Answer :
Answer:
The correct option is d
Step-by-step explanation:
From the question we are told that
The population size is [tex]N = 47000[/tex]
The sample size is [tex]n = 200[/tex]
The sample mean is [tex]\= x = 118.0 \ mmHg[/tex]
The standard deviation is [tex]\sigma = 11.0 \ mmHg[/tex]
Given that the confidence level is 95% then the level of significance can be calculated 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 z-table , the value is [tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.05 }{2} } = 1.96[/tex]
The reason we are obtaining critical value of [tex]\frac{\alpha }{2}[/tex] instead of [tex]\alpha[/tex] is because
[tex]\alpha[/tex] represents the area under the normal curve where the confidence level interval ( [tex]1-\alpha[/tex]) did not cover which include both the left and right tail while
[tex]\frac{\alpha }{2}[/tex] is just the area of one tail which what we required to calculate the margin of error .
NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)
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{11.0 }{ \sqrt{200} }[/tex]
[tex]E = 1.5245[/tex]
The 95% confidence level interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]118.0 - 1.5245 < \mu < 118.0 + 1.5245[/tex]
[tex]116.5< \mu < 119.5[/tex]
[tex][116.5 , 119.5][/tex]