A teacher wanted to estimate the proportion of students who take notes in her class. She...

A teacher wanted to estimate the proportion of students who take notes in her class. She used data from a random sample of size n = 76 and found that 45 of them took notes. The 97% confidence interval for the proportion of student that take notes is:

Solutions

Expert Solution

Solution :

Given that,

n = 76

x = 45

Point estimate = sample proportion = = x / n = 45 /76=0.469

1 - = 1-0.469=0.531

At 97% confidence level

= 1 - 97%

= 1 - 0.97 =0.03

/2 = 0.015

Z/2 = Z0.015 =2.170( Using z table )

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 2.170 ( $\sqrt$ ((0.469*0.531) /76 )

E = 0.1242

A 97% confidence interval for population proportion p is ,

- E < p < + E

0.469 - 0.1242< p < 0.469 +0.1242

0.3448< p < 0.5932

The 97% confidence interval for the population proportion p is : 0.3448, 0.5932

orchestra answered 1 year ago

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