A random sample of 50 students chosen from an elementary school with 400 students was found to have a mean allowance of $26 per week with a standard deviation of $8. Obtain a 98% confidence interval for the total amount of allowance money given to students in this school per week.
26-2.33*8/sqrt((50)), 26+ 2.33*8/sqrt((50)),
= ( , ) ?
That's what I originally did but it's not one of the correct answers
[ 23.37, 28.63]
$9,531 to $11,269
$9,347 to $11,453
$10,028 to $10,772
$9,414 to $11,386
Are my multiple choice
mean 26(400) = $10400
Standard deviation = 8(400) = 3200
10400-+ 2.3263*3200/sqrt(50))
[$9,347, 11,453]
Thank you so much !
To obtain a 98% confidence interval for the total amount of allowance money given to students in this school per week, we need to use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
First, let's calculate the critical value. Since we want a 98% confidence interval, there is 1% of the data in each tail. To find the critical value, we need to find the z-score that corresponds to the 0.01 tail probability. We can use a z-table or a statistical calculator to find the value.
In this case, the critical value for a 98% confidence interval is approximately 2.33.
Next, we need to calculate the standard error. The standard error represents the variability of the sample mean. The formula to calculate the standard error is:
Standard Error = Standard Deviation / √Sample Size
In this case, the standard deviation is $8 and the sample size is 50, so the standard error is:
Standard Error = 8 / √50 ≈ 1.131
Now we can calculate the confidence interval:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
Sample Mean = $26 (given)
Lower Limit = $26 - (2.33 * 1.131) ≈ $26 - 2.641 ≈ $23.359
Upper Limit = $26 + (2.33 * 1.131) ≈ $26 + 2.641 ≈ $28.641
Therefore, the 98% confidence interval for the total amount of allowance money given to students in this school per week is approximately $23.36 to $28.64.