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 for \(\mu\) if the sample size, \(n,\) is 20 (b) Construct a \(98 \%\) confidence interval for \(\mu\) 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 for \(\mu\) 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 information for part (a)

Given: Sample mean, \(\bar{x} = 50\)Sample standard deviation, \(s = 8\)Sample size, \(n = 20\)Confidence level = 98%Since the population standard deviation is unknown and the sample size is small, use the t-distribution.

02

- Find the critical value for part (a)

Use the t-distribution table to find the critical value (\(t_{\frac{\text{α}}{2}}\)) for a 98% confidence level and degrees of freedom ( \( \text{df} = n-1 \)). So, \(\text{df} = 20 - 1 = 19\). From the t-table, \({t_{0.01, 19}} ≈ 2.539\).

03

- Calculate the margin of error for part (a)

The margin of error, \(E = t_{\frac{\text{α}}{2}} \frac{s}{\root{n}} \), where \({t_{0.01, 19}} ≈ 2.539\), \(s = 8\), and \(\root n \)= \( \root 20\). \(E ≈ 2.539 \times \frac{8}{4.47} = 2.539 \times 1.789 \) \(E ≈ 4.54 \)

04

- Construct the confidence interval for part (a)

The confidence interval (CI) for the population mean \(\text{μ}\) is: \(\bar{x} - E < \text{μ} < \bar{x} + E\).So, \(50 - 4.54 < \text{μ} < 50 + 4.54\) \(45.46 < \text{μ} < 54.54 \)

05

- Repeat steps 1 to 4 for part (b) with n = 15

New sample size: \ n = 15 \( \text{df} = n-1 = 14 \). From the t-table, \({t_{0.01, 14}} ≈ 2.624 \). Margin of error, \(E = t_{\frac{\text{α}}{2}} \frac{s}{\root{n}}\). \ \( E ≈ 2.624 \times \frac{8}{3.87} = 2.624 \times 2.067 \) \( E ≈ 5.42 \) Confidence interval: \( 50 - 5.42 < \text{μ} < 50 + 5.42 \) \ \( 44.58 < \text{μ} < 55.42 \)

06

- Analyze the effect of decreasing sample size

Decreasing the sample size increases the margin of error. This means the confidence interval becomes wider.

07

- Repeat steps 1-4 for part (c) with a 95% confidence level and n = 20

Given: n = 20, confidence level = 95%\text{df} = n - 1 = 19.From t-table, \({t_{0.025, 19}} ≈ 2.093 \).\text{Margin of error, E} = t_{\frac{\text{α}}{2}} \frac{s}{\root{n}} )\text{E} ≈ 2.093 × \frac{8}{4.47} ≈ 2.093 × 1.791 \text{E} ≈ 3.75\text{}CI: 50 - 3.75 < \text{\text{μ}} < 50 + 3.75) \44.58 < \text{μ} < 55.42\text{E} ≈ 3.85\text{}

08

- Compare the results from part (a) and (c)

Decreasing the level of confidence reduces the margin of error. The confidence interval becomes narrower.

09

- Assess Impact of Population Distribution

If the population is not normally distributed, then it is not guaranteed that the sample means would follow a t-distribution, making these confidence intervals less valid, particularly for smaller sample sizes.

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Key Concepts

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

t-distribution

When dealing with small sample sizes and unknown population standard deviations, we use the t-distribution. This approach helps us estimate population parameters like the mean. Unlike the normal distribution, the t-distribution accounts for greater variability in smaller samples, providing wider confidence intervals. The shape of the t-distribution is similar to the normal distribution but has heavier tails, meaning there's a higher probability of values farther from the mean. The t-distribution becomes closer to the normal distribution as the sample size increases.

sample mean

The sample mean (\(\bar{x}\)) is the average of all the observations in our sample. For example, in the exercise, the sample mean is found to be 50. This value is a point estimate of the population mean (\(\text{μ}\)). By calculating the sample mean, we gain an estimate of the central tendency of the population we're studying. The formula for the sample mean is the sum of all sample observations divided by the number of observations (\( \bar{x} = \frac{1}{n} \times \text{Σ} x_i\)). The sample mean serves as the centerpiece around which the confidence intervals are constructed.

margin of error

The margin of error quantifies the uncertainty or the range within which we expect the true population mean to lie. It is calculated using the formula: \(E = t_{\frac{α}{2}} \frac{s}{\root{n}}\), where \(t_{\frac{α}{2}}\) is the critical value from the t-distribution, \(s\) is the sample standard deviation, and \(\root{n}\) is the square root of the sample size. In simpler terms, the margin of error increases or decreases based on the sample size and the level of confidence we choose. A smaller sample size or a higher confidence level results in a larger margin of error, indicating more uncertainty. In the exercise, for example, decreasing the sample size from 20 to 15 increased the margin of error.

degrees of freedom

Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. In the context of the t-distribution, it is given by \( \text{df} = n - 1 \), where \(n\) is the sample size. For example, in part (a) of the exercise, with a sample size of 20, the degrees of freedom are 19. Degrees of freedom are crucial because they affect the shape of the t-distribution, making it more accurate for smaller samples. As the degrees of freedom increase, the t-distribution approaches the shape of the normal distribution, which is why degrees of freedom are particularly important in small sample studies.