a teacher wants to estimate the mean height of seniors who attend a large high school. he randomly selects…

a teacher wants to estimate the mean height of seniors who attend a large high school. he randomly selects 35 students from this school and records the height, in inches, of these students. the average height is 65.3 inches with a standard deviation of 3.6 inches. which of the following is the 95% confidence interval for the true mean height of students at this school? find the t - table here. (63.64, 66.96) (64.06, 66.54) (64.27, 66.33) (64.51, 66.10)

a teacher wants to estimate the mean height of seniors who attend a large high school. he randomly selects 35 students from this school and records the height, in inches, of these students. the average height is 65.3 inches with a standard deviation of 3.6 inches. which of the following is the 95% confidence interval for the true mean height of students at this school? find the t - table here. (63.64, 66.96) (64.06, 66.54) (64.27, 66.33) (64.51, 66.10)

Answer

Answer:

(64.06, 66.54)

Explanation:

Step1: Determine degrees of freedom

$n - 1=35 - 1 = 34$. Since 34 is not on the t - table, we use 30 (a common approximation). For a 95% confidence interval and df = 30, the t - value $t_{\alpha/2}=2.042$.

Step2: Calculate the margin of error

The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Here, $s = 3.6$, $n = 35$. So $E=2.042\times\frac{3.6}{\sqrt{35}}\approx2.042\times\frac{3.6}{5.916}\approx2.042\times0.6085\approx1.24$.

Step3: Find the confidence interval

The confidence interval is $\bar{x}\pm E$. $\bar{x}=65.3$. Lower limit: $65.3 - 1.24 = 64.06$. Upper limit: $65.3+1.24 = 66.54$. So the 95% confidence interval is (64.06, 66.54).