91Ó°ÊÓ

A simple random sample of size \(n\) is drawn from a population that is normally distributed. The sample mean, \(\bar{x}\), is found to be \(50,\) and the sample standard deviation, \(s,\) is found to be \(8 .\) (a) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(20 .\) (b) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(15 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(20 .\) Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, \(E ?\) (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why?

Step by step solution

01

Identify the given data

Given that \(\bar{x} = 50\) and \(s = 8\). For part (a), the sample size \(n = 20\), for part (b) \(n = 15\), and for part (c) \(n = 20\). The confidence levels are 98% for parts (a) and (b), and 95% for part (c).

02

Find the t-values corresponding to the given confidence levels

Using the t-distribution table, for 98% confidence and \(n-1\) degrees of freedom: For \(n=20\), \(df = 19\), \(t_{0.01, 19} \approx 2.539\). For \(n=15\), \(df = 14\), \(t_{0.01, 14} \approx 2.624\). For 95% confidence: For \(n=20\), \(df = 19\), \(t_{0.025, 19} \approx 2.093\).

03

Calculate the standard error

The standard error (SE) is given by \(SE = \frac{s}{\root{n}}\). For \(n = 20\), \(SE = \frac{8}{\root{20}} \approx 1.789\). For \(n = 15\), \(SE = \frac{8}{\root{15}} \approx 2.065\).

04

Compute the margin of error for each part

The margin of error (E) is given by \(E = t \cdot SE\). For (a), \(E = 2.539 \cdot 1.789 \approx 4.54\). For (b), \(E = 2.624 \cdot 2.065 \approx 5.42\). For (c), \(E = 2.093 \cdot 1.789 \approx 3.74\).

05

Construct the confidence intervals

(a) 98% CI for \(n = 20\): \(50 \pm 4.54\) or \([45.46, 54.54]\). (b) 98% CI for \(n = 15\): \(50 \pm 5.42\) or \([44.58, 55.42]\). (c) 95% CI for \(n = 20\): \(50 \pm 3.74\) or \([46.26, 53.74]\).

06

Analysis and comparison

(b) Decreasing the sample size increases the margin of error. (c) Decreasing the confidence level decreases the margin of error.

07

Check for normal distribution

(d) If the population were not normally distributed, the t-distribution assumption might not hold, and accuracy of these intervals could be compromised.

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Key Concepts

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

Sample Mean

The sample mean, denoted as \(\bar{x}\), is the average of all the data points in a sample. It provides a central value that represents a typical data point within your sample. The formula for computing the sample mean is given by: \(\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\), where \(n\) is the sample size and \(x_i\) represents each data point in the sample.
In the given exercise, the sample mean \(\bar{x}=50\). This value indicates the central tendency of the sampled data. Knowing the sample mean is crucial as it serves as an estimate for the population mean, \(\mu\).

t-Distribution

The t-distribution, also known as Student's t-distribution, is used when the sample size is small, generally \(n<30\), and the population standard deviation is unknown. The t-distribution is similar to the normal distribution but has heavier tails, meaning it produces more variability in the data.
In this exercise, since the sample sizes are relatively small (\