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Chapter 6: Problem 117

Gribbles are small, pale white, marine worms that bore through wood. While sometimes considered a pest since they can wreck wooden docks and piers, they are now being studied to determine whether the enzyme they secrete will allow us to turn inedible wood and plant waste into biofuel. \(^{35}\) A sample of 50 gribbles finds an average length of \(3.1 \mathrm{~mm}\) with a standard deviation of 0.72 . Give a best estimate for the length of gribbles, a margin of error for this estimate (with \(95 \%\) confidence), and a \(95 \%\) confidence interval. Interpret the confidence interval in context. What do we have to assume about the sample in order to have confidence in our estimate?

Short Answer

Expert verified
The best estimate for the length of gribbles is 3.1mm. The margin of error for this estimate, with 95% confidence, is approximately 0.1995mm. Thus, the 95% confidence interval is approximately [2.9005 mm, 3.2995 mm]. This means we are 95% confident that the true average length of gribbles lies within this range. The assumptions made are that the gribbles were randomly chosen, the lengths are independent of each other, and the distribution of lengths is approximately normal.

Step by step solution

01

Calculate the mean

First, take the average length of the gribbles as the best estimate. This is 3.1mm.
02

Calculate the standard error of the mean

The standard error of the mean is the standard deviation divided by the square root of the sample size. Here, it's \(0.72 / \sqrt{50} \approx 0.1018 \, mm\)
03

Compute for the Margin of Error

The margin of error for a 95% confidence interval is usually calculated as 1.96 times the standard error. Here, it's \(1.96 \times 0.1018 \approx 0.1995 \, mm\)
04

Calculate the Confidence Interval

The 95% confidence interval is the mean plus or minus the margin of error. Here, it's \(3.1 \pm 0.1995\), so the interval is \([2.9005 \, mm, 3.2995 \, mm]\)
05

Interpretation and Assumptions

The interpretation is that we can be 95% confident that the true average length of gribbles lies between 2.9005 mm and 3.2995 mm. For the assumptions, we have to assume that the sample of gribbles was randomly chosen, that the lengths of different gribbles are independent of each other, and that the distribution of lengths is approximately normal.

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Key Concepts

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

Standard Error
Standard error is a crucial concept when attempting to estimate the precision of a sample mean. It is the measure of the variability or spread of sample means around the true population mean if you were to take multiple samples. To calculate the standard error, we divide the standard deviation by the square root of the sample size. In our gribble example, the standard deviation is 0.72 mm, and the sample size is 50. Thus, the standard error is calculated as \( \frac{0.72}{\sqrt{50}} \approx 0.1018 \, \text{mm} \). This tells us how much the sample mean would be expected to vary from one sample to another.
  • Smaller standard errors indicate more precise estimates of the mean.
  • Larger sample sizes result in smaller standard errors, highlighting the importance of larger samples in research.
  • The standard error is directly used to calculate the margin of error for confidence intervals.
Margin of Error
The margin of error provides a range above and below the sample mean within which you expect the true population mean to lie, with a certain level of confidence. It signifies the degree of uncertainty or potential error in the sample estimate. In this exercise, the margin of error is calculated using 1.96 times the standard error, which is appropriate for a 95% confidence interval under a normal distribution. Here, the margin of error is \( 1.96 \times 0.1018 \approx 0.1995 \, \text{mm} \).
  • This margin means that we add and subtract 0.1995 mm from the sample mean to create the confidence interval.
  • The margin of error accounts for sampling variability, reflecting how much the sample mean might deviate from the true population mean due to random chance.
  • The larger the margin of error, the less precise our estimate is.
Normal Distribution Assumptions
Calculating a confidence interval and making reliable inferences depends on certain assumptions, especially concerning the distribution of the underlying data. One critical assumption is that the data are approximately normally distributed. In the case of our sample of gribbles, this assumption allows us to apply the normal distribution's properties to estimate the population mean. To ensure the validity of this assumption:
  • The sample should be randomly chosen to avoid bias in the estimates.
  • Observations should be independent, meaning the length of one gribble should not affect another's.
  • The sample should ideally be large enough (usually n > 30 is a safe benchmark) for the Central Limit Theorem to hold, which states that the sampling distribution of the sample mean will be approximately normal regardless of the population distribution.
  • If the above conditions are met, it is reasonable to use the normal distribution to approximate confidence intervals and make inferences about the population mean.

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Most popular questions from this chapter

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.) $$ \begin{aligned} &\text { Two-Sample T-Test and Cl }\\\ &\begin{array}{lrrrr} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Late } & 32 & 22.56 & 5.13 & 0.91 \\ \text { Early } & 30 & 19.73 & 6.61 & 1.2 \end{array} \end{aligned} $$ Difference \(=\mathrm{mu}\) (Late) \(-\mathrm{mu}\) (Early) Estimate for difference: 2.83 \(95 \%\) Cl for difference: (-0.20,5.86) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=1.87\) P-Value \(=0.066 \quad \mathrm{DF}=54\)

Percent of Free Throws Made Usually, in sports, we expect top athletes to get better over time. We expect future athletes to run faster, jump higher, throw farther. One thing has remained remarkably constant, however. The percent of free throws made by basketball players has stayed almost exactly the same for 50 years. \(^{5}\) For college basketball players, the percent is about \(69 \%,\) while for players in the NBA (National Basketball Association) it is about \(75 \%\). (The percent in each group is also very similar between male and female basketball players.) In each case below, find the mean and standard deviation of the distribution of sample proportions of free throws made if we take random samples of the given size. (a) Samples of 100 free throw shots in college basketball (b) Samples of 1000 free throw shots in college basketball (c) Samples of 100 free throw shots in the \(\mathrm{NBA}\) (d) Samples of 1000 free throw shots in the \(\mathrm{NBA}\)

What Percent of Houses Are Owned vs }\end{array}\( Rented? The 2010 US Census \)^{4}\( reports that, of all the nation's occupied housing units, \)65.1 \%\( are owned by the occupants and \)34.9 \%$ are rented. If we take random samples of 50 occupied housing units and compute the sample proportion that are owned for each sample, what will be the mean and standard deviation of the distribution of sample proportions?

On page 1 1 in Section \(1.1,\) we describe studies to investigate whether there is evidence of pheromones (subconscious chemical signals) in female tears that affect sexual arousal in men. In one of the studies, \(^{71} 50\) men had a pad attached to the upper lip that contained either female tears or a salt solution dripped down the same female's face. Each subject participated twice, on consecutive days, once with tears and once with saline, randomized for order, and doubleblind. Testosterone levels were measured before sniffing and after sniffing on both days. While normal testosterone levels vary significantly between different men, average levels for the group were the same before sniffing on both days and after sniffing the salt solution (about \(155 \mathrm{pg} / \mathrm{mL}\) ) but were reduced after sniffing the tears (about \(133 \mathrm{pg} / \mathrm{mL}\) ). The mean difference in testosterone levels after sniffing the tears was 21.7 with standard deviation \(46.5 .\) (a) Why did the investigators choose a matchedpairs design for this experiment? (b) Test to see if testosterone levels are significantly reduced after sniffing tears? (c) Can we conclude that sniffing female tears reduces testosterone levels (which is a significant indicator of sexual arousal in men)?

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