Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Due to a variation in laboratory techniques, impurities in materials, and other unknown factors, the results of an experiment in a chemistry laboratory will not always yield the same numerical answer. In an electrolysis experiment, a class measured the amount of copper precipitated from a saturated solution of copper sulfate over a 30 -minute period. The \(n=30\) students calculated a sample mean and standard deviation equal to .145 and .0051 mole, respectively. Find a \(90 \%\) confidence interval for the mean amount of copper precipitated from the solution over a 30 -minute period.

Short Answer

Expert verified
Answer: The 90% confidence interval for the mean amount of copper precipitated is approximately (0.1434, 0.1466) mole.

Step by step solution

01

Identify the relevant information

From the exercise we have: Sample size: \(n = 30\) Sample mean: \(\bar{X} = 0.145\) mole Sample standard deviation: \(s = 0.0051\) mole Confidence level: \(90\%\)
02

Determine the degree of freedom and the t-score

The degree of freedom is given by \(df = n - 1\), where \(n\) is the sample size. In this case, we have: \(df = 30 - 1 = 29\). With a 90% confidence level and 29 degrees of freedom, we need to determine the t-score that corresponds to the 95th percentile (since we take half of the 10% as the area in each tail). Using a t-distribution table or calculator, we find that the t-score is approximately: \(t_{0.95,29} \approx 1.699\)
03

Calculate the margin of error

The margin of error is given by: Margin of error \(= t_{0.95,29} \times \frac{s}{\sqrt{n}}\) Plugging in the given values: Margin of error \(= 1.699 \times \frac{0.0051}{\sqrt{30}} \approx 0.0016\)
04

Find the confidence interval

Now we can find the confidence interval by adding and subtracting the margin of error from the sample mean: Lower bound: \(\bar{X} - \text{Margin of error} = 0.145 - 0.0016 = 0.1434\) Upper bound: \(\bar{X} + \text{Margin of error} = 0.145 + 0.0016 = 0.1466\) Confidence Interval \(= (0.1434, 0.1466)\) Thus, the 90% confidence interval for the mean amount of copper precipitated from the saturated solution of copper sulfate over a 30-minute period is approximately \((0.1434,\) \(0.1466)\) mole.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Sample Size
When conducting a study or an experiment, the sample size, denoted as , is crucial because it represents the number of observations or measurements taken from the population. A larger sample size can lead to more accurate estimates of population parameters due to a reduction in sampling error. In statistics, the precision of your findings, particularly in confidence interval calculations, often increases with a larger sample size as it provides a better representation of the population.

