Question

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

% ± %

Answer #1

Solution :

Given that,

Lower confidence interval = 0.357

Upper confidence interval = 0.423

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

= (0.357 + 0.423) / 2

= 0.39

= 0.39

Margin of error = E = Upper confidence interval - = 0.423 - 0.39 = 0.033

ME = **39%
0.3%**

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 18.5 % < p < 26.7 % 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
___% ± ____%

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

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

Express the confidence interval (84.6,268.8) in the form of
¯x±ME.
Express the confidence interval 587.7<μ<814.5 in the form
of ¯x±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 596.9<μ<746.1 in the form
of ¯x±ME. ¯x±ME= ±

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