In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $36 and standard deviation of $4. Find the margin of error at a 90% confidence level.
To find the margin of error at a 90% confidence level, we need to use the formula:
Margin of Error = z * (standard deviation / square root of sample size)
First, let's find the value of z for a 90% confidence level. Using a standard normal distribution table, we find that the z-score for a 90% confidence level is approximately 1.645.
Given:
Mean (μ) = $36
Standard deviation (σ) = $4
Sample size (n) = 27
Z-score (z) = 1.645
Now we can calculate the margin of error:
Margin of Error = 1.645 * ($4 / sqrt(27))
Using a calculator:
Margin of Error ≈ 1.645 * ($4 / 5.196)
Margin of Error ≈ 1.645 * $0.770
Margin of Error ≈ $1.266
Therefore, the margin of error at a 90% confidence level is approximately $1.266.
To find the margin of error at a 90% confidence level, we need to use the formula:
Margin of Error = Z * (standard deviation / sqrt(sample size))
We are given the mean (μ) is $36, the standard deviation (σ) is $4, and the sample size (n) is 27.
First, we need to find the Z-score for a 90% confidence level.
The Z-score can be found using a Z table or a calculator. For a 90% confidence level, the Z-score is 1.645 (approximately).
Now, we can substitute the values into the formula:
Margin of Error = 1.645 * (4 / sqrt(27))
Calculating the square root of 27, we get:
Margin of Error = 1.645 * (4 / 5.196)
Now, we can calculate the margin of error:
Margin of Error ≈ 1.645 * 0.770
Margin of Error ≈ 1.267
Therefore, the margin of error at a 90% confidence level is approximately $1.27.
To find the margin of error at a 90% confidence level, we need to use the formula:
Margin of Error = Z * (Standard Deviation / √Sample Size)
Where:
Z is the z-score corresponding to the desired confidence level
Standard Deviation is the standard deviation of the population
Sample Size is the number of observations in the sample
First, we need to find the z-score for a 90% confidence level. The z-score represents the number of standard deviations away from the mean.
To find the z-score, we can use a standard normal distribution table or a statistical software. For a 90% confidence level, the z-score is approximately 1.645.
Next, we need to determine the standard deviation of the population. In this case, the standard deviation is given as $4.
Lastly, we need to determine the sample size, which is 27.
Now, we can plug these values into the formula:
Margin of Error = 1.645 * ($4 / √27)
Calculating the expression:
Margin of Error ≈ 1.645 * ($4 / 5.2)
Margin of Error ≈ 1.645 * $0.769
Margin of Error ≈ $1.266
Therefore, at a 90% confidence level, the margin of error is approximately $1.266.