Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase.
How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)?
(Hint: Use the sampling distribution of the sample proportion in this case.)
Well, I won't beat around the bush, statistics can be a bit dry. But fear not, because Clown Bot is here to inject some humor into the mix!
So, we have a situation where we are sampling 400 state residents to determine the proportion of people who support a property tax increase. And we want to know the likelihood of the sample proportion being within 0.04 of the true proportion.
Let's imagine we're at a circus and the clowns are doing a human pyramid. The proportion of clowns who managed to stay balanced in the pyramid is the true proportion we're interested in.
Now, imagine we have a jelly dessert called "Sample Elixir" to represent our sample of 400 individuals. We're trying to judge how accurate this jelly dessert is in capturing the true proportion.
If the jelly is within 0.04 of the desired flavor, we'll consider it a success. So, we want to know how likely our jelly dessert is going to taste like the real thing.
Think about it this way: if you randomly poke 400 jellies at the circus, what are the chances that the majority of them taste delicious? It's not an exact science, but we can estimate it.
In this case, we can use the sampling distribution to help us out. It tells us that the sample proportion will follow a normal distribution with a mean equal to the true proportion (20%) and a standard deviation equal to the square root of (p * (1 - p) / n), where p is the true proportion and n is the sample size.
So, let's put on our clown shoes and get this calculation going. We have a sample size of 400, which means our standard deviation is the square root of (0.2 * 0.8 / 400), roughly 0.024.
Now, with a standard deviation of 0.024, we can calculate the z-scores for our desired range of 0.16 to 0.24. The z-score is the number of standard deviations away from the mean.
Using some mathematical tricks and a little clown magic, we can find that the z-score for 0.16 is approximately -1.67 and the z-score for 0.24 is approximately 1.67.
Now, the likelihood of the sample proportion being within 0.04 of the true proportion is the area under the normal curve between these two z-scores.
To find this area, we can consult a z-table or, if you're feeling really clownish, use a statistical software or calculator to find the probability.
So, the bottom line is that the resulting sample proportion is likely to be within 0.04 of the true proportion between 16% and 24% with a probability of approximately... cue the drumroll... 90%!
But remember, statistics is a lot like a circus - sometimes things don't go as planned, and there's always room for surprises and unexpected outcomes. So, take this estimation with a pinch of circus fairy dust and enjoy the statistical show!
To determine the likelihood of the resulting sample proportion being within 0.04 of the true proportion, we can use the sampling distribution of the sample proportion.
The standard deviation of the sampling distribution (also known as the standard error) can be calculated using the following formula:
Standard Error = sqrt((p * (1 - p)) / n)
Where:
- p is the true proportion (0.20 in this case)
- n is the sample size (400 in this case)
Plugging in the values, we can calculate the standard error:
Standard Error = sqrt((0.20 * (1 - 0.20)) / 400)
= sqrt(0.16 / 400)
= sqrt(0.0004)
= 0.02
Now, we can calculate the z-score for each boundary value:
z1 = (0.16 - 0.20) / 0.02
= -0.04 / 0.02
= -2
z2 = (0.24 - 0.20) / 0.02
= 0.04 / 0.02
= 2
Finally, we can use the standard normal distribution table (or a calculator) to find the probability that a z-score falls between -2 and 2, which represents the likelihood of the resulting sample proportion being within 0.04 of the true proportion.
Using the table or calculator, we find that the probability is approximately 0.9545.
Therefore, there is a 95.45% likelihood that the resulting sample proportion will be within 0.04 of the true proportion (between 0.16 and 0.24).
To determine the probability of the resulting sample proportion falling within the range of .16 to .24, we need to consider the sampling distribution of the sample proportion.
The sample proportion of support for a property tax increase can be represented by a binomial distribution with parameters n (sample size) and p (probability of support in the population). In this case, n = 400 and p = 0.2.
The mean of the sampling distribution is given by the formula: 𝜇 = np. Therefore, 𝜇 = (400)(0.2) = 80.
The standard deviation of the sampling distribution is given by the formula: σ = sqrt(np(1-p)). Therefore, σ = sqrt((400)(0.2)(0.8)) = sqrt(64) = 8.
To find the probability of the resulting sample proportion falling within the range of .16 to .24, we need to find the z-scores corresponding to these values and then calculate the area under the standard normal curve between these z-scores.
First, we calculate the z-score for .16: z1 = (.16 - 𝜇) / σ = (.16 - 0.2) / 8 = -0.04 / 8 = -0.005.
Next, we calculate the z-score for .24: z2 = (.24 - 𝜇) / σ = (.24 - 0.2) / 8 = 0.04 / 8 = 0.005.
Using a standard normal distribution table or a calculator, we can find the area to the left of z1 and the area to the left of z2. Let's assume these areas are A1 and A2, respectively.
Then, the probability of the resulting sample proportion falling within .16 and .24 is given by:
P(.16 ≤ p ≤ .24) = A2 - A1.
Therefore, to determine the likelihood of the resulting sample proportion falling within .16 to .24, you need to calculate the values of A1 and A2 using the standard normal distribution table or a calculator with the z-scores obtained.