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Chapter 6: Problem 148

Metal Tags on Penguins and Survival Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. One variable examined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a \(90 \%\) confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins \(\left(p_{M}-p_{E}\right)\). Interpret the result.

Short Answer

Expert verified
The \(90 \%\) confidence interval for the difference in survival proportions between metal-tagged and electronic-tagged penguins is \([-0.30945, -0.05054]\).

Step by step solution

01

Calculate the Proportions

First, calculate the proportions of surviving penguins for both groups. This can be done by dividing the number of survivors in each group by the total number of individuals in that group: \n\[\hat{p_{M}} = \frac{10}{50} = 0.2\] and\n\[\hat{p_{E}} = \frac{18}{50} = 0.36\]
02

Compute the Standard Error

Next, calculate the standard error using the following formula: \n\[\sqrt{\frac{\hat{p_{M}} \cdot (1-\hat{p_{M}})}{n_{M}} + \frac{\hat{p_{E}} \cdot (1-\hat{p_{E}})}{n_{E}}} = \sqrt{\frac{0.2 \cdot 0.8}{50} + \frac{0.36 \cdot 0.64}{50}} = 0.09173...\]
03

Determine the Critical Value

Then find the critical value from the standard normal distribution that corresponds to the midpoint of the desired \(90 \%\) confidence interval. The critical value \( z_{\alpha / 2} = 1.645\) for a \(90 \%\) confidence interval.
04

Construct the Confidence Interval

Finally, construct the difference's confidence interval by adding and subtracting the critical value times the standard error from the difference in survival rates: \n\[\hat{p_{M}} - \hat{p_{E}} \pm z_{\alpha /2} * SE = 0.2 - 0.36 \pm 1.645 * 0.09173 = [-0.30945, -0.05054]\]
05

Interpret the Results

Interpretation of the confidence interval says that with \(90 \%\) confidence, the true difference in survival proportion between metal-tagged penguins and electronic-tagged penguins lies between \(-0.30945\) and \(-0.05054\). This interval does not include zero, suggesting that there may be a real difference in survival rates between the two tagging groups, with the electronic-tagged penguins surviving at a higher rate.

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Key Concepts

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

Proportion
In statistics, a proportion refers to the fraction of the total that possesses a certain characteristic. It's expressed as a ratio or percentage. In the context of our penguin study, we are interested in the survival rates of the tagged penguins. For the metal tagged penguins, the proportion of survivors is calculated as \[ \hat{p_{M}} = \frac{10}{50} = 0.2 \] This means 20% of the metal tagged penguins survived. Similarly, for the electronic tagged penguins, the proportion is \[ \hat{p_{E}} = \frac{18}{50} = 0.36 \] indicating 36% survived.
  • Proportion helps us quickly understand the share of a group exhibiting a particular trait.
  • It provides a way to compare different groups, such as metal versus electronic tagged penguins.
Using proportions allows us to directly compare different groups and deduce which one performed better in terms of survival, as done in this study.
Standard Error
Standard error is a measure that quantifies the amount of variability or dispersion of a sample statistic from the population parameter. It estimates how much the sample mean (or proportion) would vary if you drew multiple samples from the same population. In our penguin study, the standard error helps us understand the variability in the difference between the two proportions.
  • A smaller standard error indicates less variability among sample statistics, meaning our sample estimate is more precise.
  • A larger standard error generally indicates more variability, leading to less confidence in the sample statistic as an estimate of the population parameter.
To calculate the standard error of the difference between two proportions, we use the formula: \[ SE = \sqrt{\frac{\hat{p_{M}} \cdot (1-\hat{p_{M}})}{n_{M}} + \frac{\hat{p_{E}} \cdot (1-\hat{p_{E}})}{n_{E}}} = 0.09173 \] This provides a measure of how much the sample proportions (0.2 and 0.36) are expected to vary due to sampling error.
Critical Value
The critical value is a key component in inferential statistics, used to denote the threshold at which the null hypothesis is rejected in a significance test. In constructing confidence intervals, it helps us determine the margin we allow around our point estimate.
  • The critical value is derived from a probability distribution, commonly the standard normal distribution in cases involving proportions.
  • For a given confidence level, the critical value indicates the z-score at which the remaining probability is equally split between the tails of the distribution.
In our penguin study, to construct a 90% confidence interval, the critical value is \( z_{\alpha / 2} = 1.645 \). This z-score corresponds to the tails containing 5% of the data each (10% of total), leaving 90% in the center. By using this value, we establish how far values can deviate from the estimated difference in proportions, \(\hat{p_{M}} - \hat{p_{E}}\), while still being within the confidence interval.

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Most popular questions from this chapter

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Raisins after 20 Days After 20 days, 275 of the 500 fruit flies eating organic raisins are still alive, while 170 of the 500 eating conventional raisins are still alive.

Exercises 6.192 and 6.193 examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DEHP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=83.6\) with \(s_{F}=194.7\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=59.1\) with \(s_{N}=152.1\)

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Do Babies Prefer Speech? Psychologists in Montreal and Toronto conducted a study to determine if babies show any preference for speech over general noise. \(^{61}\) Fifty infants between the ages of \(4-13\) months were exposed to both happy-sounding infant speech and a hummed lullaby by the same woman. Interest in each sound was measured by the amount of time the baby looked at the woman while she made noise. The mean difference in looking time was 27.79 more seconds when she was speaking, with a standard deviation of 63.18 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies prefer actual speaking to humming.

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 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$
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