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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. \({ }^{25}\) 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
From the calculations, best estimate for the length of gribbles is 3.1 mm. The margin of error for this estimate at 95% confidence level would be calculated based on SE and the Z-value for 95% confidence level. The 95% confidence interval would be the range calculated by subtracting and adding the margin of error from the mean. This interval gives us an estimation of where the true population mean lies with 95% confidence. The assumptions are that the sample is a simple random sample representative of the population and the population distribution is approximately normal.

Step by step solution

01

Calculate Standard Error (SE)

The standard error can be calculated using the formula: SE = \(\frac{s}{\sqrt{n}}\) where \(s\) is the standard deviation and \(n\) is the size of the sample. In this case \(s = 0.72\) and \(n = 50\). Plug these values into the formula to find the SE.
02

Calculate Margin of Error (ME)

The margin of error can be calculated using the formula: ME = SE * Z where the Z value corresponds to the desired confidence level. For a 95% confidence interval, the Z value is 1.96. Multiply the SE calculated in step 1 by 1.96 to find the ME.
03

Calculate 95% Confidence Interval

A confidence interval can be calculated using the formula: CI = mean ± ME. The mean given in the problem is 3.1. Add and subtract the ME calculated in step 2 from the mean to find the confidence interval.
04

Interpret the Confidence Interval

The confidence interval calculated in step 3 is an estimate of the range in which the true population parameter (the length of a gribble) lies with a 95% level of confidence. It means, if the experiment were repeated many times, 95% of the intervals would contain the true population parameter.
05

State the Assumptions

To have confidence in our estimate, we must assume that the sample of gribbles is a simple random sample - it represents a portion of the entire population of gribbles without any bias. Also, we assume that the distribution of gribble length in the population is approximately normal.

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Most popular questions from this chapter

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the sample results \(\bar{x}_{1}=15.3, s_{1}=11.6\) with \(n_{1}=100\) and \(\bar{x}_{2}=18.4, s_{2}=14.3\) with \(n_{2}=80\).

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 \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lcc} \hline \text { Case } & \text { Situation 1 } & \text { Situation 2 } \\ \hline 1 & 77 & 85 \\ 2 & 81 & 84 \\ 3 & 94 & 91 \\ 4 & 62 & 78 \\ 5 & 70 & 77 \\ 6 & 71 & 61 \\ 7 & 85 & 88 \\ 8 & 90 & 91 \\ \hline \end{array} $$

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

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Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

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