Respuesta :
Answer:
Given : Sample mean = x = 5.25
Standard deviation = 0.52
To Find : Use a 0.01 significance level and the given calculator display to test the claim that the sample is from a population with a mean less than 5.4
Solution:
Sample size = n = 58
Hypothesis : [tex]H_0:\mu = 5.4\\H_a:\mu<5.4[/tex]
Test statistic:[tex]z =\frac {x-\mu}{\frac {s}{\sqrt{n}}}[/tex]
x = 5.25
μ = 5.4
s = 0.52
n = 58
Substitute the values :
[tex]z =\frac{5.25-5.4}{\frac{0.52}{\sqrt{58}}}[/tex]
[tex]z =−2.196[/tex]
Refer the p value using z table
P-value = P(z ≤ -2.196) = 0.0143
As P-value > 0.01,
So, we accept the null hypothesis
Conclusion: There is significant statistical evidence to conclude that the claim that the sample is from a population with a mean less than 5.4 is False at alpha = 0.01, the level of significance.
