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The ages of a group of 135 randomly selected adult females have a standard deviation of 18.3 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let sigmaequals18.3 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 90​% confidence that the sample mean is within​ one-half year of the population mean.

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Answer:

Therefore, the number of female statistics student ages that must be obtained in order to estimate the mean age of all female statistics student is 3625 female statistics student

Step-by-step explanation:

Here, we have

[tex]n = \left (\frac{(z_{\alpha/2}) (\sigma) }{E} \right )^{2}[/tex]

Where:

n = Required sample size

σ = Standard deviation = 18.3

E = Margin of error = 1/2 year

[tex]z_{\alpha /2}[/tex] = Critical score from the confidence level = 1.645

Therefore,

[tex]n = \left (\frac{(1.645) (18.3) }{\frac{1}{2} } \right )^{2} = 3624.88[/tex]

Therefore, the number of female statistics student ages that must be obtained in order to estimate the mean age of all female statistics student = 3625 female statistics student.

The number of female statistics student ages that must be obtained in order to estimate the mean age of all female statistics students is; 3625 students ages.

We are given;

Standard deviation; σ = 18.3

Margin of error; E = 1/2 year

Confidence level; CL = 90%

Now, formula for margin of error is;

E = z(σ/√n)

Where;

E is margin of error

z is critical value at confidence level

σ is standard deviation

n is required sample size

The critical value of z at CL of 90% is;

z = 1.645

Let's make n the subject of the formula to get;

n = (zσ/E)²

n = (1.645 × 18.3/½)²

n = 3,624.882849

Approximating to a whole number gives;

n = 3625 females

Read more about margin of error at; https://brainly.com/question/6650225

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