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: Calculate the standard error

The formula for standard error $SE=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 1.8$ is the standard - deviation and $n = 225$ is the sample size. $SE=\frac{1.8}{\sqrt{225}}=\frac{1.8}{15}=0.12$

Step2: Calculate the margin of error

For a 90% confidence level, the $z$-score $z = 1.645$. The margin of error $ME=z\times SE$. $ME = 1.645\times0.12=0.1974$

Step3: Calculate the confidence interval

The confidence interval is given by $\bar{x}\pm ME$, where $\bar{x}=48.5$ is the sample mean. The lower limit $LL=\bar{x}-ME=48.5 - 0.1974=48.3026$ The upper limit $UL=\bar{x}+ME=48.5 + 0.1974=48.6974$

Step4: Check if 48.8 is in the interval

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

Answer:

48.8 is not within the 90% confidence interval.