A random sample of 28 fields of spring wheat has a mean yield of 44.7 bushels per acre and standard deviation of 6.96 bushels per acre. Determine the 95% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places

Answer :

Answer:

[tex]44.7-2.052\frac{6.96}{\sqrt{28}}=42.001[/tex]    

[tex]44.7+2.052\frac{6.96}{\sqrt{28}}=47.399[/tex]    

And the confidence interval would be between 42.001 and 47.399

Step-by-step explanation:

Information given

[tex]\bar X=44.7[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean

s=6.96 represent the sample standard deviation

n=28 represent the sample size  

Confidence interval

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

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

The degrees of freedom are given by:

[tex]df=n-1=28-1=27[/tex]

The Confidence interval is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex],the critical value for this case would be [tex]t_{\alpha/2}=2.052[/tex]

And replacing we got:

[tex]44.7-2.052\frac{6.96}{\sqrt{28}}=42.001[/tex]    

[tex]44.7+2.052\frac{6.96}{\sqrt{28}}=47.399[/tex]    

And the confidence interval would be between 42.001 and 47.399

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