1)for 99% CI
for 99 % CI value of z= | 2.576 |
standard deviation σ= | 16.30 |
margin of error E = | 3 |
required sample size n=(zσ/E)2 = | 197.0 |
( please try 196 if this comes wrong due to rounding)
2)for 90% CI
for 90 % CI value of z= | 1.645 |
standard deviation σ= | 16.30 |
margin of error E = | 3 |
required sample size n=(zσ/E)2 = | 80.0 |
3)
option C: the lower the confidence level ; the smaller the sample size
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A survey of college students was conducted during final exam week to assess the number of hours spent studying each day. The mean number of hours was 5 with a standard deviation of 1.5 hours. The distribution was normal. How many hours would an individual at the 60th percentile rank study?
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