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

Suppose X1,X2,,Xn is a random sample drawn from a N(μ,σ2) distribution. In this case, the pivot random variable for a confidence interval is t=X¯μS/n where X¯ and S are the sample mean and standard deviation, respectfully. Recall by Theorem 3.6.1 that t has a Student t -distribution with n1 degrees of freedom; hence, its distribution is free of all parameters for this normal situation. In the notation of this section tn1(γ) denotes the γ100% percentile of a t -distribution with n1 degrees of freedom. Using this notation show that a (1α)100% confidence interval for μ is (x¯t(1α/2)sn,x¯t(α/2)sn)

Short Answer

Expert verified
The (1-α)100% confidence interval for μ is (x¯t(1α/2)sn,x¯t(α/2)sn).

Step by step solution

01

Introduce variables

Introduce the necessary variables for the proof and understand their meaning. X1,X2,...,Xn are random variables of the sample drawn from a N(μ,σ2) (Normal) distribution. μ is the mean of the population, σ2 is the variance of the population, t is the critical value, X¯ is the sample mean, S is the sample standard deviation, and n is the sample size.
02

Apply 'Pivot random variable' formula to variables

The exercise starts from the formula for pivot random variable which is given by t=X¯μS/n. Rewriting this formula for μ, we get: μ=X¯tSn.
03

Calculate percentiles of t-distribution

The exercise asks to show the (1-α)100% confidence interval using the percentile notation for a t-distribution. Recall that the θ-th percentile is the value below which θ% of the data falls. So tn1(γ) denotes the θ 100% percentile of a t-distribution with n-1 degrees of freedom. Meaning θ% of the data (of tn1) falls below tn1(γ). Therefore, to find the (1-α)100% confidence level for the mean µ we should look at the symmetrical area around the mean that covers (1-α)100% of the data. This will give us the lower and upper bounds of the confidence interval. According to the normal distributed t-distribution, half the remaining α% of the data will fall below the lower bound and half will fall above the upper bound.
04

Insert percentiles into the pivot formula for μ

Substituting the percentiles for t in µ's formula you get μ=X¯t(1α/2)Sn for the lower bound and μ=X¯t(α/2)Sn for the upper bound. Remember half the remaining α% of the data will fall below the lower bound and half will fall above the upper bound due to the symmetric nature of a normal distribution, hence the differing values for α in each bound.
05

Present final Confidence Interval

The final confidence interval thus becomes (x¯t(1α/2)sn,x¯t(α/2)sn).

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.

Understanding Random Sample
In statistics, a random sample is a subset of individuals chosen from a larger set or population. Each individual is chosen randomly and entirely by chance, such that each individual has the same probability of being chosen at any stage during the sampling process, and each subset of individuals has the same probability of being chosen for the sample as any other subset of individuals.

When conducting experiments or surveys, it's essential to use a random sample to ensure that the results are representative of the entire population, thereby minimizing bias. This is crucial when creating confidence intervals as it helps to ensure that the interval estimated from the sample is likely to contain the true population parameter.

As in the exercise, the sample consists of variables drawn from a normal distribution, denoted by X1,X2,...,Xn. It's imperative that these variables be a random sample so that inferences derived, such as the confidence interval for the mean μ, are statistically valid and applicable to the broader population from which the sample was drawn.
T-Distribution Explained
The t-distribution, also known as the Student’s t-distribution, is a probability distribution that arises when estimating the mean of a normally distributed population when the sample size is small and the population standard deviation is unknown.

The shape of the t-distribution is similar to the standard normal distribution, but with thicker tails, which means there is a higher probability of values further from the mean. As the sample size increases, the t-distribution approaches the normal distribution.

When calculating confidence intervals for a population mean from a sample, one often uses the t-distribution if the population standard deviation is unknown and the sample size is not large. In practice, this involves using the sample standard deviation as an estimate for the population standard deviation, which introduces variability, hence the thicker tails and the use of the t-distribution. In our exercise, t is the value from the t-distribution that corresponds to the desired confidence level and reflects this additional uncertainty.
Degrees of Freedom in Detail
The concept of degrees of freedom in statistics often causes confusion, but it is a vital part of calculations involving the t-distribution. Degrees of freedom can be thought of as the number of values in a calculation that are free to vary when estimating a statistical parameter.

In the context of the t-distribution, the degrees of freedom are directly tied to sample size. Specifically, for a t-distribution used to estimate the mean of a normally distributed population, the degrees of freedom are equal to n1, where n is the sample size. This accounts for the fact that we use the sample itself to estimate the population standard deviation.

In our exercise, the critical values of the t-distribution, denoted as t(1α/2) and t(α/2), are found for n1 degrees of freedom. The degrees of freedom ensure that the t-distribution accurately reflects the variability in the estimate of the population standard deviation. The correct degrees of freedom are vital in finding the correct t-values, which in turn are essential to calculate accurate confidence intervals for the 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

Let x1,x2,,xn be the values of a random sample. A bootstrap sample, x=(x1,x2,,xn), is a random sample of x1,x2,,xn drawn with replacement. (a) Show that x1,x2,,xn are iid with common cdf F^n, the empirical cdf of x1,x2,,xn (b) Show that E(xi)=x¯. (c) If n is odd, show that median Missing \left or extra \right. (d) Show that V(xi)=n1i=1n(xix¯)2.

Let a random sample of size 17 from the normal distribution N(μ,σ2) yield x¯=4.7 and s2=5.76. Determine a 90 percent confidence interval for μ.

Let Y be b(300,p). If the observed value of Y is y=75, find an approximate 90 percent confidence interval for p.

Among the data collected for the World Health Organization air quality monitoring project is a measure of suspended particles in μg/m3. Let X and Y equal the concentration of suspended particles in μg/m3 in the city center (commercial district) for Melbourne and Houston, respectively. Using n=13 observations of X and m=16 observations of Y, we shall test H0:μX=μY against H1:μX<μY. (a) Define the test statistic and critical region, assuming that the unknown variances are equal. Let α=0.05. (b) If x¯=72.9,sx=25.6,y¯=81.7, and sy=28.3, calculate the value of the test statistic and state your conclusion.

. Let X1,X2,,Xn be random sample from a distribution of either type. A measure of spread is Gini's mean difference G=j=2ni=1j1|XiXj|/(n2) (a) If n=10, find a1,a2,,a10 so that G=i=110aiYi, where Y1,Y2,,Y10 are the order statistics of the sample. (b) Show that E(G)=2σ/π if the sample arises from the normal distribution N(μ,σ2)

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