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Chapter 8: Problem 113

If 36 measurements of the specific gravity of aluminum had a mean of 2.705 and a standard deviation of .028 , construct a \(98 \%\) confidence interval for the actual specific gravity of aluminum.

Short Answer

Expert verified
Answer: The 98% confidence interval for the actual specific gravity of aluminum is [2.69226, 2.71774].

Step by step solution

01

Identify the appropriate statistics

First, note down the given values from the problem: Sample mean (x虅) = 2.705 Standard deviation (s) = 0.028 Sample size (n) = 36 Confidence level = 98%
02

Find the t-score

In order to construct a 98% confidence interval, we must determine the appropriate t-distribution score for the given confidence level and sample size. We can either use a t-score table or statistical software to find the t-score. We want the value of t that captures 99% of the distribution (half of the remaining 2% is in the left tail, and the other half is in the right tail). With a sample size of 36, we have 35 degrees of freedom (\(df = n -1\)). Using a t-score table or software, we find that \(t_{0.01, 35} = 2.728\).
03

Calculate the standard error

Next, we need to calculate the standard error of the mean (SEM). The formula for SEM is given by: $$SEM = \frac{s}{\sqrt{n}}$$ Plug in the values from Step 1: $$SEM = \frac{0.028}{\sqrt{36}}$$ and calculate SEM to get: $$SEM = 0.00467$$
04

Calculate the margin of error

Now we need to calculate the margin of error (ME) using the t-score that we found in Step 2 and the standard error from Step 3. The formula for ME is given by: $$ME = t * SEM$$ Plug in the values from Steps 2 and 3: $$ME = 2.728 * 0.00467$$ and calculate ME to get: $$ME = 0.01274$$
05

Calculate the confidence interval

Finally, we can calculate the 98% confidence interval using the sample mean, and the margin of error. The formula for the confidence interval is given by: $$CI = x虅 \pm ME$$ Plug in the values from Steps 1 and 4: $$CI = 2.705 \pm 0.01274$$ and calculate the lower and upper bound for the interval: $$CI = [2.705 - 0.01274, 2.705 + 0.01274] = [2.69226, 2.71774]$$ Now we have constructed the 98% confidence interval for the actual specific gravity of aluminum, which is [2.69226, 2.71774].

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Key Concepts

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

Specific Gravity
Specific gravity is a term used to describe the ratio of the density of a substance to the density of a reference substance. In most cases, the reference substance is water for liquids and air for gases. For example, the specific gravity of aluminum in this problem is measured in comparison to water. This is a dimensionless quantity meaning it has no units.
  • A specific gravity greater than 1 means the substance is denser than the reference and will sink, like aluminum in water.
  • A specific gravity less than 1 indicates the substance is less dense than the reference and will float.
  • Because specific gravity is a ratio, it provides a useful measure to identify substances and compare their densities without the need for units.
T-Distribution
The t-distribution is a type of probability distribution that is helpful when working with small sample sizes or when the population standard deviation is unknown. It resembles the standard normal distribution (bell curve) but has heavier tails, meaning it is more prone to producing values that fall far from its mean.
  • The shape of the t-distribution changes with the degrees of freedom, which is typically the sample size minus one (\( n - 1 \).
  • As the sample size increases, the t-distribution approaches a normal distribution.
  • The value we derive from the t-distribution, known as the t-score, helps in constructing confidence intervals and conducting hypothesis tests.
In the exercise, we found the t-score for a 98% confidence interval with 35 degrees of freedom to be 2.728. This score is used to calculate the margin of error, critical for determining the confidence interval.
Margin of Error
The margin of error (ME) represents the range within which we can be fairly certain the true population parameter lies. It is a critical component in constructing confidence intervals, such as the interval for the specific gravity of aluminum in this exercise.
  • The formula for the margin of error combines the t-score and standard error: \( ME = t \times SEM \).
  • A larger margin of error indicates a wider confidence interval, suggesting less precision in the estimate.
  • Conversely, a smaller margin of error implies greater precision and a narrower confidence interval.
In the example, the margin of error was calculated as 0.01274. This means that we are confident the actual specific gravity of aluminum is within 0.01274 above or below the sample mean of 2.705.
Standard Error
The standard error (SE) measures the variability of a sample mean estimate from the population mean. It is key in calculating the margin of error and, consequently, the confidence interval.
  • The formula for standard error is: \( SEM = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • The smaller the SEM, the more representative the sample mean is of the population mean.
  • For a fixed sample size, a smaller standard deviation reduces the SEM, improving the sample estimate's accuracy.
In this exercise, with a sample standard deviation of 0.028 and sample size of 36, the SEM was calculated to be 0.00467. This low value reflects that there is not much variation in the sample mean, suggesting an accurate estimation of the population mean for the specific gravity of aluminum.

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Most popular questions from this chapter

Samples of 400 printed circuit boards were selected from each of two production lines \(A\) and \(B\). Line A produced 40 defectives, and line B produced 80 defectives. Estimate the difference in the actual fractions of defectives for the two lines with a confidence coefficient of \(.90 .\)

In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24 -hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error.

An entomologist wishes to estimate the average development time of the citrus red mite correct to within .5 day. From previous experiments it is known that \(\sigma\) is approximately 4 days. How large a sample should the entomologist take to be \(95 \%\) confident of her estimate?

Do you think that the United States should pursue a program to send humans to Mars? An opinion poll conducted by the Associated Press indicated that \(49 \%\) of the 1034 adults surveyed think that we should pursue such a program. \(^{5}\) a. Estimate the true proportion of Americans who think that the United States should pursue a program to send humans to Mars. Calculate the margin of error. b. The question posed in part a was only one of many questions concerning our space program that were asked in the opinion poll. If the Associated Press wanted to report one sampling error that would be valid for the entire poll, what value should they report?

What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{9}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) and a standard deviation of \(0.73^{\circ} .\) a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ},\) the usual average temperature cited by physicians and others? If not, what conclusions can you draw?
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