Stat Question 1

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Statistics
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A telecommunications company wants to estimate the mean length of time (in minutes) that 18 to 24
year olds spend text messaging each day. In a random sample of twenty seven 18 to 24 year olds, the
mean length of time spent texting was 29 minutes. From past studies the standard deviationis 4.5
minutes. Find the 90% confidence interval.
1. Find the critical value
2. Find the margin of error: E =
3. Find the confidence interval: CI =
Write the conclusion.
1-According to the table the critical value for 90% or .90 is 1.645.
2-The formula for finding margin of error is E = z
α/2
* σ /(√ n) where α is the confidence level, Z
α/2
is
the critical value σ is the standard deviation, and n is the sample size. When we plug the values of
Z
α/2
and n we get E=1.645*4.5/√27.When we further solve it we get E=1.43.
3-For Confidence interval we will add and subtract mean from margin of error from mean i.e. 29-
1.43=27.57 and 29+1.43=30.43 to get the lower and upper bounds of the confidence interval.
4- We conclude that for the critical value of 1.645 margin of error is 1.43 given if the values of σ
and n is 4.5 and 27.The values of upper and lower bounds for mean 29 and margin of error 1.43 are
27.57 and 30.43 respectively.

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A telecommunications company wants to estimate the mean length of time (in minutes) that 18 to 24 year olds spend text messaging each day. In a random sample of twenty seven 18 to 24 year olds, the mean length of time spent texting was 29 minutes. From past studies the standard deviationis 4.5 minut ...
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