Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9, s_{d}=\) \(143.6, n_{d}=100\)

What Gives a Small P-value? In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

Metal Tags on Penguins and Arrival Dates Data 1.3 on page 10 discusses a study designed to test whether applying a metal tag is detrimental to a penguin, as opposed to applying an electronic tag. One variable examined is the date penguins arrive at the breeding site, with later arrivals hurting breeding success. Arrival date is measured as the number of days after November \(1^{\text {st }}\). Mean arrival date for the 167 times metal-tagged penguins arrived was December \(7^{\text {th }}\left(37\right.\) days after November \(\left.1^{\text {st }}\right)\) with a standard deviation of 38.77 days, while mean arrival date for the 189 times electronic-tagged penguins arrived at the breeding site was November \(21^{\text {st }}(21\) days after November \(\left.1^{\text {st }}\right)\) with a standard deviation of \(27.50 .\) Do these data provide evidence that metal tagged penguins have a later mean arrival time? Show all details of the test.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free