Question

A company offering online speed reading courses claims that students who take their courses show a 5 times (500%) increase in the number of words they can read in a minute without losing comprehension. A random sample of 100 students yielded an average increase of 415% with a standard deviation of 220%. Calculate a 95% confidence interval for the average increase in number of words students can read in a minute without losing comprehension. Choose the closest answer.

Answer 1

Answer:

C.I = (371.88  , 458.12)

Step-by-step explanation:

Given that:

sample size n = 100

sample mean $\overline x =$ 415

standard deviation = 220

The objective is to calculate the  95% confidence interval for the average increase in number of words students who can read in a minute without losing comprehension.

At 95% confidence interval; level of significance ∝ = 1 - 0.95

level of significance ∝ = 0.05

$z_{\alpha/2} = 0.05/2$

$z_{\alpha/2} = 0.025$

The critical value at $z_{\alpha/2} = 0.025$ is 1.96

C.I = $\overline x \pm M.O,E$

C.I = $\overline x \pm z_{\alpha/2} \dfrac{\sigma }{\sqrt{n}}$

C.I = $415\pm 1.96 \dfrac{220 }{\sqrt{100}}$

C.I = $415\pm 1.96 *\dfrac{220 }{10}$

C.I = $415\pm 1.96 *22$

C.I = $415\pm 43.12$

C.I = (371.88  , 458.12)