Answer:
Step-by-step explanation:
Given that:
population mean = 150
Sample size = 10
Sample mean = 136
Standard deviation = 28
Significance level = 0.05
The null hypothesis:
[tex]H_o : \mu \ge 150[/tex]
The alternative hypothesis
[tex]H_a : \mu < 150[/tex]
This is a left-tailed test.
The t-test statistics can be computed as:
[tex]t = \dfrac{\overline x - \mu}{ \dfrac {s} {\sqrt{n}} }[/tex]
[tex]t = \dfrac{136 - 150}{ \dfrac {28} {\sqrt{150}} }[/tex]
t = -1.58
The P-value = P(t < -1.58)
Using the t tables
P-value = 0.0743
Decision rule: To reject the null hypothesis if the P-value is lesser than the level of significance.
Conclusion: We fail to reject the null hypothesis; since P-value is greater than (∝). Thus, there is sufficient evidence to support Fred's Claim.
No, we can not reject Fred's claim.
