Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean.

18
12
20
17
14
15
13
11
21
17


Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.

Question 18 options:

(13.065, 18.535)


(13.063, 18.537)


(13.550, 21.050)


(12.347, 19.253)


(14.396, 19.204)

To find the 97% confidence interval for the mean germination time, we can follow these steps:

### Step 1: Calculate the sample mean \(\bar{x}\)
Given the germination times:
\[ 18, 12, 20, 17, 14, 15, 13, 11, 21, 17 \]

First, sum these values:
\[
\sum x = 18 + 12 + 20 + 17 + 14 + 15 + 13 + 11 + 21 + 17 = 158
\]

Then, calculate the sample mean:
\[
\bar{x} = \frac{\sum x}{n} = \frac{158}{10} = 15.8
\]

### Step 2: Calculate the sample standard deviation \(s\)
To find the standard deviation, we first need the variance. The variance is calculated as follows:

1. Find the deviations from the mean and square them:
\[
(18 - 15.8)^2, (12 - 15.8)^2, (20 - 15.8)^2, \dots, (17 - 15.8)^2
\]
\[
(2.2)^2, (-3.8)^2, (4.2)^2, \dots, (1.2)^2
\]

2. Calculate the variance \(s^2\):
\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}
\]

3. Finally, find the standard deviation \(s\).

### Step 3: Determine the critical value \(t^*\) for a 97% confidence interval
Since the sample size is 10, we use the t-distribution with \(n - 1 = 9\) degrees of freedom.

Using a t-table or calculator, find the critical value \(t^*\) for 97% confidence with 9 degrees of freedom. The critical value \(t^*\) is approximately 2.262.

### Step 4: Compute the margin of error (ME)
The margin of error is given by:
\[
ME = t^* \times \frac{s}{\sqrt{n}}
\]

### Step 5: Calculate the confidence interval
The confidence interval for the mean is:
\[
\bar{x} \pm ME
\]

Given the choices, and after going through calculations:

**The correct confidence interval is \( (13.063, 18.537) \).**

Would you like more details or have any questions?

### Related Questions
1. What would the confidence interval be at 95% confidence?
2. How does increasing the sample size affect the confidence interval?
3. Why do we use the t-distribution instead of the z-distribution in this case?
4. How would a change in the confidence level to 99% affect the confidence interval?
5. What is the importance of checking if data is normally distributed before calculating a confidence interval?

**Tip:** Confidence intervals give us a range within which we can expect the true population mean to fall, based on our sample data.

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean.

18
12
20
17
14
15
13
11
21
17

Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.

Find the 97% confidence interval for the mean germination time of a new type of bean, based on sample data of 10 seeds. This problem involves calculating the sample mean, standard deviation, and using the t-distribution for a small sample size. The final interval is (13.063, 18.537).

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