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 :
![H_0:\mu = 5.4\\H_a:\mu<5.4](https://img.qammunity.org/2020/formulas/mathematics/college/wd40u9s1mwicdezy5bljwwe4q8e3g3u9xn.png)
Test statistic:
![z =\frac {x-\mu}{\frac {s}{√(n)}}](https://img.qammunity.org/2020/formulas/mathematics/college/g2d3jlpsb8kpemf3t4er517cnm00l9uqxu.png)
x = 5.25
μ = 5.4
s = 0.52
n = 58
Substitute the values :
![z =(5.25-5.4)/((0.52)/(√(58)))](https://img.qammunity.org/2020/formulas/mathematics/college/gk43kityrburxg8zqydaj9f87e23cq4gmb.png)
![z =−2.196](https://img.qammunity.org/2020/formulas/mathematics/college/qsetzd26gu88z688xum53dqclgflajkkbo.png)
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.