Test 4 Question 7 of 32 (1 point) | Question Artempt 1 of 1 Time Remaining: $3: 20: 13$; 1 $=2$ 3 $=4$ $=5$ 6 8 9 10 11 12 13 An ecologist randomly samples 19 plants of a specific species and measures their heights. He finds that this sample has a mean of $18 \mathrm{~cm}$ and a standard deviation of $4 \mathrm{~cm}$. If we assume that the height measurements are normally distributed, find a $95 \%$ confidence interval for the mean height of all plants of this species. Give the lower limit and upper limit of the $95 \%$ confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit: $\times$ 5

Step 1 ：The problem provides us with a sample size of 19 plants, a sample mean of 18 cm, and a sample standard deviation of 4 cm. We are asked to find a 95% confidence interval for the mean height of all plants of this species.

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：For a 95% confidence interval, the z-score is approximately 1.96.

Step 4 ：Substituting the given values into the formula, we get \(18 \pm 1.96 \frac{4}{\sqrt{19}}\).

Step 5 ：Calculating the margin of error, we get approximately 1.798619353545204.

Step 6 ：Subtracting this margin of error from the sample mean, we get the lower limit of the confidence interval, which is approximately 16.2 cm.

Step 7 ：Adding the margin of error to the sample mean, we get the upper limit of the confidence interval, which is approximately 19.8 cm.

Step 8 ：Thus, the 95% confidence interval for the mean height of all plants of this species is \(\boxed{16.2}\) cm to \(\boxed{19.8}\) cm.