Question

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

______% ± % _______

Answer #1

Solution:

Given that,

Lower confidence interval = 27.5% = 0.275

Upper confidence interval = 42.7% = 0.427

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

= ( 0.275 + 0.427 )/2

= 0.702 / 2

= 0.351

Margin of error E = Upper confidence interval -

= 0.427 - 0.351

= 0.076

In the form of

ME

0.351 0.076

35.1% 7.6%

Express the confidence interval 35.7% < p < 42.3% 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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