Sampling And Estimations.

For the second stratum, which has 30% of the population, you should select 30% of the total sample size.

To calculate how many individuals to select from the second stratum: $0.30\times 100=30\text{ individuals}$

To calculate the minimum sample size required to estimate the proportion of voters supporting a particular candidate with a 90% confidence level and a margin of error of 3%, you can use the formula for sample size in a confidence interval:

n=Z^2⋅p⋅(1−p)/E^2​

Where:

• $n$ is the required sample size.
• $Z$ is the Z-score corresponding to the desired confidence level (for a 90% confidence level, Z ≈ 1.645).
• $p$ is the estimated proportion of supporters (45% or 0.45).
• $E$ is the margin of error (3% or 0.03).

Now, plug in the values:

$n\approx 301.823$

Since you can’t have a fraction of a person in your sample, you should round up to the nearest whole number to ensure your sample is large enough. So, the minimum sample size required is 302.

To calculate the minimum sample size required to estimate the average household income with a 99% confidence level and a margin of error of $1,000, you can use the formula for sample size in a confidence interval:

n=(Z^2⋅σ^2​)/E^2

Where:

• $n$ is the required sample size.
• $Z$ is the Z-score corresponding to the desired confidence level (for a 99% confidence level, Z ≈ 2.576).
• $\sigma$ is the estimated standard deviation of household income ($15,000).
• $E$ is the margin of error ($1,000).

Now, plug in the values:

n=(2.576^2⋅15,000^2)/1,000^2

=1495.75

To calculate the 95% confidence interval for the population mean age, you can use the t-distribution since the population standard deviation is not known, and you have a small sample size (30 employees). The formula for the confidence interval is:

Confidence Interval=xˉ ± t α / 2 ​(​s/sqrt(n)​)

Where:

• xˉ is the sample mean age (35 years).
• tα/2​ is the critical t-value for a 95% confidence interval with $n-1$ degrees of freedom (29 degrees of freedom for 30 samples).
• $s$ is the sample standard deviation (5 years).
• $n$ is the sample size (30 employees).

You’ll need to find the critical t-value for a 95% confidence interval with 29 degrees of freedom, which you can find using a t-table or calculator. For a 95% confidence interval, the critical t-value is approximately 2.045.

Now, plug in the values:

Confidence Interval=35±2.045(530)Confidence Interval=35±2.045(30​5​)

Calculate the margin of error:

Margin of Error=2.045(530)Margin of Error=2.045(30​5​)

$\text{Margin of Error}\approx 1.869$

Now, calculate the lower and upper bounds of the confidence interval:

Lower Bound = $35-1.869\approx 33.13$

Upper Bound = $35+1.869\approx 36.87$

So, the 95% confidence interval for the population mean age is approximately $33.13$ years to $36.87$ years.

To calculate the minimum sample size required to estimate the proportion of defective items with a 95% confidence level and a margin of error of 2%, you can use the formula:

n=Z^2⋅p⋅(1−p)/E^2​

Where:

• $n$ is the required sample size.
• $Z$ is the Z-score corresponding to the desired confidence level (for a 95% confidence level, Z ≈ 1.96).
• $p$ is the estimated proportion of defects (10% or 0.10).
• $E$ is the margin of error (2% or 0.02).

Now, plug in the values:

n=0.00040.038416​

$n=96.04$

Since you can’t have a fraction of a person in your sample, you should round up to the nearest whole number to ensure your sample is large enough. So, the minimum sample size required is 97.