This problem does not make any sense to me!!! I would greatly appreciate your assistance, thanks.
A recent survey showed 4.6 percent of the sample of 250 had suffered some kind of identity theft in the past 12 months.
a. Construct a 90 percent confidence interval for the true proportion of those suffering identity theft.
b. Can we assume the distribution of the population is normal? Explain your answer.
a. To construct a confidence interval for the true proportion of those suffering identity theft, we can use the formula:
Confidence Interval = Sample Proportion ± (Critical Value) * Standard Error
First, let's calculate the sample proportion:
Sample Proportion = Number of cases in sample / Sample size
Sample Proportion = 4.6% = 0.046
Next, let's calculate the standard error:
Standard Error = sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size)
Standard Error = sqrt((0.046 * (1 - 0.046)) / 250)
Now, we need to find the critical value corresponding to a 90% confidence level. Since we have a large enough sample size, we can use the z-score table to find the critical value. For a 90% confidence level, the critical value is 1.645.
Finally, we can substitute all the values into the formula to calculate the confidence interval:
Confidence Interval = 0.046 ± (1.645 * Standard Error)
b. To determine if we can assume the distribution of the population is normal, we need to check two conditions:
1. Random sample: We are told that the survey was conducted on a sample of 250 people, which implies it was a random sample. This condition is satisfied.
2. Sample size: In general, for a normal distribution assumption to be valid, the sample size should be large enough (n * p ≥ 10, n * (1 - p) ≥ 10, where n is the sample size and p is the sample proportion). In this case, 250 * 0.046 ≈ 12 and 250 * (1 - 0.046) ≈ 238, which suggest that the sample size is large enough.
Both conditions are satisfied, so we can assume the distribution of the population is approximately normal.
Of course! I'd be happy to help you with this problem.
For part (a), constructing a confidence interval can provide us with a range of values within which the true proportion of people suffering from identity theft is likely to fall. To do this, we can use a formula for confidence intervals for proportions.
The formula for a confidence interval for proportions is:
CI = p̂ ± Z * sqrt((p̂ * (1 - p̂)) / n)
Where:
- CI represents the confidence interval
- p̂ is the sample proportion of people suffering from identity theft (4.6% or 0.046 in decimal form)
- Z is the critical value associated with the chosen confidence level (90% corresponds to a Z-value of 1.645)
- sqrt represents the square root
- n is the sample size (250)
Using the given values, we can now calculate the confidence interval:
CI = 0.046 ± 1.645 * sqrt((0.046 * (1 - 0.046)) / 250)
Now, we can calculate the lower and upper bounds of the confidence interval:
Lower bound: 0.046 - 1.645 * sqrt((0.046 * (1 - 0.046)) / 250)
Upper bound: 0.046 + 1.645 * sqrt((0.046 * (1 - 0.046)) / 250)
This will give us the range within which we can be 90% confident that the true proportion lies.
For part (b), to determine whether we can assume the population distribution is normal, we need to consider the conditions for using a normal distribution. In this case, we are dealing with proportions, and when the sample size is large enough (greater than or equal to 30), we can assume that the distribution of sample proportions is approximately normal.
Since the sample size of 250 is larger than 30, we can assume that the distribution of the sample proportion of identity theft is approximately normal.
I hope this explanation helps! Let me know if you have any further questions.