Suppose you are interested in determining the probability that a college student will own a personal computer. You poll 8 of your classmates from randomly selected classes and find that 5 of them own their own computers. Find a 90% confidence interval estimate for the probability than any randomly selected college student will own his or her own computer. Round E and the interval limits to the nearest thousandth.
I want to know what are the rules of thumb?
To find a confidence interval estimate for the probability that a college student owns a personal computer, we can use the binomial distribution and apply the normal approximation to the sampling distribution of sample proportions.
Step 1: Determine the sample proportion
In this case, you polled 8 classmates and found that 5 of them own computers. So the sample proportion is:
p̂ = 5/8 = 0.625
Step 2: Determine the standard error
The standard error measures the average amount by which the sample proportion differs from the true population proportion. It is calculated using the formula:
SE = √(p̂(1-p̂)/n)
Where p̂ is the sample proportion and n is the sample size.
SE = √((0.625)(1-0.625)/8) = √(0.153/8) = 0.196
Step 3: Determine the critical value
The critical value is determined based on the desired confidence level. Since we want a 90% confidence interval, we need to find the critical value corresponding to a 5% tail on each side of the distribution.
For a normal distribution, a 90% confidence level corresponds to a critical value (z*) of approximately 1.645.
Step 4: Determine the margin of error
The margin of error identifies the range in which we expect the population proportion to fall within. It is calculated as:
E = z* * SE
Where E is the margin of error and z* is the critical value.
E = (1.645)(0.196) = 0.322
Step 5: Calculate the confidence interval
Finally, we can construct the confidence interval estimate for the probability that a randomly selected college student will own a personal computer using the formula:
CI = (p̂ - E, p̂ + E)
CI = (0.625 - 0.322, 0.625 + 0.322) = (0.303, 0.947)
Rounded to the nearest thousandth, the 90% confidence interval estimate for the probability that any randomly selected college student owns his or her own computer is (0.303, 0.947).