Question

Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a period when typing. 440 people responded that they leave one space. Create a 90% confidence interval for the proportion of people who leave one space after a period.

Answer 1

Answer:

The confidence interval is (0.81, 0.87).

Step-by-step explanation:

There's 90% confidence that population proportion is within the interval obtained from the following formula:

$\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}$

Knowing that the sample size, $n=525$ we obtain the proportion of people from the sample who leave one space after a period as $\hat{p}=\frac{440}{525}=0.8381\approx 0.84$.

We then look for the critical value:

$z_{\alpha/2}=1.645$

Now we can replace in the formula to obtain the confidence interval:

$0.84\pm 1.645\sqrt{\frac{0.84*(1-0.84)}{525}}= (0.8137; 0.8663)$

Therefore we can say that there's 90% probability that the population proportion of people who leave one space after a period lies between the values (0.8137; 0.8663).

Answer 2

Answer:  $(0.81155,\ 0.86445)$

Step-by-step explanation:

The confidence interval for population proportion (p) is given by :_

$\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

z* = Critical value.

$\hat{p}$ = Sample proportion.

Let p  be the true population proportion of people who leave one space after a period.

As per given , we have

n= 525

$\hat{p}=\dfrac{440}{525}=0.838$

By z-table , the critical value for 90% confidence interval : z* = 1.645

Now , 90% confidence interval for the proportion of people who leave one space after a period:

$0.838\pm (1.645) \sqrt{\dfrac{0.838(1-0.838)}{525}}$

$0.838\pm (1.645) \sqrt{0.00025858}$

$0.838\pm (1.645) (0.0160805117)$

$\approx0.838\pm(0.02645)$

$=(0.838-0.02645,\ 0.838+0.02645)=(0.81155,\ 0.86445)$

Hence, a 90% confidence interval for the proportion of people who leave one space after a period. $=(0.81155,\ 0.86445)$