a social worker at a large high school wants to estimate the number of days students are absent. he randomly…
a social worker at a large high school wants to estimate the number of days students are absent. he randomly selects 45 students and records the number of days each one is absent. the social worker constructs a 95% confidence interval for the true mean number of days students at this school are absent. which of the following would decrease the margin of error?\nusing a sample of size 30\nusing a sample of size 40\nconstructing a 90% confidence interval\nconstructing a 99% confidence interval
Answer
Explanation:
Step1: Recall margin - of - error formula
The margin of error $E = z\cdot\frac{\sigma}{\sqrt{n}}$ for a confidence interval of the population mean (when $\sigma$ is known, or $E = t\cdot\frac{s}{\sqrt{n}}$ when $\sigma$ is unknown, where $z$ or $t$ is the critical value, $\sigma$ or $s$ is the standard deviation and $n$ is the sample size).
Step2: Analyze sample size effect
Increasing the sample size $n$ decreases the margin of error because $E$ is inversely proportional to $\sqrt{n}$. So reducing the sample size from 45 to 30 or 40 will increase the margin of error.
Step3: Analyze confidence - level effect
The critical value $z$ (or $t$) increases as the confidence level increases. A 90% confidence interval has a smaller critical value than a 95% confidence interval, and a 99% confidence interval has a larger critical value than a 95% confidence interval. Since $E$ is directly proportional to the critical value, decreasing the confidence level (from 95% to 90%) will decrease the margin of error, and increasing the confidence level (from 95% to 99%) will increase the margin of error.
Answer:
constructing a 90% confidence interval