Answer:
Step-by-step explanation:
Given that:
[tex]H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55[/tex]
(a) For x = 54 and s = 5.3
The test statistics can be computed as:
[tex]Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }[/tex]
[tex]Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }[/tex]
[tex]Z = \dfrac{-1}{\dfrac{5.3}{6} }[/tex]
Z = -1.132
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(-1.132,35,1) = 0.1326
Decision: p-value is greater than significance level; do not reject [tex]\mathbf{H_o}[/tex]
[tex]Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \[/tex] [tex]\mu < 55[/tex]
b
For x = 53 and s = 4.6
The test statistics can be computed as:
[tex]Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }[/tex]
[tex]Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }[/tex]
[tex]Z = \dfrac{-2}{\dfrac{4.6}{6} }[/tex]
Z = -2.6087
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(-2.6087,35,1) =0.0066
Decision: p-value is < significance level; we reject the null hypothesis.
[tex]Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that[/tex] [tex]\mu < 55[/tex]
c)
For x = 56 and s = 5.0
The test statistics can be computed as:
[tex]Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }[/tex]
[tex]Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }[/tex]
[tex]Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }[/tex]
Z = 1.2
degree of freedom = n - 1
= 36 - 1
= 35
Using the Excel Formula:
P-Value = T.DIST(1.2,35,1) = 0.88009
Decision: p-value is greater than significance level; do not reject [tex]\mathbf{H_o}[/tex]
[tex]Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \[/tex] [tex]\mu < 55[/tex]
