Question

Hi, can you help me answer this question please, thank you

Answer 1

Confidence interval is written as

point estimate ± margin of error

In this case, the point estimate is the sample mean since we want to estimate the population mean. We wwould find the margin of error by applying the formula,

margin of error =

`\begin{gathered} \text{margin of error = z}_(\varphi)*\frac{s}{\sqrt[]{n}} \\ \text{where} \\ z_(\varphi)\text{ is the z score of the confidence level given} \\ n\text{ is the number of samples} \\ s\text{ is the standard deviation} \end{gathered}`

From the information given,

s = 4

n = 19

the z score for a 95% confidence level is 1.96

Thus,

`\text{margin of error = 1.96 x }\frac{4}{\sqrt[]{19}}\text{ = }1.799`

The lower limit would be 46 - 1.799 = 44.201

The upper limit would be 46 + 1.799 = 47.799

Thus, we have

44.201 < mean < 47.799