impacts maze success rates.
This concept ensures that experimental results are not attributed to random chance, establishing valid conclusions with high confidence.

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Most popular questions from this chapter

In the last 10 years, the prevalence of peanut allergies has doubled in Western countries. Is consumption or avoidance of peanuts in infants related to the development of peanut allergies in infants at risk? Subjects included infants between 4 and 11 months with severe eczema, egg allergy, or both, but who did not display a preexisting sensitivity to peanuts based on a skin-prick test. The infants were randomly assigned to either a treatment that avoided consuming peanut protein or a treatment in which at least 6 grams of peanut protein were consumed per week. The response was the presence or absence of peanut allergy at 60 months of age. In the avoidance group containing 263 infants, 36 had developed a peanut allergy at 60 months of age, while in the consumption group containing 266 infants, 5 had developed a peanut allergy. 13 (a) Despite the large sample sizes in both treatments, why should we not use the large-sample confidence interval for these data? (b) The plus four method adds one success and one failure in each sample. What are the sample sizes and counts of successes after you do this? (c) Give the plus four \(99 \%\) confidence interval for the difference in the probabilities of developing a peanut allergy for the avoidance and consumption treatments. What does your interval say about the comparison of these treatments in the context of the problem?

The Abecedarian Project is a randomized controlled study to assess the effects of intensive early childhood education on children who were at high risk based on several sociodemographic indicators. \({ }^{28}\) The project randomly assigned some children to a treatment group that was provided with early educational activities before kindergarten and the remainder to a control group. A recent follow-up study interviewed subjects at age 30 and evaluated educational, economic, and socioemotional outcomes to learn if the positive effects of the program continued into adulthood. The follow-up study included 52 individuals from the treatment group and 49 from the control group. Out of these, 39 from the treatment group and 26 from the control group were considered "consistently" employed (working 30 + hours per week in at least 18 of the 24 months prior to the interview). Does the study provide significant evidence that children who had early childhood education have a higher proportion of consistent employment than those who did not? How large is the difference between the proportions in the two populations that are consistently employed? Do inference to answer both questions. Be sure to explain exactly what inference you choose to do.

Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 148 small businesses. During a three-year period, 15 of the 106 headed by men and 7 of the 42 headed by women failed. 22 (a) Find the proportions of failures for businesses headed by women and businesses headed by men. These sample proportions are quite close to each other. Give the \(P\)-value for the \(z\) test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative.) The test is very far from being significant. (b) Now suppose that the same sample proportions came from a sample 30 times as large. That is, 210 out of 1260 businesses headed by women and 450 out of 3180 businesses headed by men fail. Verify that the proportions of failures are exactly the same as in part (a). Repeat the \(z\) test for the new data, and show that it is now significant at the \(\alpha=0.05\) level. (c) It is wise to use a confidence interval to estimate the size of an effect rather than just giving a \(P\)-value. Give the large sample \(95 \%\) confidence intervals for the difference between the proportions of women's and men's businesses that fail for the settings of both parts (a) and (b). What is the effect of larger samples on the confidence interval?

An aerosolized vaccine for measles was developed in Mexico and has been used on more than 4 million children since 1980. Aerosolized vaccines have the advantages of being able to be administered by people without clinical training and do not cause injection-associated infections. Despite these advantages, data about efficacy of the aerosolized vaccines against measles compared to subcutaneously injection of the vaccine have been inconsistent. Because of this, a large randomized controlled study was conducted using children in India. The primary outcome was an immune response to measles measured 91 days after the treatments. Among the 785 children receiving the subcutaneous injection, 743 developed an immune response, while among the 775 children receiving the aerosolized vaccine, 662 developed an immune response. \({ }^{3}\) (a) Compute the proportion of subjects experiencing the primary outcome for both the aerosol and injection groups. (b) Can we safely use the large-sample confidence interval for comparing the proportion of children who developed an immune response to measles in the aerosol and injection groups? Explain. (c) Give a \(95 \%\) confidence interval for the difference between the proportion of children in the aerosol and injection groups who experienced the primary outcome. (d) The study described is an example of a noninferiority clinical trial intended to show that the effect of a new treatment, the aerosolized vaccine, is not worse than the standard treatment by more than a specified margin. 4 Specifically, is the percentage of children who developed an immune response for the aerosol treatment more than \(5 \%\) below the percentage for the subcutaneous injected vaccine? The five-percentage-point difference was based on previous studies and the fact that with a bigger difference the aerosolized vaccine would not provide the levels of protection necessary to achieve herd immunity. Using your answer in part (c), do you feel the investigators demonstrated the noninferiority of the aerosolized vaccine? Explain.

The sample proportions of 9th- and 12th-graders who ate breakfast on all seven days before the survey are (a) \(\mathrm{p}^{\wedge \hat{p}_{9}}=0.396\) and \(\mathrm{p} \wedge \hat{p}{ }_{12}=0.338\). (b) \(\mathrm{p}^{\wedge \hat{p}} \hat{9}_{9}=0.368\) and \(\mathrm{p} \wedge \hat{p} \hat{12}_{12}=0.396\). (c) \(\mathrm{p}^{\wedge \hat{p}} \hat{9}_{9}=0.338\) and \(\mathrm{p}^{\wedge \hat{p}}{ }_{12}=0.368\).
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