(f) The United States has ten times the population of Canada. Assuming equal standard deviations, how many American teenagers must be sampled in order to estimate the true mean amount of time American teenagers spend on the internet per day to within 15 minutes with 96% confidence?
Thank you!
| [ ]The required sample size decreases. | | | | |
*All numerical answers should be integers (no decimals). We would like to estimate the true mean amount of time (in minutes) Canadian teenagers spend on the internet per day. The population standard deviation of daily internet usage time is known to be 41 minutes. (a) What sample size is required in order to estimate to within 15 minutes with 98% confidence? (e) For a fixed margin of error, when we increase the desired confidence level, what happens to the required sample size? [ ]The required sample size decreases. [ ]The required sample size increases. (f) The United States has ten times the population of Canada. Assuming equal standard deviations, how many American teenagers must be sampled in order to estimate the true mean amount of time American teenagers spend on the internet per day to within 15 minutes with 96% confidence? Thank you! to within 5 minutes with 96% confidence? (c) For a fixed confidence level, when we decrease the desired margin of error to one third its original value, we require ____ times the sample size. (d) What sample size is required in order to estimate to within 15 minutes with 96% confidence?
a)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z?/2 at 0.04% LOS is = 2.05 ( From Standard Normal Table )
Standard Deviation ( S.D) = 41
ME =15
n = ( 2.05*41/15) ^2
= (84.05/15 ) ^2
= 31.397 ~ 32
b)
Z?/2 at 0.04% LOS is = 2.05 ( From Standard Normal Table )
Standard Deviation ( S.D) = 41
ME =5
n = ( 2.05*41/5) ^2
= (84.05/5 ) ^2
= 282.576 ~ 283
d)
Z?/2 at 0.02% LOS is = 2.33 ( From Standard Normal Table )
Standard Deviation ( S.D) = 41
ME =15
n = ( 2.33*41/15) ^2
= (95.53/15 ) ^2
= 40.56 ~ 41
e)
The required sample size increases.
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