Confidence interval for the population mean: Use of the t distribution
Recorded here are the germination times (in days) for 14 randomly chosen seeds of a new type of bean.

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Assuming that germination times are normally distributed, find a confidence Interval for the mean germination time for all beans of this type. Give the
lower limit and upper limit of the 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.)
Lower limit:
Upper limit:

Ask by Wells Mccarthy. in the United States

Mar 27,2025

Question

Confidence interval for the population mean: Use of the t distribution
Recorded here are the germination times (in days) for 14 randomly chosen seeds of a new type of bean.

Send data to calculator Send data to Excel
Assuming that germination times are normally distributed, find a confidence Interval for the mean germination time for all beans of this type. Give the
lower limit and upper limit of the 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.)
Lower limit:
Upper limit:

Ask by Wells Mccarthy. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

Let the data be  
$$
18,\;17,\;16,\;19,\;21,\;20,\;18,\;15,\;19,\;18,\;11,\;22,\;16,\;16.
$$

1. First, compute the sample mean $$ \bar{x} $$.  
   The sum of the data is  
   $$
   18 + 17 + 16 + 19 + 21 + 20 + 18 + 15 + 19 + 18 + 11 + 22 + 16 + 16 = 246,
   $$  
   and the number of observations is $$ n = 14 $$. Thus,  
   $$
   \bar{x} = \frac{246}{14} \approx 17.571.
   $$

2. Compute the sample standard deviation $$ s $$.  
   First, calculate the squared deviations from the mean. For each observation $$ x_i $$, compute  
   $$
   (x_i - \bar{x})^2.
   $$
   The deviations and their squares (rounded to four decimals) are approximately:
   
   - For $$ x_1=18 $$: $$ (18 - 17.5714)^2 \approx 0.1837 $$
   - For $$ x_2=17 $$: $$ (17 - 17.5714)^2 \approx 0.3265 $$
   - For $$ x_3=16 $$: $$ (16 - 17.5714)^2 \approx 2.4694 $$
   - For $$ x_4=19 $$: $$ (19 - 17.5714)^2 \approx 2.0408 $$
   - For $$ x_5=21 $$: $$ (21 - 17.5714)^2 \approx 11.7551 $$
   - For $$ x_6=20 $$: $$ (20 - 17.5714)^2 \approx 5.8980 $$
   - For $$ x_7=18 $$: $$ (18 - 17.5714)^2 \approx 0.1837 $$
   - For $$ x_8=15 $$: $$ (15 - 17.5714)^2 \approx 6.6122 $$
   - For $$ x_9=19 $$: $$ (19 - 17.5714)^2 \approx 2.0408 $$
   - For $$ x_{10}=18 $$: $$ (18 - 17.5714)^2 \approx 0.1837 $$
   - For $$ x_{11}=11 $$: $$ (11 - 17.5714)^2 \approx 43.1837 $$
   - For $$ x_{12}=22 $$: $$ (22 - 17.5714)^2 \approx 19.5918 $$
   - For $$ x_{13}=16 $$: $$ (16 - 17.5714)^2 \approx 2.4694 $$
   - For $$ x_{14}=16 $$: $$ (16 - 17.5714)^2 \approx 2.4694 $$
   
   Summing these squared deviations gives  
   $$
   \text{SS} \approx 99.4082.
   $$
   
   The sample variance is  
   $$
   s^2 = \frac{\text{SS}}{n-1} = \frac{99.4082}{13} \approx 7.6460,
   $$  
   and hence the sample standard deviation is  
   $$
   s \approx \sqrt{7.6460} \approx 2.766.
   $$

3. For a $$ 99\% $$ confidence interval, the confidence level corresponds to a two-tailed significance level $$ \alpha = 0.01 $$, so $$ \alpha/2 = 0.005 $$. With $$ n - 1 = 13 $$ degrees of freedom, the $$ t $$-critical value is approximately  
   $$
   t_{0.005,13} \approx 3.012.
   $$

4. The margin of error (ME) is calculated as  
   $$
   \text{ME} = t_{0.005,13} \cdot \frac{s}{\sqrt{n}} \approx 3.012 \cdot \frac{2.766}{\sqrt{14}}.
   $$
   Since  
   $$
   \sqrt{14} \approx 3.742,
   $$  
   we have  
   $$
   \text{ME} \approx 3.012 \times \frac{2.766}{3.742} \approx 3.012 \times 0.739 \approx 2.225.
   $$

5. The $$ 99\% $$ confidence interval for the mean is then given by  
   $$
   \bar{x} \pm \text{ME} \approx 17.571 \pm 2.225.
   $$
   This gives:
   - Lower limit:  
     $$
     17.571 - 2.225 \approx 15.346,
     $$
   - Upper limit:  
     $$
     17.571 + 2.225 \approx 19.796.
     $$

Rounded to one decimal place, the $$ 99\% $$ confidence interval is:
- Lower limit: $$ 15.3 $$
- Upper limit: $$ 19.8 $$

Thus, the $$ 99\% $$ confidence interval for the mean germination time is  
$$
\left(15.3,\; 19.8\right) \text{ days.}
$$
The 99% confidence interval for the mean germination time is (15.3, 19.8) days.

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