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Chapter 10: Problem 77

An experimenter was interested in determining the mean thickness of the cortex of the sea urchin egg. The thickness was measured for \(n=10\) sea urchin eggs. These measurements were obtained: $$ \begin{array}{lllll} 4.5 & 6.1 & 3.2 & 3.9 & 4.7 \\ 5.2 & 2.6 & 3.7 & 4.6 & 4.1 \end{array} $$ Estimate the mean thickness of the cortex using a \(95 \%\) confidence interval.

Short Answer

Expert verified
Answer: The estimated mean thickness of the cortex for sea urchin eggs is between 3.544 and 4.976 with a 95% confidence level.

Step by step solution

01

Find the sample mean and sample standard deviation.

First, we need to calculate the sample mean (\(\bar{x}\)) which is the sum of thickness measurements divided by the number of measurements, and the sample standard deviation (s) which is the square root of the sum of squared deviations of measurements from the mean, divided by the degrees of freedom (n-1). We will use the following formulas: $$ \bar{x} = \frac{\sum x_i}{n} $$ $$ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} $$ Using the given data, we compute the sample mean and sample standard deviation as follows. $$ \bar{x} = \frac{4.5+6.1+3.2+3.9+4.7+5.2+2.6+3.7+4.6+4.1}{10} = \frac{42.6}{10} = 4.26 $$ The squared deviations from the mean are: \begin{align*} (4.5-4.26)^2 &= 0.0576 \\ (6.1-4.26)^2 &= 3.3684 \\ (3.2-4.26)^2 &= 1.1236 \\ (3.9-4.26)^2 &= 0.1296 \\ (4.7-4.26)^2 &= 0.1936 \\ (5.2-4.26)^2 &= 0.8836 \\ (2.6-4.26)^2 &= 2.7456 \\ (3.7-4.26)^2 &= 0.3136 \\ (4.6-4.26)^2 &= 0.1156 \\ (4.1-4.26)^2 &= 0.0256 \end{align*} The sum of squared deviations is \(0.0576+3.3684+1.1236+0.1296+0.1936+0.8836+2.7456+0.3136+0.1156+0.0256=8.9564\). Therefore, the sample standard deviation is \(s = \sqrt{\frac{8.9564}{9}} = \sqrt{0.9952} = 0.997596\)
02

Find the t-value for a 95% confidence interval.

Use a t-table to look up the t-value corresponding to a 95% confidence interval for a sample size of 10 (which has 9 degrees of freedom). We find that the t-value is about \(2.262\).
03

Calculate the margin of error.

We will now calculate the margin of error (ME) using the t-value, the sample standard deviation, and the sample size. The formula for the margin of error is: $$ ME = t \times \frac{s}{\sqrt{n}} $$ Using our t-value of \(2.262\), the sample standard deviation of \(0.997596\) and a sample size of \(10\), the margin of error is: $$ ME = 2.262 \times \frac{0.997596}{\sqrt{10}} = 2.262 \times \frac{0.997596}{3.162} = 0.716 $$
04

Calculate the 95% confidence interval.

To calculate the 95% confidence interval, we subtract the margin of error from the sample mean and add the margin of error to the sample mean. This gives us the lower and upper bounds for the confidence interval. $$ CI = (\bar{x} - ME, \bar{x} + ME) = (4.26 - 0.716, 4.26 + 0.716) = (3.544, 4.976) $$ The estimated mean thickness of the cortex for sea urchin eggs is between \(3.544\) and \(4.976\) with a \(95\%\) confidence level.

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Key Concepts

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

Sample Mean
The sample mean is a key concept in statistics, serving as an estimate of the true mean of a population. In this exercise, the sample mean represents the average thickness of the sea urchin egg cortex based on the measurements collected from the sample.

To find the sample mean, we divide the sum of all measured values by the number of observations. It is mathematically represented as \[\bar{x} = \frac{\sum x_i}{n}\]where \(\bar{x}\) is the sample mean, \(\sum x_i\) is the sum of all measurements, and \(n\) is the total number of samples.

This gives us a point estimate of the mean, which, while informative, is subject to sampling variation.
  • In our example, \( n = 10 \) measurements resulted in a sample mean of \( \bar{x} = 4.26 \).
Understanding the sample mean helps us comprehend how close our sample's average value is to the true population mean, assuming the sample is representative.
Sample Standard Deviation
Sample standard deviation provides insight into the variability or spread of the sample data around the sample mean. It is a measure commonly used to quantify the amount of variation or dispersion in a set of values.

The formula for sample standard deviation is:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]This formula takes the square root of the average of the squared deviations of the sample values from the sample mean.
  • \( s \) denotes the sample standard deviation,
  • \( x_i \) are individual sample points,
  • \( \bar{x} \) is the sample mean,
  • \( n \) is the number of observations.
In this exercise, the standard deviation helps quantify how much the thickness measurements of the sea urchin egg cortex vary around the mean of 4.26.

