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Can Malaria Parasites Control Mosquito Behavior? Are malaria parasites able to control mosquito behavior to their advantage? A study \(^{43}\) investigated this question by taking mosquitos and giving them the opportunity to have their first "blood meal" from a mouse. The mosquitoes were randomized to either eat from a mouse infected with malaria or an uninfected mouse. At several time points after this, mosquitoes were put into a cage with a human and it was recorded whether or not each mosquito approached the human (presumably to bite, although mosquitoes were caught before biting). Once infected, the malaria parasites in the mosquitoes go through two stages: the Oocyst stage in which the mosquito has been infected but is not yet infectious to others and then the Sporozoite stage in which the mosquito is infectious to others. Malaria parasites would benefit if mosquitoes sought risky blood meals (such as biting a human) less often in the Oocyst stage (because mosquitos are often killed while attempting a blood meal) and more often in the Sporozoite stage after becoming infectious (because this is one of the primary ways in which malaria is transmitted). Does exposing mosquitoes to malaria actually impact their behavior in this way? (a) In the Oocyst stage (after eating from mouse but before becoming infectious), 20 out of 113 mosquitoes in the group exposed to malaria approached the human and 36 out of 117 mosquitoes in the group not exposed to malaria approached the human. Calculate the Z-statistic. (b) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Oocyst stage approaching the human is lower in the group exposed to malaria. (c) In the Sporozoite stage (after becoming infectious), 37 out of 149 mosquitoes in the group exposed to malaria approached the human and 14 out of 144 mosquitoes in the group not exposed to malaria approached the human. Calculate the z-statistic. (d) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Sporozoite stage approaching the human is higher in the group exposed to malaria. (e) Based on your p-values, make conclusions about what you have learned about mosquito behavior, stage of infection, and exposure to malaria or not. (f) Can we conclude that being exposed to malaria (as opposed to not being exposed to malaria) causes these behavior changes in mosquitoes? Why or why not?

Short Answer

Expert verified
The Z-statistics and p-values show that there can be a change in the behavior for the group of mosquitoes exposed to malaria in both Oocyst and Sporozoite stages. However, while there is statistical association, it does not imply causation and further study is needed.

Step by step solution

01

Calculating the Z-Statistic (Oocyst stage)

First, calculate the proportion of mosquitoes approaching humans in both Exposure Groups (Malaria and non-Malaria) as \(p_1=20/113\) and \(p_2=36/117\), respectively. Then, calculate the standard error as \(\sqrt{[(p_1*(1-p_1)/113)+(p_2*(1-p_2)/117)]}\). Lastly, compute the Z- statistic as \((p_1-p_2) / standard \ error\)
02

Calculating the p-value (Oocyst stage)

Use the calculated Z-statistic and a standard normal (Z) table or a software to find the p-value which is the probability that the Z-statistic is lower than the calculated value given that the null hypothesis is true.
03

Calculating the Z-Statistic (Sporozoite stage)

Using the same methodology as the Oocyst stage, calculate the proportion of mosquitoes approaching humans in both Exposure Groups (Malaria and non-Malaria) as \(p_1=37/149\) and \(p_2=14/144\), respectively. Then, calculate the standard error as \(\sqrt{[(p_1*(1-p_1)/149)+(p_2*(1-p_2)/144)]}\). Lastly, compute the Z- statistic as \((p_1-p_2) / standard \ error\)
04

Calculating the p-value (Sporozoite stage)

Use the calculated Z-statistic and a standard normal (Z) table or software to find the p-value which is the probability that the Z-statistic is higher than the calculated value given that the null hypothesis is true.
05

Drawing conclusions from p-values

Use the calculated p-values from Steps 2 and 4 to draw conclusions. If the p-value is less than a predetermined significance level of 0.05, we reject the null hypothesis and conclude that there is statistical evidence showing that the proportion of mosquitoes approaching the human is lower in the Oocyst stage for the group exposed to malaria and higher in the Sporozoite stage for the group exposed to malaria.
06

Establishing causality or correlation

Causality should not be assumed simply through statistical evidence. Further study is needed to determine whether the exposure to malaria causes these changes in behavior.

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Key Concepts

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

z-statistic calculation
To understand whether the exposure to malaria affects mosquito behavior, we can use the z-statistic calculation. A z-statistic is a criterion used to determine if two sample proportions are significantly different from each other.

For the Oocyst stage, we start by calculating the proportion of mosquitoes that approached humans in the malaria-exposed group and the unexposed group. These proportions are calculated as follows:
  • Malaria-exposed group: \( p_1 = \frac{20}{113} \)
  • Unexposed group: \( p_2 = \frac{36}{117} \)
The next step is to compute the standard error, which is essential for understanding the variability between the proportions:
  • Standard error = \( \sqrt{\left( \frac{p_1(1-p_1)}{113} + \frac{p_2(1-p_2)}{117} \right)} \)
Finally, the z-statistic is calculated using the formula \( \frac{p_1 - p_2}{\text{standard error}} \). This z-statistic helps us figure out how many standard deviations away the observed difference is from the null hypothesis of no difference.

By repeating these steps for the Sporozoite stage, we compare different proportions, which sheds light on behavioral changes in mosquitoes due to malaria exposure.
p-value interpretation
Once the z-statistic is calculated, the p-value helps us interpret whether the results are statistically significant. The p-value is a probability that explains how likely it is to observe our data if the null hypothesis is true.

In the context of mosquito behavior, the null hypothesis is that there is no difference in the proportion of mosquitoes approaching humans whether they are exposed to malaria or not.
  • If the p-value is less than 0.05, we reject the null hypothesis, indicating that there is statistical evidence to suggest a real difference in behavior.
For example, in the experiment during the Oocyst stage, a p-value less than 0.05 would suggest that malaria exposure likely decreases the frequency of mosquitoes approaching humans compared to those not exposed.

Understanding p-value interpretation is crucial because it helps researchers make informed decisions about their hypothesis, backing up conclusions with evidence from the data.
statistical significance
Statistical significance is a concept that helps determine whether the results of a study are likely due to a specific cause or simply occurred by chance.

In this mosquito study, evaluating statistical significance means determining if behaviors like approaching humans are genuinely influenced by malaria exposure. When the p-value is below the significance threshold (commonly set at 0.05), it suggests statistical significance.
  • For the Oocyst stage: A significant result implies that malaria-exposed mosquitoes are less likely to approach humans, as opposed to non-exposed ones, during this stage.
  • For the Sporozoite stage: Statistical significance indicates that malaria-exposed mosquitoes are more likely to approach humans when they become infectious.
Statistical significance gives researchers confidence in their findings, suggesting that observed patterns are likely not random, but potentially caused by the experimental conditions.
causality in experiments
While statistical significance indicates a potential association, it does not establish causality. Causality refers to a direct cause-and-effect relationship, showing that one's actions directly result in another's reaction.

In the malaria and mosquito study, although the results may suggest changes in behavior due to malaria exposure, we cannot automatically conclude causality. There must be further studies to determine whether these behavior modifications are truly caused by the parasites.
  • Randomization, as used in this study, helps reduce bias and allows for more reliable results.
  • However, despite statistical evidence, other factors may influence mosquito behavior, necessitating further experiments to confirm causality.
Ultimately, understanding causality in experiments is crucial for developing deeper scientific insights and making well-supported scientific claims.

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Most popular questions from this chapter

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether males are less likely than females to support a ballot initiative, if \(24 \%\) of a random sample of 50 males plan to vote yes on the initiative and \(32 \%\) of a random sample of 50 females plan to vote yes.

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of \(\pm 2 \%\) with \(95 \%\) confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

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