Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
