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1 year ago

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A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The

98% confidence interval for the true average age of all students in the university is:________
a. 20.0 to 30.0
b. 20.5 to 26.5
c. 23.0 to 27.0
d. 24.4 to 25.6

1 answer:

satela [25.4K]1 year ago

6 0

Answer:

The correct option is;

d. 24.2 to 25.6

Step-by-step explanation:

Here we have a sample with unknown population standard deviation, we therefore apply the student t distribution at 64 - 1 degrees of freedom

Therefore, we have

$CI=\bar{x}\pm t\frac{s}{\sqrt{n}}$

Where:

$\bar x$ = Mean = 25

σ = Standard deviation = 2

n = Sample size = 64

t = T value at 98% = $\pm 2.387$

Which gives

$CI=25\pm t_{63}\frac{2}{\sqrt{64}}$

That is the value is from

24.40325 to 25.59675 which gives,by rounding to one decimal place, is

24.4 to 25.6.

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Data given

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$\alpha$ Significance level provided

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