A random sample of 10 fields of durum wheat has a mean yield of 39.8 bushels per acre and standard deviation of 3.87 bushels per acre. Determine the $$ 90 \% $$ 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.

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The first step is to note that since the population standard deviation is unknown and the sample size is small ($$ n = 10 $$), we use the $$ t $$-distribution. The degrees of freedom are

$$
df = n - 1 = 10 - 1 = 9.
$$

For a $$ 90\% $$ confidence interval, the area in each tail is

$$
\alpha/2 = \frac{1 - 0.90}{2} = 0.05.
$$

Thus, we need to find the critical value $$ t_{0.05,9} $$.

Using a $$ t $$-table or calculator, we find:

$$
t_{0.05,9} \approx 1.833.
$$

So, the critical value is approximately $$ 1.833 $$.
The critical value for a 90% confidence interval with 9 degrees of freedom is approximately 1.833.

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