Statistics Quiz with Solutions (5 Questions)

To construct a 99.9% confidence interval for the mean difference between the marks scored last week and those scored this week by all the students, we'll use the following formula for the confidence interval of the mean:

\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \]

Where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(s\) is the sample standard deviation
- \(n\) is the sample size

Given:
- Sample mean difference, \(\bar{x} = 12\)
- Sample standard deviation, \(s = 3\)
- Sample size, \(n = 35\)
- Confidence level, \(99.9\%\)

For a 99.9% confidence level, the z-score (from the standard normal distribution table) is approximately \(z = 3.291\).

Now, calculate the standard error:

\[ \text{Standard Error} = \frac{s}{\sqrt{n}} = \frac{3}{\sqrt{35}} \approx \frac{3}{5.92} \approx 0.507 \]

Next, calculate the margin of error:

\[ \text{Margin of Error} = z \times \text{Standard Error} = 3.291 \times 0.507 \approx 1.669 \]

Finally, construct the confidence interval:

\[ \text{Lower Limit} = \bar{x} - \text{Margin of Error} = 12 - 1.669 = 10.331 \]
\[ \text{Upper Limit} = \bar{x} + \text{Margin of Error} = 12 + 1.669 = 13.669 \]

Therefore, the 99.9% confidence interval for the mean difference between the marks scored last week and those scored this week by all the students is approximately \((10.331, 13.669)\).

As the sample size of a random sample increases, the standard deviation of the sample mean decreases.

This phenomenon occurs because the standard deviation of the sample mean, often referred to as the standard error of the mean, is calculated as the population standard deviation divided by the square root of the sample size (\(n\)). Mathematically, this relationship is expressed as:

\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \]

where \(\sigma\) is the population standard deviation and \(n\) is the sample size. As \(n\) increases, the denominator increases, resulting in a smaller standard error. This reduction in the standard error signifies that the sample mean becomes a more precise estimate of the population mean as the sample size grows.

A simple linear regression model is a sum of two parts:

1. **Deterministic Part (Predicted Value)**: This is the linear component that represents the relationship between the independent variable \(x\) and the dependent variable \(y\). It is expressed as \( \hat{y} = \beta_0 + \beta_1 x \), where:
- \( \beta_0 \) is the y-intercept.
- \( \beta_1 \) is the slope of the regression line.
- \( x \) is the independent variable.
- \( \hat{y} \) is the predicted value of the dependent variable.

2. **Stochastic Part (Error Term or Residual)**: This is the random component that accounts for the variability in the dependent variable \(y\) that cannot be explained by the linear relationship with \(x\). It is represented as \( \epsilon \).

Combining these two parts, the simple linear regression model can be written as:

\[ y = \beta_0 + \beta_1 x + \epsilon \]

Here, \(y\) is the actual observed value of the dependent variable, and \(\epsilon\) is the error term that captures the deviation of the observed value from the predicted value.

In this context, the sample consists of the 1,200 people who were randomly selected to ascertain their income. These individuals represent a subset of the total population of employed people in Omaha, Nebraska, and their incomes are used to estimate the average income of the entire population.

Let's calculate the probability using the z-score and the standard normal distribution table.

Given:
- Mean (\(\mu\)) = 15 hours
- Standard deviation (\(\sigma\)) = 3 hours

We want to find the probability that Noel plays video games between 15 and 18 hours a week.

First, calculate the z-score for 18 hours:

\[ z = \frac{18 - 15}{3} = \frac{3}{3} = 1 \]

Now we need to find the probability corresponding to a z-score of 1. Using the standard normal distribution table, the cumulative probability for a z-score of 1 is approximately 0.8413.

This value represents the probability that Noel plays video games up to 18 hours a week. Since the mean (15 hours) is the center of the distribution, the cumulative probability up to the mean (z = 0) is 0.5.

To find the probability that Noel plays video games between 15 and 18 hours, subtract the cumulative probability at the mean from the cumulative probability at the z-score of 1:

\[ P(15 < X < 18) = P(z < 1) - P(z < 0) \]
\[ P(15 < X < 18) = 0.8413 - 0.5 = 0.3413 \]

Thus, the probability that Noel plays video games between 15 and 18 hours a week is approximately 34.13%.