Question

Express the confidence interval ( 0.038 , 0.104 ) in the form of p - E < p< p+e

Answer #1

Solution :

Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = ( 0.038 + 0.104 )/ 2=0.071

Margin of error = E = Upper confidence interval - = 0.104-0.071=0.033

In the form of :

- E < p < + E is,

0.071 - 0.033 < p <0.071+0.033

0.038<p<0.104

Express the confidence interval (0.074,0.162) in the form of
Ep−E<p<p+E.

express the confidence interval (o.026,0.100) in the form of p
(hat) - E < p < p (hat) + E

Express the confidence interval
61.9%<p<76.9%61.9%<p<76.9% in the form of ˆp±E?
__________% ± _____________ %

Show all work.
1. Express the confidence interval (0.412,0.789) in the form of
p +- E

Express the confidence interval ( 0.076, 0.122) in the form p
with caret minus Upper E less than p less than p with caret plus
Upper E p-E<P<p+E

Express the confidence interval 0.528 < p < 0.746 in the
form

Express the confidence interval 18.5 % < p < 26.7 % in the
form of ˆp ± ME.
% ±. %

Express the confidence interval (0.054, 0.110 ) in the form of
ModifyingAbove p with caret minus Upper E less than p less than
ModifyingAbove p with caret plus Upper E.
_________< p< _______________(Type integers or
decimals.)

A) Express the confidence interval
66.4%<p<81.2%66.4%<p<81.2% in the form of
ˆp±Ep^±E.
_____% ± ± %_____
B) Express the confidence interval 0.69±0.0510.69±0.051 in open
interval form (i.e., (0.155,0.855)).
C) In a survery of 181 households, a Food Marketing Institute found
that 141 households spend more than $125 a week on groceries.
Please find the 98% confidence interval for the true proportion of
the households that spend more than $125 a week on groceries.
Enter your answer as an open-interval
(i.e., parentheses)...

Express the confidence interval
(0.055,0.147)
in the form of
Modifying Above p with caret minus Upper E less than p less than
Modifying Above p with caret plus Upper Ep−E<p<p+E.

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