Statistics Quiz with Solutions (5 Questions)

Given a continuous random variable \( X \) that is normally distributed with a mean (\( \mu \)) of 50, and the probability \( P(X > 55) = 0.25 \), we can use the symmetry of the normal distribution to answer the query about \( P(X < 45) \).

Since the normal distribution is symmetric around its mean, the distribution of values below the mean mirrors the distribution of values above the mean. Specifically, the value 55 is 5 units above the mean of 50, and thus 45 is 5 units below the mean of 50.

Given that \( P(X > 55) = 0.25 \), the equivalent area on the left side of the mean (i.e., below 45) should have the same probability due to the symmetry of the normal distribution. Therefore, the probability \( P(X < 45) \) is also \( 0.25 \).

The observed significance of a test of hypotheses is the area of the tail of the distribution cut off by the **test statistic**. This area represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming that the null hypothesis is true. This probability is commonly referred to as the p-value.

To estimate the minimum sample size required to achieve a desired confidence interval width for the mean time a child takes to arrange a jigsaw puzzle, we use the formula related to the sample size estimation for a mean:

\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]

Where:
- \( n \) is the sample size.
- \( Z \) is the z-value corresponding to the desired confidence level (here, 90%).
- \( \sigma \) is the population standard deviation (here, estimated as 4.2 minutes from another class).
- \( E \) is the margin of error (the desired precision, here ±3 minutes).

For a 90% confidence interval, the critical z-value (since the normal distribution table or calculator provides the critical values) corresponds approximately to 1.645 for a one-sided interval of 95% (as it splits the remaining 10% into 5% on each tail in a two-tailed test). We're looking at half the interval width as the margin of error \( E \), so we use ±3 minutes.

Calculate the necessary sample size using these inputs.

To estimate the mean time a child takes to arrange a jigsaw puzzle to within 3 minutes with 90% confidence, the kindergarten teacher needs a minimum sample size of 6 children.

Among the options provided, the one that is not a random variable is:

b. Speed of light in a vacuum

The speed of light in a vacuum is a constant physical quantity, not a variable that changes and has different values in different situations. It is universally recognized and fixed at approximately 299,792 kilometers per second. All other options describe quantities that can vary and therefore are considered random variables.

To construct an 80% confidence interval for the population mean when the sample size is \( n = 25 \), the sample mean is \( \bar{x} = 12 \), and the sample standard deviation is \( s = 5 \), we can use the following steps:

1. **Determine the Confidence Level and Critical Value**: The confidence level is 80%. We need to find the critical value from the t-distribution because we are estimating the mean from a sample and the sample size is relatively small.

2. **Calculate the Margin of Error**: The margin of error is computed using the formula:
\[
\text{Margin of Error} = t^* \times \frac{s}{\sqrt{n}}
\]
where \( t^* \) is the critical value from the t-distribution for \( df = n - 1 = 24 \) degrees of freedom and a two-tailed test at the 80% confidence level.

3. **Construct the Confidence Interval**: The confidence interval for the population mean \( \mu \) is then given by:
\[
\text{CI} = \left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right)
\]

Calculate this, beginning with finding the appropriate t-value for an 80% confidence level.

The t-critical value for an 80% confidence interval with 24 degrees of freedom is approximately 1.318. The margin of error is calculated as approximately 1.318.

Thus, the 80% confidence interval for the population mean is approximately \( (10.68, 13.32) \). This interval suggests that we can be 80% confident that the population mean lies within this range.