/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Basic Computation: Confidence In... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Basic Computation: Confidence Interval Suppose \(x\) has a mound-shaped symmetric distribution. A random sample of size 16 has sample mean 10 and sample standard deviation 2 (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu\) ? Explain. (b) Find a \(90 \%\) confidence interval for \(\mu\) (c) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
Yes, use a \(t\) distribution. The 90% confidence interval for \(\mu\) is [9.1235, 10.8765]. This interval means we are 90% confident the true mean is within this range.

Step by step solution

01

Check Sample Size Requirement

The sample size is 16, which is less than 30. This means the sample size is small, making the use of the Student's \(t\) distribution appropriate for small sample sizes to estimate population mean.
02

Symmetry Requirement

Since the distribution is described as mound-shaped symmetric, it satisfies the requirement for using a \(t\) distribution, as it does not demand normality but requires symmetric bell shape.
03

T-Score Selection for 90% Confidence

For a 90% confidence interval with 15 degrees of freedom (which is sample size minus one, \(n-1=16-1\)), look up the two-tailed \(t\) score in the \(t\) distribution table. The critical value for 90% confidence level is approximately 1.753.
04

Confidence Interval Calculation

The formula to calculate the confidence interval is \[ \text{CI} = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \], where \(\bar{x} = 10\), \(s = 2\), \(n = 16\), and \(t_{\alpha/2} = 1.753\). Plugging in the values, we have \[ \text{CI} = 10 \pm 1.753 \times \frac{2}{\sqrt{16}} = 10 \pm 0.8765. \] Thus, the 90% confidence interval is approximately \([9.1235, 10.8765]\).
05

Interpret the Confidence Interval

The interval \([9.1235, 10.8765]\) suggests that we are 90% confident that the true population mean \(\mu\) lies within this range, indicating that repeated sampling would produce a sample mean that falls within this interval 90% of the time.

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 91Ó°ÊÓ!

Key Concepts

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

Student's t-distribution
The Student's t-distribution is a probability distribution that is particularly useful when dealing with small sample sizes. It is similar to the normal distribution in shape but has heavier tails. This means it has a higher probability of values further away from the mean, which accounts for the higher variability expected with smaller samples.

You might wonder why it's called the Student's t-distribution. Well, it got its name from a statistician named William Sealy Gosset who published under the pseudonym "Student". The t-distribution is vital because it allows us to make inferences about a population mean when the population standard deviation is unknown and the sample size is small.

In our case, with a sample size of 16, using the Student's t-distribution helps accurately estimate the confidence interval, ensuring our results are reliable despite not having a large sample.
Sample Size Requirement
Sample size plays a critical role in statistical confidence and reliability. When the sample size is small, typically less than 30, we often rely on the Student's t-distribution for estimating the population mean. This is because the t-distribution adapts to the smaller sample size's higher variability.

In this exercise, the sample size is 16, which is considered small. For a small sample, the central limit theorem (which states that the sampling distribution of the sample mean will be normal) does not adequately apply unless the sample comes from a normal distribution. Hence, the use of the t-distribution is appropriate to capture the true variability.

If our sample were larger, say over 30, we might opt to use the standard normal distribution (z-distribution) as it gives similarly reliable estimates in larger sample sizes.
Symmetric Distribution
A symmetric distribution is one where the left and right sides of the distribution are mirror images of each other. This characteristic is crucial when using the Student's t-distribution because it does not assume normality but does assume a symmetric bell shape similar to the normal distribution.

In the given problem, it's stated that the distribution is mound-shaped and symmetric. This description ensures that the t-distribution can be applied without issue. When the distribution is skewed or not symmetric, the estimates from a t-distribution might become biased.

Symmetry helps in keeping the assumptions of the statistical methods intact, lending more reliability to the confidence interval derived from such data.
T-Score Selection
Selecting the correct t-score is a vital step in calculating the confidence interval. The t-score corresponds to the desired confidence level and the degrees of freedom, which is one less than the sample size (i.e., -1).

For a 90% confidence interval and 15 degrees of freedom (since our sample size is 16), we refer to the t-distribution table to find the required t-score. In our example, this value is approximately 1.753. This critical value accounts for the tails of the distribution being heavier, catering to the variability in smaller samples.

Choosing the right t-score ensures that our confidence interval appropriately covers the desired proportion of the data, representing a 90% likelihood that the true population mean is within this interval.
Confidence Interval Calculation
Calculating the confidence interval involves combining all the elements we've discussed so far. The formula used is \[ \text{CI} = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \], where \( \bar{x} \) is the sample mean, \( t_{\alpha/2} \) is the t-score, \( s \) is the sample standard deviation, and \( n \) is the sample size.

In our example:
  • The sample mean \( \bar{x} \) is 10.
  • The sample standard deviation \( s \) is 2.
  • The t-score (for 90% confidence and 15 degrees of freedom) is 1.753.
  • The confidence interval then becomes \( 10 \pm 1.753 \times \frac{2}{\sqrt{16}} = 10 \pm 0.8765 \).
Thus, the confidence interval is approximately \([9.1235, 10.8765]\).

This range indicates that we have a 90% confidence that the true population mean \( \mu \) lies within \( 9.1235 \) to \( 10.8765 \). This means, if we repeated the sampling process many times, about 90% of the intervals we compute would contain the true population mean.

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

Answer true or false. Explain your answer. A larger sample size produces a longer confidence interval for \(\mu\).

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther (Reference: Hummingbinds by K. Long and W. Alther). A small group of 15 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is \(\bar{x}=3.15\) grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with \(\sigma=0.33\) gram. (a) Find an \(80 \%\) confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for an \(80 \%\) confidence level with a maximal margin of error \(E=0.08\) for the mean weights of the hummingbirds.

Sam computed a \(90 \%\) confidence interval for \(\mu\) from a specific random sample of size \(n .\) He claims that at the \(90 \%\) confidence level, his confidence interval contains \(\mu .\) Is his claim correct? Explain.

Air Temperature How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds by Wirth and Young (Random House) claims that the air in the crown should be an average of \(100^{\circ} \mathrm{C}\) for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly \(100^{\circ} \mathrm{C}\). What is a reasonable and safe range of temperatures? This range may vary with the size and (decorative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 56 readings (for a balloon in equilibrium) gave a mean temperature of \(\bar{x}=97^{\circ} \mathrm{C} .\) For this balloon, \(\sigma \approx 17^{\circ} \mathrm{C}\) (a) Compute a \(95 \%\) confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (b) Interpretation If the average temperature in the crown of the balloon goes above the high end of your confidence interval, do you expect that the balloon will go up or down? Explain.

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

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.