Question

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 1

Answer:

$44.7-2.052\frac{6.96}{\sqrt{28}}=42.001$    

$44.7+2.052\frac{6.96}{\sqrt{28}}=47.399$    

And the confidence interval would be between 42.001 and 47.399

Step-by-step explanation:

Information given

$\bar X=44.7$ represent the sample mean for the sample  

$\mu$ 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:

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

The degrees of freedom are given by:

$df=n-1=28-1=27$

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

And replacing we got:

$44.7-2.052\frac{6.96}{\sqrt{28}}=42.001$    

$44.7+2.052\frac{6.96}{\sqrt{28}}=47.399$    

And the confidence interval would be between 42.001 and 47.399