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Sagot :
Answer:
The p-value is approximately 0.0508
Step-by-step explanation:
The given parameters are;
The expected mass of banana chips bag filled by the machine, μ = 421 gams
The number of bags in the sample of bags filled by the machine, n = 26 bags
The mean mass of the bags in the sample, [tex]\overline x[/tex] = 413 grams
The standard deviation of the mass in the bags, s = 24 grams
The level of significance, α = 0.02
The null hypothesis, H₀; μ = 421 grams
The alternative hypothesis, Hₐ; μ < 421 grams
The test statistic is given as follows;
[tex]t=\dfrac{\bar{x}-\mu }{\dfrac{s}{\sqrt{n}}}[/tex]
We get;
[tex]t=\dfrac{413-421 }{\dfrac{24}{\sqrt{26}}} \approx -1.6996731712[/tex]
The t-value = -1.6996731712
The degrees of freedom df = n - 1 = 26 - 1 = 25
From an online source, we get the critical-t = 2.166587
The p-value is obtained using an online source as p(t ≤ -1.6996731712) = 0.0508
From the p-value table, we get
0.05 < p < 0.1
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