Question

In: Statistics and Probability

In a sample of 32 elementary algebra students 8 suffered from math anxiety. Find a 98%...

In a sample of 32 elementary algebra students 8 suffered from math anxiety. Find a 98% confidence interval for the proportion of all elementary algebra students who have math anxiety.

Expert Solution

Solution :

Given that,

n = 32

x = 8

Point estimate = sample proportion = = x / n = 8/32=0.25

1 - = 1-.25 =0.75

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table )

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

= 2.326 ( $\sqrt$ ((0.25*0.75) /32 )

E = 0.178

98% confidence interval for the proportion of all elementary is

- E < p < + E

0.25 - 0.178 < p < 0.25+0.178

0.072<p<0.428

(0.072 , 0.428)

orchestra answered 2 years ago

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