If 200 of the adults surveyed were in the age category of 24 and under and they provided a standard deviation of $14.50, construct a 95% confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under. Assume fast food weekly expenditures are normally distributed.
95% = mean ± 1.96 SEm
SEm = SD/√n
Do you know the mean?
To construct a confidence interval, we need to use the formula:
Confidence Interval = sample mean ± margin of error
To calculate the margin of error, we need to find the critical value associated with the desired confidence level. Since we want a 95% confidence interval, we need to find the critical value for a two-tailed test with an alpha level of 0.05/2 = 0.025.
The critical value can be found using a z-table or by using a statistical software or calculator. For a 95% confidence level, the z-score is approximately 1.96.
Now, let's calculate the sample mean and the margin of error:
1. Sample Mean:
The sample mean is the average expenditure of the adults in the age category of 24 and under. Since the sample mean is not provided in the question, we cannot calculate it. We will assume that the sample mean is represented by "x̄".
2. Margin of Error:
The margin of error is calculated by multiplying the standard deviation by the critical value and dividing it by the square root of the sample size (n). In this case, the standard deviation is given as $14.50, and the sample size is 200.
Margin of Error = (1.96 * standard deviation) / √n
= (1.96 * $14.50) / √200
Now, you can calculate the margin of error using a calculator or software.
Once you have the sample mean and the margin of error, you can construct the confidence interval by adding and subtracting the margin of error from the sample mean.
Confidence Interval = x̄ ± Margin of Error
Please provide the sample mean (x̄) to calculate the confidence interval for the weekly average expenditure on fast food for adults 24 years of age and under.