you are told that in a sample of size 225 the mean is 48.5 and the standard deviation is 1.8. the study is…

Explanation:

Step1: Identify the formula

The formula for the confidence - interval is $\bar{x}\pm z^\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z^$ is the z - score, $\sigma$ is the standard deviation, and $n$ is the sample size.

Step2: Determine the values

We have $\bar{x} = 48.5$, $z^*=1.645$ (for 90% confidence level), $\sigma = 1.8$, and $n = 225$.

Step3: Calculate the margin of error

The margin of error $E=z^*\frac{\sigma}{\sqrt{n}}=1.645\times\frac{1.8}{\sqrt{225}}=1.645\times\frac{1.8}{15}=1.645\times0.12 = 0.1974$.

Step4: Calculate the confidence interval

The lower limit of the confidence interval is $48.5−0.1974 = 48.3026$ and the upper limit is $48.5 + 0.1974=48.6974$.

Answer:

Since $48.8>48.6974$, 48.8 is not within the 90% confidence interval.