A random sample of 20 purchases showed the sample mean of $48.77 and the sample standard deviation s is $17.58. How large is the margin of error for a 80 % confidence interval for the mean purchases of all customers

Step-by-step explanation:

Given data          

Sample Mean (M):  $48.77  

Sample Size (n): 20

Standard Deviation (σ) :  $17.58

Confidence Level: 80%

 we know that z*-Values for 80% Confidence  Levels is 1.28

the expression for margin of error is given bellow\

MOE= z*σ/√n

We can now substitute into the expression and solve for the MOE as

MOE= 1.28*17.58/√20

MOE= 22.502/4.47

MOE= 22.502/4.47

MOE= 5.03

The margin of error for a 80 % confidence interval is $5.03

a

step-by-step explanation:

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.