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

`44.7-2.052(6.96)/(√(28))=42.001`

`44.7+2.052(6.96)/(√(28))=47.399`

And the confidence interval would be between 42.001 and 47.399

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)(s)/(√(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(6.96)/(√(28))=42.001`

`44.7+2.052(6.96)/(√(28))=47.399`

And the confidence interval would be between 42.001 and 47.399