Conflaence Intervals and Hypothesle Testing
Choosing an appropriate sample slze
Espanol

A personality test has a subsection designed to assess the “honesty” of the test-taker. Suppose that you’re interested in the mean score, , on this subsection among the general population. You decide that you’ll use the mean of a random sample of scores on this subsection to estimate . What is the minimum sample size needed in order for you to be confident that your estimate is within 5 of ? Use the value 21 for the population standard deviation of scores on this subsection.

Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).
(If necessary, consult a list of formulas.)

Ask by Guzman Bob. in the United States

Mar 27,2025

Question

Conflaence Intervals and Hypothesle Testing
Choosing an appropriate sample slze
Espanol

A personality test has a subsection designed to assess the “honesty” of the test-taker. Suppose that you’re interested in the mean score, , on this subsection among the general population. You decide that you’ll use the mean of a random sample of scores on this subsection to estimate . What is the minimum sample size needed in order for you to be confident that your estimate is within 5 of ? Use the value 21 for the population standard deviation of scores on this subsection.

Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).
(If necessary, consult a list of formulas.)

Ask by Guzman Bob. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

We begin with the formula for the margin of error in estimating a mean when the population standard deviation is known:
$$
E = z^* \frac{\sigma}{\sqrt{n}}
$$
where
- $$ E = 5 $$ is the desired margin of error,
- $$ \sigma = 21 $$ is the population standard deviation,
- $$ z^* $$ is the critical value from the standard normal distribution,
- $$ n $$ is the sample size.

For a $$ 90\% $$ confidence level, the significance level is $$ \alpha = 0.10 $$ and each tail has $$ \alpha/2 = 0.05 $$. The corresponding $$ z^* $$ value is approximately:
$$
z^* \approx 1.645
$$

The formula for the margin of error becomes:
$$
5 = 1.645 \frac{21}{\sqrt{n}}
$$

Solve for $$\sqrt{n}$$:
$$
\sqrt{n} = 1.645 \frac{21}{5}
$$

Calculate the numerator:
$$
1.645 \times 21 = 34.545
$$

Thus,
$$
\sqrt{n} = \frac{34.545}{5} \approx 6.909
$$

Now, square both sides to solve for $$ n $$:
$$
n \approx (6.909)^2 \approx 47.725
$$

Since $$ n $$ must be a whole number and the sample size must be large enough so that the margin of error is within 5, we round up to the next whole number:
$$
n = 48
$$

The minimum sample size needed is $$ 48 $$.
The minimum sample size needed is 48.

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