Question

1. a) a nutritionist recorded the daily sugar intake (in teaspoons) of 120 adolescents and found a mean of 25.2 teaspoons. construct a 99% confidence interval for the population mean daily sugar intake of all adolescents. assume the population standard deviation is 4.8 teaspoons. interpret your result.

Explanation:

Step1: Find the critical value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. Then $\alpha/2 = 0.005$. The critical value $z_{\alpha/2}$ is the $z$-score such that the area to the right of $z_{\alpha/2}$ is $0.005$ and the area to the left is $0.995$. From the standard normal table, $z_{0.005} = 2.576$.

Step2: Calculate the margin of error

The formula for the margin of error $E$ when the population standard deviation $\sigma$ is known is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 4.8$, $n = 120$.
$E=2.576\times\frac{4.8}{\sqrt{120}}$.
First, calculate $\sqrt{120}\approx10.954$.
Then $\frac{4.8}{10.954}\approx0.438$.
$E = 2.576\times0.438\approx1.13$.

Step3: Construct the confidence interval

The confidence interval for the population mean $\mu$ is given by $\bar{x}-E<\mu <\bar{x} + E$.
Given $\bar{x}=25.2$.
The lower limit is $25.2-1.13 = 24.07$.
The upper limit is $25.2 + 1.13=26.33$.

Interpretation:

We are 99% confident that the true population mean daily sugar intake of all adolescents lies between $24.07$ teaspoons and $26.33$ teaspoons.

Answer:

The 99% confidence interval is $(24.07, 26.33)$. We are 99% confident that the population mean daily sugar intake of all adolescents is within this interval.