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A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.64 years and the standard deviation is 9.39 years.

a) Construct a 80% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the interval have been met.

b) How large is the margin of error?


a) What is the confidence interval?


b) What is the margin of error?

The margin of error is

A Survey Of 25 Randomly Selected Customers Found The Ages Shown In Years The Mean Is 3264 Years And The Standard Deviation Is 939 Years A Construct A 80 Confide class=

Sagot :

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

Assuming that conditions have been met for the interval, we use the formula [tex]\displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}}[/tex] where [tex]\bar{x}[/tex] represents the sample mean, [tex]t[/tex] represents the critical value, [tex]s[/tex] represents the sample standard deviation, and [tex]n[/tex] is the sample size.

The critical value of [tex]t[/tex] for an 80% confidence level with degrees of freedom [tex]df=n-1=25-1=24[/tex] is equivalent to [tex]t=1.317836[/tex]

Thus, we can compute the confidence interval:

[tex]\displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}}\\\\CI=32.64\pm1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\\\\CI\approx\{30.17,35.11\}[/tex]

Therefore, we are 80% confident that the true mean age of all customers is between 30.17 and 35.11 years.

Part B

The margin of error is [tex]\displaystyle t\frac{s}{\sqrt{n}}=1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\approx2.47[/tex]

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