Exercise 8.92

Tony's Pizza guarantees all pizza deliveries within 30 minutes of the placement of orders. An agency wants to estimate the proportion of all pizzas delivered within 30 minutes by Tony's. What is the most conservative estimate of the sample size that would limit the margin of error to within .02 of the population proportion for a 99% confidence interval?

Exercise 8.93
Refer to Exercise 8.92. Assume that a preliminary study has shown that 93% of all Tony's pizzas are delivered within 30 minutes. How large should the sample size be so that the 99% confidence interval for the population proportion has a margin of error of .02?

To calculate the sample size needed for Exercise 8.92, we can use the formula:

n = (z^2 * p * (1-p)) / E^2

Where:
n = sample size
z = z-value from the standard normal distribution corresponding to the desired confidence level (99% confidence interval)
p = preliminary estimate of the proportion (from Exercise 8.93, it is 0.93)
E = margin of error

Now let's plug in the values:

n = (z^2 * p * (1-p)) / E^2
n = (2.576^2 * 0.93 * (1-0.93)) / 0.02^2
n = (6.634176 * 0.93 * 0.07) / 0.0004
n = 0.044910816 / 0.0004
n = 112.27804

Since the sample size needs to be a whole number, we round up to the nearest integer:

n = 113

Therefore, the most conservative estimate of the sample size that would limit the margin of error to within 0.02 of the population proportion for a 99% confidence interval is 113.

Now let's move on to Exercise 8.93.

To calculate the sample size needed for Exercise 8.93, we can still use the same formula as before:

n = (z^2 * p * (1-p)) / E^2

But this time, we will use the preliminary estimate of the proportion (p) as 0.93.

Now let's plug in the values:

n = (z^2 * p * (1-p)) / E^2
n = (2.576^2 * 0.93 * (1-0.93)) / 0.02^2
n = (6.634176 * 0.93 * 0.07) / 0.0004
n = 0.044910816 / 0.0004
n = 112.27804

Again, round up to the nearest whole number:

n = 113

Therefore, the sample size needed for a 99% confidence interval for the population proportion with a margin of error of 0.02 is also 113.

To find the most conservative estimate of the sample size in Exercise 8.92, which limits the margin of error to within 0.02 of the population proportion for a 99% confidence interval, we need to use the formula for sample size calculation for estimating proportions.

The formula for calculating the sample size for estimating proportions is:

n = (Z^2 * p * (1-p)) / E^2

Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level (99% in this case)
- p is the estimated proportion from the preliminary study (if available)
- E is the desired margin of error

In Exercise 8.93, we are given the preliminary study result that 93% of all Tony's pizzas are delivered within 30 minutes, and we want to find the sample size for a 99% confidence interval with a margin of error of 0.02.

Substituting the given values into the sample size formula:

n = (Z^2 * p * (1-p)) / E^2
n = ((Z * p) / E)^2

We can calculate the Z-score for a 99% confidence level using a Z-score table or a statistical calculator. The Z-score corresponding to a 99% confidence level is approximately 2.576.

Substituting the values into the formula:

n = ((2.576 * 0.93) / 0.02)^2
n = (2.396/0.02)^2
n = 119.8^2
n ≈ 14,351

Therefore, in Exercise 8.93, the sample size should be approximately 14,351 in order to have a 99% confidence interval for the population proportion with a margin of error of 0.02.

Margin of error = 0.02

z = 2.58

Unknown p
We are going use 50%

n = (za/2 /E ) ^2 *p(1-p)

n = (2.58/0.02)^2*.5* .5 = 4160.25
n = 4161