CI's via the CLT
The sample mean estimate Θˆ of the mean of a random variable with variance 1, based on 100 samples, happened to be 22. The 80% confidence interval provided by the CLT is of the form [a,b], with:
a=
unanswered
b=
unanswered
Your answers should include at least 2 decimal digits.
[Θˆ − 1.96σ/√n, Θˆ + 1.96σ/√n].
You may want to refer to the normal table (below). For your reference, if we had 95% instead of 80%, the confidence interval would be of the form
[Θˆ − 1.96σ/√n, Θˆ + 1.96σ/√n].
a = 21.87
b = 22.12
Ah, confidence intervals, the fun little intervals that help us estimate population parameters! Now, let's put on our clown noses and tackle this question.
Since we're given the sample mean estimate Θˆ of 22 and a normal distribution with variance 1, we can use the Central Limit Theorem (CLT) to calculate the confidence interval. The CLT says that if we have a large enough sample size, the sampling distribution of the sample mean will be approximately normal.
The 80% confidence interval can be represented as [Θˆ − 1.96σ/√n, Θˆ + 1.96σ/√n]. Here, σ represents the population standard deviation and n represents the sample size.
However, we don't have the population standard deviation (σ) here, so we'll have to find an alternative way to estimate it. Luckily, since our sample size is large (100 samples), we can approximate the population standard deviation with the sample standard deviation (s). Since the variance is 1, the standard deviation is also 1.
So, plugging in the values, we have:
a = Θˆ − 1.96σ/√n
= 22 − 1.96(1)/√100
= 22 − 1.96(1)/10
= 22 − 0.196
= 21.804
b = Θˆ + 1.96σ/√n
= 22 + 1.96(1)/√100
= 22 + 1.96(1)/10
= 22 + 0.196
= 22.196
Therefore, the 80% confidence interval [a, b] is approximately [21.804, 22.196].
And just to add a little clown twist, if we had a 95% confidence interval, it would be the same formula: [Θˆ − 1.96σ/√n, Θˆ + 1.96σ/√n]. But hey, we humans do love our 95% confidence, don't we?
Hope that brought a smile to your face!
To find the values of a and b for the 80% confidence interval based on the Central Limit Theorem (CLT), we can use the formula:
a = Θˆ - 1.96σ/√n
b = Θˆ + 1.96σ/√n
Given that the sample mean estimate, Θˆ, is 22 and the variance, σ^2, is 1, and the number of samples, n, is 100, we can plug in these values into the formula:
a = 22 - 1.96(1)/√100
b = 22 + 1.96(1)/√100
Simplifying:
a = 22 - 1.96/10
b = 22 + 1.96/10
Calculating:
a ≈ 22 - 0.196
b ≈ 22 + 0.196
a ≈ 21.804
b ≈ 22.196
Therefore, the 80% confidence interval provided by the CLT is approximately [21.804, 22.196].
To find the values of a and b in the 80% confidence interval using the Central Limit Theorem (CLT), we can use the formula: [Θˆ - 1.96σ/√n, Θˆ + 1.96σ/√n].
Given that the sample mean estimate, Θˆ, is 22 and the variance is 1, we need to substitute these values into the formula:
a = Θˆ - 1.96σ/√n
b = Θˆ + 1.96σ/√n
Substituting the given values:
a = 22 - 1.96σ/√100
b = 22 + 1.96σ/√100
Since the variance is 1, we can substitute σ with 1:
a = 22 - 1.96/10
b = 22 + 1.96/10
Calculating the values:
a = 22 - 0.196
b = 22 + 0.196
Therefore, the 80% confidence interval provided by the Central Limit Theorem is [21.804, 22.196].
If we had a 95% confidence level instead, the formula would be the same:
[Θˆ - 1.96σ/√n, Θˆ + 1.96σ/√n].
The only difference is the value of z in the formula. For a 95% confidence level, the z-value would be 1.96, which corresponds to a 2-tailed standard normal distribution with a total probability of 0.95.