Question

If it can be assumed that S.D= 60 , then how large a sample is needed so that one will be able to assert with 90% confidence that sample mean is off by atmost 10
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Answer 1

A sample of 97 must be needed so that one will be able to assert with 90% confidence that the sample mean is off by at most 10.

Step-by-step explanation:

We are given that it can be assumed that S.D = 60 and we have to find that how large a sample is needed so that one will be able to assert with 90% confidence that sample mean is off by at most 10.

As we know that the margin of error is given by the following formula;

The Margin of error =  $Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }$

Here, $\sigma$ = standard deviation = 60

n = sample size

$\alpha$ = level of significance = 1 - 90% = 10% or 0.10

Now, in the z table the critical value of x at $(\frac{0.10}{2}= 0.05)$ level of significance is given as 1.645.

So,   $10 = Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }$

        $10 = 1.645 \times \frac{60}{\sqrt{n} }$

        $\sqrt{n} = \frac{1.645 \times 60}{10}$

        $\sqrt{n} =9.87$

Now, squaring both sides we get;

         n = $9.87^{2}$ = 97.42 ≈ 97

Hence, a sample of 97 must be needed so that one will be able to assert with 90% confidence that sample mean is off by at most 10.