"\\sum x = (5.1+4.9+...+6.0+6.1)=107.9 \\\\\n\n\\sum x^2 = (5.1^2 + 4.9^2 + \u2026 + 6.0^2 + 6.1^2) = 587.91 \\\\\n\nn=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} } \\\\\n\n= \\sqrt{\\frac{587.91 -(107.9)^2\/20}{20-1} } = 0.5520 \\\\\n\nH_0: \\mu = 6 \\\\\n\nH_1: \\mu \u2260 6"
Test-statistic:
"t = \\frac{\\bar{x} - \\mu}{ s\/ \\sqrt{n}} \\\\\n\n\nt = \\frac{5.395 -6.0}{0.552 \/ \\sqrt{20}} = -4.9015 \\\\\n\n\ndf = 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.
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