/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 The following statistics are the... [FREE SOLUTION] | 91Ó°ÊÓ

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The following statistics are the result of an experiment conducted by P. I. Ward to investigate a theory concerning the molting behavior of the male Gammarus pulex, a small crustacean. \(^{\star}\) If a male needs to molt while paired with a female, he must release her, and so loses her. The theory is that the male \(G\). pulex is able to postpone molting, thereby reducing the possibility of losing his mate. Ward randomly assigned 100 pairs of males and females to two groups of 50 each. Pairs in the first group were maintained together (normal); those in the second group were separated (split). The length of time to molt was recorded for both males and females, and the means, standard deviations, and sample sizes are shown in the accompanying table. (The number of crustaceans in each of the four samples is less than 50 because some in each group did not survive until molting time.) a. Find a \(99 \%\) confidence interval for the difference in mean molt time for "normal" males versus those "split" from their mates. b. Interpret the interval.

Short Answer

Expert verified
Compute the confidence interval for the difference in mean molt times using the provided data and interpret the results regarding the theory.

Step by step solution

01

Identify the Statistical Information

From the problem, we have data for two groups: 'normal' males (males with female) and 'split' males (males without female). We are given the means, standard deviations, and sample sizes for these groups. Let's assume: Mean A = \( \bar{x}_1 \), SD A = \( s_1 \), sample size A = \( n_1 \) for 'normal' males; and Mean B = \( \bar{x}_2 \), SD B = \( s_2 \), sample size B = \( n_2 \) for 'split' males.
02

Formula for Confidence Interval

The formula to calculate the confidence interval for the difference between two means is given by: \[ (\bar{x}_1 - \bar{x}_2) \pm z \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Where \( z \) is the z-score corresponding to the desired confidence level, which is \(99\%\) in our case.
03

Find the Z-score for 99% Confidence

For a 99% confidence interval, the z-score (critical value) can be found from a standard normal distribution table. The z-score corresponding to a 99% confidence level is approximately 2.576.
04

Calculate the Standard Error

Use the given standard deviations \( s_1 \) and \( s_2 \), and the sample sizes \( n_1 \) and \( n_2 \) to calculate the standard error:\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
05

Calculate the Confidence Interval

Substitute \( \bar{x}_1 \), \( \bar{x}_2 \), the z-score value, and the calculated standard error into the confidence interval formula:\[ (\bar{x}_1 - \bar{x}_2) \pm 2.576 \times SE \]
06

Interpret the Interval

Once the confidence interval is computed, it gives the range within which the true difference in mean molt times for 'normal' versus 'split' males lies with 99% confidence. If this interval does not include zero, it suggests a significant difference in molt times between the two groups, affirming or refuting the crustaceans' ability to postpone molting.

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Key Concepts

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

Understanding the Difference of Means
When comparing two distinct groups, it's essential to understand the concept of "difference of means." In this context, it involves comparing the average molt times of 'normal' males (those with a female) to 'split' males (those without a female). The difference between these two means helps us assess whether the group condition (being with or without a mate) affects the time to molt.

The formula for calculating the difference of means is simple: subtract one group's mean from the other:
  • Difference of Means = Mean of Group A - Mean of Group B
This difference is a crucial part of inference to determine whether any observed difference is statistically significant or likely due to random variation.

By creating a confidence interval around this difference, researchers can assess whether the difference is representative of the entire population of crustaceans.
Basics of Statistical Hypothesis Testing
Statistical hypothesis testing is a method used to determine if there are significant differences between groups, like our 'normal' and 'split' males. A hypothesis test involves a null hypothesis, typically stating that there is no difference between the groups, and an alternative hypothesis, suggesting a difference exists.

Here's a simple outline of hypothesis testing:
  • Null Hypothesis (H0): There is no difference in mean molt times between 'normal' and 'split' males.
  • Alternative Hypothesis (H1): There is a difference in mean molt times.
  • Determine the significance level (often 0.05 or 0.01).
  • Use statistical tools to analyze the data and calculate a p-value or construct a confidence interval.
If the calculated confidence interval for the difference of means does not include zero, it suggests that the difference observed is statistically significant, thus rejecting the null hypothesis.
The Z-Score in Confidence Intervals
The z-score plays a critical role in constructing confidence intervals, especially when the population standard deviations are known or the sample size is large. It reflects how many standard deviations away a data point is from the mean in a standard normal distribution.

For our problem, a 99% confidence interval is required. The z-score for this confidence level is approximately 2.576. This value is derived from standard normal distribution tables and indicates that the constructed interval will capture the true difference of means 99% of the time.

Including the z-score in the confidence interval formula helps set the range within which we believe the true difference lies. It accounts for variability in the data and helps estimate the precision and reliability of our observed difference.
Calculating the Standard Error
The standard error is a measure of the variability or dispersion of the sample mean from the true population mean. In the context of comparing two means, it gives insight into how much the means of the samples are expected to fluctuate.

For our exercise, the standard error (SE) is calculated using the formula:
  • SE = \( \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \)
Here, \( s_1 \) and \( s_2 \) represent the standard deviations of the 'normal' and 'split' male groups, while \( n_1 \) and \( n_2 \) represent the sample sizes.

Calculating the standard error is a vital step in determining the confidence interval. It accommodates the sample sizes and variability of the data, providing a clearer picture of how confident we can be in our results. A smaller standard error indicates that the sample means are close to the true difference, enhancing the reliability of our confidence interval.

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Most popular questions from this chapter

Industrial light bulbs should have a mean life length acceptable to potential users and a relatively small variation in life length. If some bulbs fail too early in their life, users become annoyed and are likely to switch to bulbs produced by a different manufacturer. Large variations above the mean reduce replacement sales; in general, variation in life lengths disrupts the user's replacement schedules. A random sample of 20 bulbs produced by a particular manufacturer produced the following lengths of life (in hours): $$\begin{array}{lllllllll}2100 & 2302 & 1951 & 2067 & 2415 & 1883 & 2101 & 2146 & 2278 & 2019 \\ 1924 & 2183 & 2077 & 2392 & 2286 & 2501 & 1946 & 2161 & 2253 & 1827\end{array}$$ Set up a \(99 \%\) upper confidence bound for the standard deviation of the lengths of life for the bulbs produced by this manufacturer. Is the true population standard deviation less than 150 hours? Why or why not?

In a poll taken among college students, 300 of 500 fraternity men favored a certain proposition whereas 64 of 100 nonfraternity men favored it. Estimate the difference in the proportions favoring the proposition and place a 2 -standard-deviation bound on the error of estimation.

Given a random sample of size \(n\) from a normal population with unknown mean and variance, we developed a confidence interval for the population variance \(\sigma^{2}\) in this section. What is the formula for a confidence interval for the population standard deviation \(\sigma ?\)

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample from a population with probability density function $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta+1}\right) e^{-y /(\theta+1)}, & y>0, \theta>-1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Suggest a suitable statistic to use as an unbiased estimator for \(\theta\)

We noted in Section 8.3 that if $$S^{\prime 2}=\frac{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}{n} \text { and } S^{2}=\frac{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}{n-1}$$ then \(S^{\prime 2}\) is a biased estimator of \(\sigma^{2}\), but \(S^{2}\) is an unbiased estimator of the same parameter. If we sample from a normal population, a. find \(V\left(S^{\prime 2}\right)\) b. show that \(V\left(S^{2}\right)>V\left(S^{\prime 2}\right)\)

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