Recorded here are the germination times (in days) for 15 randomly chosen seeds of a new type of bean.
\[
14,10,15,19,21,22,19,14,18,11,21,15,13,18,11
\]
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Assuming that germination times are normally distributed, find a $90 \%$ confidence interval for the mean germination time for all beans of this type. Give the lower limit and upper limit of the $90 \%$ confidence interval.

Carry your intermediate computations to at least three decimal places, Round your answers to one decimal place. (If necessary, consult a list of formulas.)
\begin{tabular}{ll}
\hline Lower limit: [] \\
Upper limit:
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$-\times$ \\
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\end{tabular}

Step 1 ：First, we need to calculate the mean and standard deviation of the given data. The data is [14, 10, 15, 19, 21, 22, 19, 14, 18, 11, 21, 15, 13, 18, 11].

Step 2 ：The mean of the data is calculated as \( \frac{14+10+15+19+21+22+19+14+18+11+21+15+13+18+11}{15} = 16.07 \).

Step 3 ：The standard deviation of the data is calculated as \( \sqrt{\frac{(14-16.07)^2+(10-16.07)^2+(15-16.07)^2+(19-16.07)^2+(21-16.07)^2+(22-16.07)^2+(19-16.07)^2+(14-16.07)^2+(18-16.07)^2+(11-16.07)^2+(21-16.07)^2+(15-16.07)^2+(13-16.07)^2+(18-16.07)^2+(11-16.07)^2}{14}} = 3.94 \).

Step 4 ：We also need to calculate the standard error, which is the standard deviation divided by the square root of the sample size. The standard error is calculated as \( \frac{3.94}{\sqrt{15}} = 1.02 \).

Step 5 ：Then, we can calculate the confidence interval using the formula \( \text{mean} \pm \text{z-score} \times \text{standard error} \). The z-score for a 90% confidence interval is 1.645.

Step 6 ：The lower limit of the confidence interval is calculated as \( 16.07 - 1.645 \times 1.02 = 14.4 \).

Step 7 ：The upper limit of the confidence interval is calculated as \( 16.07 + 1.645 \times 1.02 = 17.7 \).

Step 8 ：The final answer is the lower limit and the upper limit of the 90% confidence interval for the mean germination time for all beans of this type. So, the 90% confidence interval is \(\boxed{[14.4, 17.7]}\).