What percentage of college students are attending a college in the state where they grew up? Let p be the proportion of college students from the same state as that in which the college resides. If no preliminary study is made to estimate p, how large a sample is needed to be 90% sure that the point estimate (p hat) will be within a distance of 0.07 from p?
139
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(1.645)^2 * .5 * .5]/0.07^2
= ? (round to the next highest whole number)
I'll let you finish the calculation.
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is 0.07 in the problem. Z-value is found using a z-table (for 90%, the value is 1.645).
I hope this will help get you started.
r=-.96 between craving for pizza and ability to concentrate on studying
To determine the required sample size, we need to use the formula for sample size estimation in a proportion:
n = (Z^2 * p * (1 - p)) / E^2
Where:
- n is the required sample size
- Z is the Z-value corresponding to the desired level of confidence (in this case, 90% confidence)
- p is the estimated proportion of college students from the same state as their college
- E is the desired margin of error (in this case, 0.07)
Since no preliminary study has been conducted to estimate p, we can use the worst-case scenario where p = 0.5. This is because a proportion of 0.5 provides the most conservative estimate, requiring the largest sample size.
Using a Z-value of 1.645 for a 90% confidence level (obtained from a standard normal distribution table), and substituting the values into the formula, we have:
n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.07^2
n = (2.706025 * 0.25) / 0.0049
n = 0.67650625 / 0.0049
n ≈ 138.1769
Therefore, you would need a sample size of at least 139 (rounded up) college students to be 90% confident that the point estimate (p hat) will be within a distance of 0.07 from the true proportion, assuming the worst-case scenario where p = 0.5.