For example, in the provided exercise, the sample size was = 30, meaning that 30 students took measurements. This sample size is considered sufficient for estimating the population mean with a reasonable level of precision, but it is also small enough to require the use of the t-distribution, which is specifically designed for smaller sample sizes.
Sample Mean
The sample mean is the average of all the measurements in the sample and is represented by \(\bar{X}\). It is a point estimate of the true population mean (though we don't expect it to be exactly the same). The sample mean is easily affected by extreme values, especially when the sample size is small, which is why it's important to look at other statistics alongside it. In our exercise, the sample mean of the amount of copper precipitated was \(\bar{X} = 0.145\) moles. We used this as a central point around which we build our confidence interval.
Sample Standard Deviation
The sample standard deviation (s) quantifies the amount of variation or dispersion in a set of sample values. It provides insight into how much individual measurements deviate from the sample mean. A high standard deviation indicates that the data points are spread out over a wider range of values, while a low standard deviation suggests that they are clustered close to the mean. In the context of the exercise, the standard deviation was \(s = 0.0051\) moles, suggesting that the measurements of the copper precipitation were quite consistent, with little variation around the mean.
Degrees of Freedom
The concept of degrees of freedom (df) is essential in the context of statistical estimation and hypothesis testing. It refers to the number of independent pieces of information that went into the calculation of a statistic. In the calculation of a sample variance or standard deviation, one degree of freedom is lost because the sample mean is used in the calculation. Thus, the degrees of freedom associated with a sample's variance or standard deviation are typically the sample size minus one \(df = n - 1\). In this exercise, with 30 measurements, there were 29 degrees of freedom (\(df = 30 - 1 = 29\)). Degrees of freedom are also used to determine the appropriate critical value from the t-distribution when estimating confidence intervals.
T-Distribution
The t-distribution is a probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but has heavier tails. It is used instead of the normal distribution when the sample size is small and the population standard deviation is unknown. The t-distribution also varies depending on the degrees of freedom. As the degrees of freedom increase, the t-distribution approaches the normal distribution. When calculating confidence intervals for small sample sizes, statisticians use a t-score, which is based on the t-distribution, to account for the additional uncertainty. This is what was done in the exercise with a 90% confidence interval and a sample size of 30, leading to the use of the t-distribution due to the small sample size.
Margin of Error
In confidence interval calculations, the margin of error defines the range above and below the sample mean in which we can be confident that the population mean lies. It is affected by the sample size, the sample standard deviation, and the desired confidence level. A larger margin of error signals less confidence in the precision of the sample mean as an estimator of the population mean. In the provided solution, the margin of error was calculated using the critical value from the t-distribution, the standard deviation, and the sample size. This calculation gave us the interval where we can be 90% confident that the true mean of copper precipitation lies.
Statistics
The field of statistics is the science of collecting, analyzing, and interpreting data. It provides tools for making predictions and decisions based on data, often in the face of uncertainty. Statistics are divided into descriptive statistics, which summarize data, such as the sample mean and standard deviation, and inferential statistics, which make predictions or inferences about a population based on a sample. The calculation of a confidence interval is an example of inferential statistics, allowing us to estimate the mean amount of substance precipitated in a chemistry experiment, based on the sample collected by the students.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the margin of error in estimating a binomial proportion for each of the following values of \(n\). Use \(p=.5\) to calculate the standard error of the estimator. a. \(n=30\) b. \(n=100\) c. \(n=400\) d. \(n=1000\)

Independent random samples of \(n_{1}=800\) and \(n_{2}=640\) observations were selected from binomial populations 1 and \(2,\) and \(x_{1}=337\) and \(x_{2}=374\) successes were observed. a. Find a \(90 \%\) confidence interval for the difference \(\left(p_{1}-p_{2}\right)\) in the two population proportions. Interpret the interval. b. What assumptions must you make for the confidence interval to be valid? Are these assumptions met?

In developing a standard for assessing the teaching of precollege sciences in the United States, an experiment was conducted to evaluate a teacher-developed curriculum, "Biology: A Community Context" (BACC) that was standards-based, activity-oriented, and inquiry-centered. This approach was compared to the historical presentation through lecture, vocabulary, and memorized facts. Students were tested on biology concepts that featured biological knowledge and process skills in the traditional sense. The perhaps not-so-startling results from a test on biology concepts, published in The American Biology Teacher, are shown in the following table. \({ }^{11}\) $$\begin{array}{lccc} & & \text { Sample } & \text { Standard } \\\& \text { Mean } & \text { Size } & \text { Deviation } \\\\\hline \text { Pretest: All BACC Classes } & 13.38 & 372 & 5.59 \\\\\text { Pretest: All Traditional } & 14.06 & 368 & 5.45 \\\\\text { Posttest: All BACC Classes } & 18.5 & 365 & 8.03 \\\\\text { Posttest: All Traditional } & 16.5 & 298 & 6.96\end{array}$$ a. Find a \(95 \%\) confidence interval for the mean score for the posttest for all BACC classes. b. Find a \(95 \%\) confidence interval for the mean score for the posttest for all traditional classes. c. Find a \(95 \%\) confidence interval for the difference in mean scores for the posttest BACC classes and the posttest traditional classes. d. Does the confidence interval in c provide evidence that there is a real difference in the posttest BACC and traditional class scores? Explain.

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 degrees and a standard deviation of 0.73 degrees. 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 degrees, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

A researcher classified his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classified as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a \(95 \%\) confidence interval.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free