an owner of a miniature golf course wants to know how many strokes it takes to get the ball in the last…

Explanation:

Step1: Recall margin - of - error formula

The margin of error $E = z\cdot\frac{\sigma}{\sqrt{n}}$ for a confidence interval of the mean (when population standard - deviation $\sigma$ is known), where $z$ is the z - score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Analyze the effect of sample size

The margin of error is inversely proportional to the square root of the sample size $n$. That is, as $n$ increases, $E$ decreases. When we go from $n = 40$ to $n = 30$, the sample size decreases, and the margin of error will increase. When we go from $n = 40$ to $n = 75$, the sample size increases, and the margin of error will decrease.

Step3: Analyze the effect of confidence level

The z - score $z$ increases as the confidence level increases. For a 90% confidence interval, the z - score $z_{90}$ is smaller than the z - score $z_{95}$ for a 95% confidence interval, and $z_{95}$ is smaller than the z - score $z_{99}$ for a 99% confidence interval. As the confidence level increases, the z - score increases, and the margin of error increases. So, constructing a 95% or 99% confidence interval will increase the margin of error compared to a 90% confidence interval.

Answer:

using a sample of size 75