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A statistics pratitioner took a random sample of 30 observations from a population whose standard deviation is 15 and computed the sample mean to be 50. The 90% confidence inverval for the mean will be ____________ and ______________.

Answer :

Answer:

[tex]50-1.64\frac{15}{\sqrt{30}}=45.509[/tex]    

[tex]50+1.64\frac{15}{\sqrt{30}}=54.491[/tex]    

The 90% confidence inverval for the mean will be 45.509 and 54.491

Step-by-step explanation:

Information given

[tex]\bar X= 50[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma= 15[/tex] represent the sample standard deviation

n= 30 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], the critical value would be [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]50-1.64\frac{15}{\sqrt{30}}=45.509[/tex]    

[tex]50+1.64\frac{15}{\sqrt{30}}=54.491[/tex]    

The 90% confidence inverval for the mean will be 45.509 and 54.491

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