Answer to Question #246041 in Statistics and Probability for boo

Question

A consumer group is investigating a producer of diet meals to examine if its prepackaged meals
actually contain the advertised 6 ounces of protein in each package. Based on the following
data, is there any evidence that the meals do not contain the advertised amount of protein? Run
the appropriate test at a 5% level of significance. 5.1 4.9 6.0 5.1 5.7 5.5 4.9 6.1 6.0 5.8
5.2 4.8 4.7 4.2 4.9 5.5 5.6 5.8 6.0 6.1

Answer 1

$\sum x = (5.1+4.9+...+6.0+6.1)=107.9 \\ \sum x^2 = (5.1^2 + 4.9^2 + … + 6.0^2 + 6.1^2) = 587.91 \\ n=20$

Mean

$\bar{x} = \frac{\sum x}{n} = \frac{107.9}{20} = 5.395$

Standard deviation

$s = \sqrt{ \frac{\sum x^2 - ((\sum x)^2/n)}{n-1} } \\ = \sqrt{\frac{587.91 -(107.9)^2/20}{20-1} } = 0.5520 \\ H_0: \mu = 6 \\ H_1: \mu ≠ 6$

Test-statistic:

$t = \frac{\bar{x} - \mu}{ s/ \sqrt{n}} \\ t = \frac{5.395 -6.0}{0.552 / \sqrt{20}} = -4.9015 \\ df = n-1 = 19$

p-value = T.DIST.2T(ABS(-4.9015), 19) = 0.0001

Decision: p-value < α, Reject the null hypothesis.

Conclusion: There is enough evidence to conclude that the meals do not contain the advertised amount of protein at 0.05 significance level.