Question

A survey of 350 students is selected randomly on a large campus. They are asked if they use a laptop in class to take notes. Suppose that based on the survey, 140 of the 350 students responded "yes."
A. What is the value of the sample proportion ^p?

B. What is the standard error of the sample Proportation?

C. Construct an approximate 95% confidence interval for the true proportion p by taking +- SEs from the sample proportion.

Answer 1

Answer:

A. The value of the sample proportion is 0.4

B. The standard error of the sample proportion is 0.02619

C. 0.3487 ≤ p ≤ 0.4513

Step-by-step explanation:

The value of the sample proportion p' is calculated as:

$p'=\frac{x}{n} = \frac{140}{350}=0.4$

Where x is the number of success in the sample or the number of students that use a laptop in class to take notes and n is the size of the sample or 350 students.

On the other hand, the standard error SE of the sample proportion is calculated as:

$SEs=\sqrt{\frac{p'(1-p')}{n}}$

so, replacing the values, we get:

$SE=\sqrt{\frac{0.4(1-0.4)}{350}}=0.02619$

Finally, an approximate 95% confidence interval for the true proportion p is calculate as:

$p'-z_{\alpha/2} SEs \leq p\leq p'+z_{\alpha/2} SEs$

Where 1-α is equal to 95%, so $z_{\alpha/2}$ is equal to 1.96. Then, replacing the values we get:

$0.4-1.96(0.02619) \leq p\leq 0.4+1.96(0.02619)$

0.4 - 0.0513 ≤ p ≤ 0.4 + 0.0513

0.3487 ≤ p ≤ 0.4513