# An agricultural researcher is interested in estimating the mean length of the growing season in a region. Treating the last 10 years as a simple random sample, he obtains the following data, which represent the number of days of the growing season.

153 164 147 141 172 183 195 178 163 151

A) Construct a 95% confidence interval for the mean lenght of the growing season in the region. _______

(use asending order. round to two decimal places as needed)

## Here the population variance is not known, so use student-t distribution.

A 95% (two-tail) confidence interval would include the limits.

The value 1.96 comes from a T-value for 97.5% (half tail).

x̄-Ts ≤ μ ≤ x̄+Ts

where x̄=sample mean

s=sample variance.

T=value from student-t distribution for 97.5% and ν=n-1=9 (degrees of freedom)

## To construct a confidence interval for the mean length of the growing season in the region, we can use the sample data provided. Here are the steps to calculate the confidence interval:

1. Calculate the mean (x̄) of the sample data. Add up all the values and divide by the total number of observations (in this case, 10).

x̄ = (153 + 164 + 147 + 141 + 172 + 183 + 195 + 178 + 163 + 151) / 10 = 164.7

2. Calculate the standard deviation (s) of the sample data. This measures the spread or variability of the data.

To calculate the standard deviation, we need to find the deviations from the mean for each observation, square them, sum them up, divide by (n-1), and then take the square root.

Deviations from the mean:

(153 - 164.7) = -11.7

(164 - 164.7) = -0.7

(147 - 164.7) = -17.7

(141 - 164.7) = -23.7

(172 - 164.7) = 7.3

(183 - 164.7) = 18.3

(195 - 164.7) = 30.3

(178 - 164.7) = 13.3

(163 - 164.7) = -1.7

(151 - 164.7) = -13.7

Squared deviations:

(-11.7)^2 = 136.89

(-0.7)^2 = 0.49

(-17.7)^2 = 312.9

(-23.7)^2 = 561.69

(7.3)^2 = 53.29

(18.3)^2 = 334.89

(30.3)^2 = 918.09

(13.3)^2 = 176.89

(-1.7)^2 = 2.89

(-13.7)^2 = 187.69

Sum of squared deviations:

136.89 + 0.49 + 312.9 + 561.69 + 53.29 + 334.89 + 918.09 + 176.89 + 2.89 + 187.69 = 2796.31

Variance (s^2):

s^2 = 2796.31 / (n-1) = 2796.31 / 9 = 310.7011

Standard deviation (s):

s = √(s^2) = √310.7011 = 17.6173

3. Calculate the standard error (SE), which is the standard deviation divided by the square root of the sample size (n).

SE = s / √n = 17.6173 / √10 = 5.5705

4. Determine the critical value based on the desired confidence level. For a 95% confidence level, the critical value can be found using a t-distribution table with n-1 degrees of freedom (in this case, 9) or using statistical software. The critical value for a 95% confidence level and 9 degrees of freedom is approximately 2.262.

5. Calculate the margin of error (ME), which is the critical value multiplied by the standard error.

ME = critical value * SE = 2.262 * 5.5705 = 12.5984

6. Construct the confidence interval by subtracting and adding the margin of error to the sample mean.

Confidence Interval = x̄ ± ME

Lower bound = x̄ - ME

Upper bound = x̄ + ME

Lower bound = 164.7 - 12.5984 = 152.1016

Upper bound = 164.7 + 12.5984 = 177.2984

Therefore, the 95% confidence interval for the mean length of the growing season in the region is (152.10, 177.30) (rounded to two decimal places in ascending order).

## To construct a confidence interval for the mean length of the growing season in the region, we can use the formula:

Confidence Interval = x̄ ± (z * (σ/√n))

Where:

- x̄ is the sample mean

- z is the z-score corresponding to the desired confidence level

- σ is the standard deviation of the population (if unknown, we can use the sample standard deviation)

- n is the sample size

Given data: 153, 164, 147, 141, 172, 183, 195, 178, 163, 151

Step 1: Calculate the sample mean (x̄):

x̄ = (153 + 164 + 147 + 141 + 172 + 183 + 195 + 178 + 163 + 151) / 10

x̄ = 163.7

Step 2: Calculate the sample standard deviation (s):

To calculate the sample standard deviation, we can use the formula:

s = √(∑(x - x̄)² / (n-1))

First, calculate the sum of squared differences from the mean:

(153-163.7)² + (164-163.7)² + (147-163.7)² + (141-163.7)² + (172-163.7)²

+ (183-163.7)² + (195-163.7)² + (178-163.7)² + (163-163.7)² + (151-163.7)²

Simplifying the above expression:

(-10.7)² + (0.3)² + (-16.7)² + (-22.7)² + (8.3)² + (19.3)² + (31.3)² + (14.3)² + (-0.7)² + (-12.7)²

Calculating the sum:

115.0 + 0.09 + 278.9 + 514.29 + 68.89 + 372.49 + 979.69 + 204.49 + 0.49 + 161.29 = 2694.72

Then divide by (n-1):

s = √(2694.72 / 9)

s ≈ √299.41

s ≈ 17.29

Step 3: Determine the z-score corresponding to the desired confidence level.

For a 95% confidence level, the z-score would be 1.96 (obtained from the standard normal distribution table).

Step 4: Calculate the confidence interval:

Confidence Interval = x̄ ± (z * (σ/√n))

Confidence Interval = 163.7 ± (1.96 * (17.29/√10))

Calculating the square root of the sample size:

√10 ≈ 3.16

Calculating the confidence interval:

Lower Limit = 163.7 - (1.96 * (17.29/3.16))

Lower Limit ≈ 163.7 - 10.72

Lower Limit ≈ 153.98

Upper Limit = 163.7 + (1.96 * (17.29/3.16))

Upper Limit ≈ 163.7 + 10.72

Upper Limit ≈ 174.42

Therefore, the 95% confidence interval for the mean length of the growing season in the region is approximately 153.98 to 174.42 days.