survey of 5050 young professionals found that they spent an average of $20.5620.56 when dining out, with a standard deviation of $11.4111.41. Can you conclude statistically that the population mean is greater than $2323? Use a 95% confidence interval. Question content area bottom Part 1 The 95% confidence interval is left bracket nothing comma nothing right bracketenter your response here, enter your response here. As $2323 is ▼ of the confidence interval, we ▼ can cannot conclude that the population mean is greater than $2323. (Use ascending order. Round to four decimal places as needed.)
survey of 5050 young professionals found that they spent an average of $20.5620.56 when dining out, with a standard deviation of $11.4111.41. Can you conclude statistically that the population mean is greater than $2323? Use a 95% confidence interval. Question content area bottom Part 1 The 95% confidence interval is left bracket nothing comma nothing right bracketenter your response here, enter your response here. As $2323 is ▼ of the confidence interval, we ▼ can cannot conclude that the population mean is greater than $2323. (Use ascending order. Round to four decimal places as needed.)
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 23PFA
Related questions
Question
survey of
mean is greater than
5050
young professionals found that they spent an average of
$20.5620.56
when dining out, with a standard deviation of
$11.4111.41.
Can you conclude statistically that the population $2323?
Use a 95% confidence interval.Question content area bottom
Part 1
The 95% confidence interval is
of the confidence interval, we
conclude that the population mean is greater than
left bracket nothing comma nothing right bracketenter your response here, enter your response here.
As
$2323
is
▼
▼
can
cannot
$2323.
(Use ascending order. Round to four decimal places as needed.)
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