the manager of an aquatic center is interested in the average number of fish people keep in their saltwater…

the manager of an aquatic center is interested in the average number of fish people keep in their saltwater tanks. she randomly selects 38 people who own saltwater fish tanks and asks how many fish they have. the mean number of fish is 48.2 with a standard deviation of 22.1 fish. which of the following is the 99% confidence interval for the true mean number of fish in saltwater tanks? find the t - table here. (38.47, 57.94) (40.94, 55.64) (42.15, 54.25) (42.93, 53.47)

the manager of an aquatic center is interested in the average number of fish people keep in their saltwater tanks. she randomly selects 38 people who own saltwater fish tanks and asks how many fish they have. the mean number of fish is 48.2 with a standard deviation of 22.1 fish. which of the following is the 99% confidence interval for the true mean number of fish in saltwater tanks? find the t - table here. (38.47, 57.94) (40.94, 55.64) (42.15, 54.25) (42.93, 53.47)

Answer

Explanation:

Step1: Identify degrees of freedom

The sample size $n = 38$. Degrees of freedom $df=n - 1=38 - 1 = 37$. Since the t - table may not have 37, we use the closest value which is 40 for a 99% confidence interval. The critical value $t_{\alpha/2}$ for a 99% confidence interval with $df = 40$ is approximately $t_{0.005}= 2.704$.

Step2: Calculate the margin of error

The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $s = 22.1$ (standard deviation) and $n = 38$. $E=2.704\times\frac{22.1}{\sqrt{38}}\approx2.704\times\frac{22.1}{6.164}\approx2.704\times3.585\approx9.69$.

Step3: Calculate the confidence interval

The confidence interval is given by $\bar{x}-E<\mu<\bar{x} + E$, where $\bar{x}=48.2$. Lower limit: $48.2-9.69 = 38.51$ (approx). Upper limit: $48.2 + 9.69=57.89$ (approx). The closest interval to our calculation is $(38.47,57.94)$.

Answer:

A. $(38.47,57.94)$