Question

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

__________% ± _____________ %

Answer #1

**Solution:**

Given: the confidence interval 61.9%<p<76.9%

We have to express it in the form of

Add both limits and divide it by 2 to get value of

Subtract Lower from Upper limit and divide it by 2 to get value of E:

Thus we get:

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

Express the confidence interval 35.7% < p < 42.3% in the
form of ˆp ± ME
% ± %

Express the confidence interval 27.5% <p< 42.7% in the
form of ˆp±ME.
______% ± % _______

Express the confidence interval 73%<p<88%73%<p<88%
in the form of ˆp±ME

Express the confidence interval
70.8%<p<77.6%70.8%<p<77.6% in the form of
ˆp±MEp^±ME

Express the confidence interval (11.2%,29.2%)in the form of
ˆp±ME
___% ± ____%

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

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

Express the confidence interval (8.6%,21.4%)(8.6%,21.4%) in the
form of ˆp±MEp^±ME.
% ±± %

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