Answer:
a
The point estimate of the population mean is [tex]\= x = 56[/tex]
b
The 80% confidence level is [tex]50.57 < \mu < 61.43[/tex]
c
There is 80% confidence that the true population mean lies within the confidence interval.
Step-by-step explanation:
From the question we are told that
The sample size is n = 18
The standard deviation is [tex]\sigma = 18 \ L[/tex]
The sample mean is [tex]\= x = 56[/tex]
Generally the point estimate of the population mean is equivalent to the sample mean whose value is [tex]\= x = 56[/tex]
Given that the confidence interval is 80% then the level of significance is mathematically represented as
[tex]\alpha = 100 - 80[/tex]
[tex]\alpha = 20 \%[/tex]
[tex]\alpha = 0.20[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table
The value is [tex]Z_{\frac{ \alpha }{2} } = 1.28[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.28 * \frac{18 }{\sqrt{18} }[/tex]
=> [tex]E = 5.43[/tex]
Generally the 80% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
=> [tex]56 - 5.43 < \mu < 56 + 5.43[/tex]
=> [tex]50.57 < \mu < 61.43[/tex]
The interpretation is that there is 80% confidence that the true population mean lies within the limit
