Question

Suppose $X_1,X_2,\dots ,X_n$ is a random sample drawn from a $N(\mu ,{\sigma }^2)$ distribution. In this case, the pivot random variable for a confidence interval is $$t=\frac{\overline{X}-\mu }{S/\sqrt{n}}$$ where $\overline{X}$ and $S$ are the sample mean and standard deviation, respectfully. Recall by Theorem $3.6.1$ that $t$ has a Student $t$ -distribution with $n-1$ degrees of freedom; hence, its distribution is free of all parameters for this normal situation. In the notation of this section $t_{n-1}^{(\gamma )}$ denotes the $\gamma 100\mathrm{\%}$ percentile of a $t$ -distribution with $n-1$ degrees of freedom. Using this notation show that a $(1-\alpha )100\mathrm{\%}$ confidence interval for $\mu$ is $$(\overline{x}-t^{(1-\alpha /2)}\frac{s}{\sqrt{n}},\overline{x}-t^{(\alpha /2)}\frac{s}{\sqrt{n}})$$

Short answer

The (1-α)100% confidence interval for $\mu$ is $(\overline{x}-t^{(1-\alpha /2)}\frac{s}{\sqrt{n}},\overline{x}-t^{(\alpha /2)}\frac{s}{\sqrt{n}})$.

Step-by-step solution

Introduce variables

Introduce the necessary variables for the proof and understand their meaning. $X_1,X_2,...,X_n$ are random variables of the sample drawn from a $N(\mu ,{\sigma }^2)$ (Normal) distribution. $\mu$ is the mean of the population, ${\sigma }^2$ is the variance of the population, $t$ is the critical value, $\overline{X}$ is the sample mean, $S$ is the sample standard deviation, and $n$ is the sample size.

Apply 'Pivot random variable' formula to variables

The exercise starts from the formula for pivot random variable which is given by $t=\frac{\overline{X}-\mu }{S/\sqrt{n}}$. Rewriting this formula for $\mu$, we get: $\mu =\overline{X}-t\frac{S}{\sqrt{n}}$.

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 $t_{n-1}^{(\gamma )}$ denotes the θ 100% percentile of a t-distribution with n-1 degrees of freedom. Meaning θ% of the data (of $t_{n-1}$) falls below $t_{n-1}^{(\gamma )}$. 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.

Insert percentiles into the pivot formula for $\mu$

Substituting the percentiles for t in µ's formula you get $\mu =\overline{X}-t^{(1-\alpha /2)}\frac{S}{\sqrt{n}}$ for the lower bound and $\mu =\overline{X}-t^{(\alpha /2)}\frac{S}{\sqrt{n}}$ 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.

Present final Confidence Interval

The final confidence interval thus becomes $(\overline{x}-t^{(1-\alpha /2)}\frac{s}{\sqrt{n}},\overline{x}-t^{(\alpha /2)}\frac{s}{\sqrt{n}})$.

Key concepts

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 $X_1,X_2,...,X_n$. It's imperative that these variables be a random sample so that inferences derived, such as the confidence interval for the mean $\mu$, 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 $n-1$, 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 $t^{(1-\alpha /2)}$ and $t^{(\alpha /2)}$, are found for $n-1$ 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.