a simple random sample of size n is drawn from a normally distributed population, and the mean of the sample…

a simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is $\bar{x}$, while the standard deviation is s. what is the 90% confidence interval for the population mean? use the table below to help you answer the question.\n| confidence level | 90% | 95% | 99% |\n| ---- | ---- | ---- | ---- |\n| $z^{*}$ -score | 1.645 | 1.96 | 2.58 |\n$\bar{x}pm\frac{0.90cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{1.645cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{1.96cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{2.58cdot s}{sqrt{n}}$

a simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is $\bar{x}$, while the standard deviation is s. what is the 90% confidence interval for the population mean? use the table below to help you answer the question.\n| confidence level | 90% | 95% | 99% |\n| ---- | ---- | ---- | ---- |\n| $z^{*}$ -score | 1.645 | 1.96 | 2.58 |\n$\bar{x}pm\frac{0.90cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{1.645cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{1.96cdot s}{sqrt{n}}$\n$\bar{x}pm\frac{2.58cdot s}{sqrt{n}}$

Answer

Answer:

$\bar{x}\pm\frac{1.645\cdot s}{\sqrt{n}}$

Explanation:

Step1: Recall confidence - interval formula

For a normally - distributed population with sample mean $\bar{x}$, sample standard deviation $s$, and sample size $n$, the confidence interval for the population mean $\mu$ is given by $\bar{x}\pm z^\frac{s}{\sqrt{n}}$, where $z^$ is the critical value.

Step2: Identify the $z^*$ value for 90% confidence level

From the provided table, the $z^*$ value corresponding to a 90% confidence level is 1.645.

Step3: Substitute the $z^*$ value into the formula

Substituting $z^* = 1.645$ into the confidence - interval formula $\bar{x}\pm z^*\frac{s}{\sqrt{n}}$, we get $\bar{x}\pm\frac{1.645\cdot s}{\sqrt{n}}$.