Our calculation yields a sample standard deviation of approximately \( s = 0.997596 \). The standard deviation's usefulness is in providing a measure of variability or uncertainty within our sample data.
T-Distribution
When dealing with small sample sizes, such as our sample of 10 sea urchin thickness measurements, the T-distribution is more appropriate than the normal distribution for estimation purposes. The T-distribution accounts for the increased variability found in smaller sample sizes.

The T-distribution is broader with heavier tails compared to the normal distribution, allowing for the uncertainty and increased variability associated with smaller samples.
  • This exercise makes use of the T-distribution to calculate the confidence interval for the sample mean.
  • Given our sample size of 10, there's \( n-1 = 9 \) degrees of freedom.
  • Using a statistical table, the T-value corresponding to a 95\% confidence level and 9 degrees of freedom is approximately 2.262.
The chosen T-value critically influences the calculation of the margin of error, which impacts the final confidence interval range.
Margin of Error
The margin of error (ME) quantifies the uncertainty surrounding the sample estimate. It provides a range that, with a certain level of confidence, likely includes the true population mean.

To calculate the margin of error, we use the following formula:\[ME = t \times \frac{s}{\sqrt{n}}\]Where:
  • \( t \) is the T-value for the desired confidence level, in this example, \( t = 2.262 \).
  • \( s = 0.997596 \) is the sample standard deviation.
  • \( n = 10 \) is the sample size.
The margin of error for our exercise is calculated as \( ME = 0.716 \).

Finally, to construct the confidence interval, we add and subtract the margin of error from the sample mean:\[CI = (\bar{x} - ME, \bar{x} + ME)\]This results in a confidence interval of \((3.544, 4.976)\), indicating that the true mean thickness of the sea urchin egg cortex is likely within this range with 95\% confidence.

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Most popular questions from this chapter

Chronic anterior compartment syndrome is a condition characterized by exercise-induced pain in the lower leg. Swelling and impaired nerve and muscle function also accompany this pain, which is relieved by rest. Susan Beckham and colleagues conducted an experiment involving 10 healthy runners and 10 healthy cyclists to determine whether there are significant differences in pressure measurements within the anterior muscle compartment for runners and cyclists. \(^{7}\) The data summary - compartment pressure in millimeters of mercury \((\mathrm{Hg})\) -is as follows: a. Test for a significant difference in the average compartment pressure between runners and cyclists under the resting condition. Use \(\alpha=.05 .\) b. Construct a \(95 \%\) confidence interval estimate of the difference in means for runners and cyclists under the condition of exercising at \(80 \%\) of maximal oxygen consumption. c. To test for a significant difference in the average compartment pressures at maximal oxygen consumption, should you use the pooled or unpooled \(t\) -test? Explain.

Independent random samples of \(n_{1}=16\) and \(n_{2}=13\) observations were selected from two normal populations with equal variances: $$ \begin{array}{lrr} & & {\text { Population }} \\ { 2 - 3 } & 1 & 2 \\ \hline \text { Sample Size } & 16 & 13 \\ \text { Sample Mean } & 34.6 & 32.2 \\ \text { Sample Variance } & 4.8 & 5.9 \end{array} $$ a. Suppose you wish to detect a difference between the population means. State the null and alternative hypotheses for the test. b. Find the rejection region for the test in part a for \(\alpha=.01\) c. Find the value of the test statistic. d. Find the approximate \(p\) -value for the test. e. Conduct the test and state your conclusions.

The object of a general chemistry experiment is to determine the amount (in milliliters) of sodium hydroxide ( \(\mathrm{NaOH}\) ) solution needed to neutralize 1 gram of a specified acid. This will be an exact amount, but when the experiment is run in the laboratory, variation will occur as the result of experimental error. Three titrations are made using phenolphthalein as an indicator of the neutrality of the solution (pH equals 7 for a neutral solution). The three volumes of \(\mathrm{NaOH}\) required to attain a \(\mathrm{pH}\) of 7 in each of the three titrations are as follows: \(82.10,75.75,\) and 75.44 milliliters. Use a \(99 \%\) confidence interval to estimate the mean number of milliliters required to neutralize 1 gram of the acid.

Find the critical value(s) of \(t\) that specify the rejection region in these situations: a. A two-tailed test with \(\alpha=.01\) and \(12 d f\) b. A right-tailed test with \(\alpha=.05\) and \(16 d f\) c. A two-tailed test with \(\alpha=.05\) and \(25 d f\) d. A left-tailed test with \(\alpha=.01\) and \(7 d f\)

A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces helmets transmit to wearers when subjected to a standard external force. The manufacturer desires the mean force transmitted by helmets to be 800 pounds (or less), well under the legal 1000 -pound limit, and \(\sigma\) to be less than \(40 .\) A random sample of \(n=40\) helmets was tested, and the sample mean and variance were found to be equal to 825 pounds and 2350 pounds \(^{2}\), respectively. a. If \(\mu=800\) and \(\sigma=40,\) is it likely that any helmet, subjected to the standard external force, will transmit a force to a wearer in excess of 1000 pounds? Explain. b. Do the data provide sufficient evidence to indicate that when the helmets are subjected to the standard external force, the mean force transmitted by the helmets exceeds 800 pounds?
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