Respuesta :
Answer:
The 90% confidence interval for population mean is [tex]14.7 < \mu < 19.1[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 16.9[/tex]
The confidence level is [tex]C = 0.90[/tex]
The sample size is [tex]n = 45[/tex]
The standard deviation
Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as
[tex]\alpha = 1-0.90[/tex]
[tex]\alpha = 0.10[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the standardized normal distribution table. The values is [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
The reason we are obtaining critical values for [tex]\frac{\alpha }{2}[/tex] instead of that of [tex]\alpha[/tex] is because [tex]\alpha[/tex] represents the area under the normal curve where the confidence level 1 - [tex]\alpha[/tex] (90%) did not cover which include both the left and right tail while [tex]\frac{\alpha }{2}[/tex] is just considering the area of one tail which is what we required calculate the margin of error
Generally the margin of error is mathematically evaluated as
[tex]MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]MOE = 1.645* \frac{ 9 }{\sqrt{45} }[/tex]
[tex]MOE = 2.207[/tex]
The 90% confidence level interval is mathematically represented as
[tex]\= x - MOE < \mu < \= x + MOE[/tex]
substituting values
[tex]16.9 - 2.207 < \mu < 16.9 + 2.207[/tex]
[tex]16.9 - 2.207 < \mu < 16.9 + 2.207[/tex]
[tex]14.7 < \mu < 19.1[/tex]