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When the sample size \( n \) is large, the 90% confidence interval for \(\mu_Y\) is:

A. \( \bar{Y} \pm 1.64 \, \text{SE}(\bar{Y}) \)

B. \( \bar{Y} \pm 1.96 \)

C. \( \bar{Y} \pm 1.64 \sigma_Y \)

D. \( \bar{Y} \pm 1.96 \, \text{SE}(\bar{Y}) \)

Answer :

When the sample size (n) is large, the 90% confidence interval for the population means (μY) is Yˉ ± 1.96SE(Yˉ).

The 90% confidence interval represents a range of values within which the true population mean is likely to fall. When the sample size is large (considered to be sufficiently large, typically greater than 30), we can approximate the sampling distribution of the sample mean (Yˉ) as approximately normally distributed.

To construct the confidence interval, we use the standard error of the sample mean (SE(Yˉ)), which is calculated by dividing the standard deviation of the population (σY) by the square root of the sample size (n).

For a 90% confidence interval, we look up the critical value associated with a 90% confidence level, which is approximately 1.96 when the sample size is large. This critical value corresponds to the z-score that leaves 5% of the area in the tails of the normal distribution.

Multiplying the standard error by the critical value gives us the margin of error, which is added and subtracted from the sample mean to create the confidence interval.

Therefore, the correct answer is D. Yˉ ± 1.96SE(Yˉ), where Yˉ represents the sample mean and SE(Yˉ) represents the standard error of the sample mean.

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