Question

A 99% confidence interval for the sugar content (per 5 oz. serving) of a shipment of strawberries is 5.9 to 8.3 grams.

(8) The sample mean and margin of error for the
confidence interval are

Answer 1

Sample mean = 7.1

Margin of error = 0.465

Explanation:

Formula for confidence interval is;

CI = x¯ ± zE

Where;

x¯ is sample mean

z is critical value at confidence level

E is margin of error.

z for 99% Cl is 2.58

We are told the CI is 5.9 to 8.3.

Thus;

5.9 = x¯ - 2.58E - - - - (1)

8.3 = x¯ + 2.58E - - - - (2)

Add both equations together to get;

14.2 = 2x¯

x¯ = 14.2/2

x¯ = 7.1

Put 7.1 for x¯ in eq 1 to get;

5.9 = 7.1 - 2.58E

7.1 - 5.9 = 2.58E

E = 1.2/2.58

E = 0.465