Complete Question
The complete question is shown on the first uploaded image
Answer:
The 95% confidence interval is [tex]49.85 < \mu < 54.15[/tex]
This means that there is 95% chance that the true population mean is within this interval
Step-by-step explanation:
From the question we are told that
The sample size is n = 30
The sample mean is [tex]\= x = 52[/tex]
The population standard deviation is [tex]\sigma = 6[/tex]
Given that the confidence level is 95% then the level of confidence is evaluate as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table , the values is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]E = 1.96 * \frac{ 6 }{\sqrt{30} }[/tex]
[tex]E = 2.147[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]52 - 2.147 < \mu < 52 + 2.147[/tex]
[tex]49.85 < \mu < 54.15[/tex]
