Question

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):Brand A: 34.36 31.26 37.36 28.52 33.14 32.74 34.34 34.33 34.95Brand B: 41.08 38.22 39.59 38.82 36.24 37.73 35.03 39.22 34.13 34.33 34.98 29.64 40.60 Let formula415.mml represent the population mean for Brand B and let formula417.mml represent the population mean for Brand A.Find a 98% confidence interval for the difference formula419.mml. Round down the degrees of freedom to the nearest integer and round the answers to three decimal places.

Answer 1

Explanation:

For brand A,

Mean, x1 = (34.36 + 31.26 + 37.36 + 28.52 + 33.14 + 32.74 + 34.34 + 34.33 + 34.95)/9 = 33.44

standard deviation, s1 = √(summation(x - mean)²/n

Summation(x - mean)² = (34.36 - 33.44)^2 + (31.26 - 33.44)^2 + (37.36 - 33.44)^2 + (28.52 - 33.44)^2 + (33.14 - 33.44)^2 + (32.74 - 33.44)^2 + (34.34 - 33.44)^2 + (34.33 - 33.44)^2 + (34.95 - 33.44)^2 = 49.6338

Standard deviation = √(49.6338/9

s1 = 2.35

For brand B,

Mean, x2 = (41.08 + 38.22 + 39.59 + 38.82 + 36.24 + 37.73 + 35.03 + 39.22 + 34.13 + 34.33 + 34.98 + 29.64 + 40.60 )/13 = 36.89

Summation(x - mean)² = (41.08 - 36.89)^2 + (38.22 - 36.89)^2 + (39.59 - 36.89)^2 + (38.82 - 36.89)^2 + (36.24 - 36.89)^2 + (37.73 - 36.89)^2 + (35.03 - 36.89)^2 + (39.22 - 36.89)^2 + (34.13 - 36.89)^2 + (34.33 - 36.89)^2 + (34.98 - 36.89)^2 + (29.64 - 36.89)^2 + (40.60 - 36.89)^2 = 124.5024

Standard deviation = √(124.5024/13

s2 = 3.09

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 98% confidence interval, we would determine the z score from the t distribution table because the number of samples are small.

Degree of freedom =

(n1 - 1) + (n2 - 1) = (9 - 1) + (13 - 1) = 20

z = 2.528

x1 - x2 = 33.44 - 36.89 = - 3.45

Margin of error = z√(s1²/n1 + s2²/n2) = 2.528√(2.35²/9 + 3.09²/13) = 2.935

The 98% confidence interval is

- 3.45 ± 2.935