A study was performed on green sea turtles inhabiting a certain area. Time-depth recorders were deployed on 6 of the 76 captured turtles. The time depth recorders allowed the scientists to track the movement of turtles in the area. The 6 turtles had a mean shel length of 51.2 and std dev of 6.7..
what is the true mean shell length of all green sea turtles in the area? Use the info on the 6 tracked turtles to estimate with 99% confidence.
also this is provided:
33.98
30.43
32.51
31.37
36.56
35.36
36.08
35.47
35.84
39.61
44.17
42.64
42.27
42.63
49.76
46.02
48.52
47.94
45.65
48.96
49.56
49.78
54.16
51.97
50.91
54.34
52.51
53.07
54.21
50.34
53.53
51.13
54.33
54.74
55.02
57.57
56.16
55.55
58.24
57.78
57.52
56.43
56.79
57.69
58.99
65.06
61.99
60.18
62.49
63.74
60.75
64.68
60.37
62.72
63.12
63.17
64.75
60.31
64.87
60.05
68.81
64.98
66.05
65.06
68.41
65.03
67.57
68.42
67.29
65.86
70.36
70.95
70.44
71.96
73.99
81.45
What is the list of numbers measuring? Depth?
For shell length,
99% = sample mean ± Z (SEm)
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.005) and its Z score.
To estimate the true mean shell length of all green sea turtles in the area with 99% confidence, we can use a confidence interval.
The formula for the confidence interval is:
Confidence Interval = sample mean ± (critical value * standard deviation / square root of sample size)
Given that we have a sample of 6 turtles with a mean shell length of 51.2 and a standard deviation of 6.7, we need to determine the critical value for a 99% confidence level.
The critical value can be found using a z-table or a statistical software. For a 99% confidence level, the critical value is approximately 2.576.
Now let's calculate the confidence interval:
Confidence Interval = 51.2 ± (2.576 * 6.7 / √6)
Calculating the values within the parentheses first:
(2.576 * 6.7) / √6 ≈ 8.573
Now substitute that value into the confidence interval formula:
Confidence Interval = 51.2 ± 8.573
Calculating the upper and lower limits:
Upper limit = 51.2 + 8.573 = 59.773
Lower limit = 51.2 - 8.573 = 42.627
Therefore, with 99% confidence, the true mean shell length of all green sea turtles in the area is estimated to be between 42.627 and 59.773.
To estimate the true mean shell length of all green sea turtles in the area using the information from the 6 tracked turtles, we can calculate a confidence interval. Since we have a small sample size (n < 30) and know the standard deviation of the population, we can use a t-distribution.
Here are the steps to calculate the confidence interval:
1. Determine the sample mean (x̄) and the sample standard deviation (s).
- In this case, the sample mean is 51.2 and the sample standard deviation is 6.7.
2. Determine the sample size (n).
- The sample size is 6.
3. Determine the desired confidence level.
- In this case, the confidence level is 99%, so the alpha level (α) is 1 - 0.99 = 0.01.
4. Find the t-value for a two-tailed test with the desired confidence level and degrees of freedom (df = n - 1).
- Since n = 6, df = 6 - 1 = 5.
- Using a t-distribution table or a statistical calculator, the t-value for a 99% confidence level with 5 degrees of freedom is approximately 4.032.
5. Calculate the standard error (SE) using the formula:
- SE = s / √n
- In this case, SE = 6.7 / √6 ≈ 2.734.
6. Calculate the margin of error (ME) using the formula:
- ME = t-value * SE
- In this case, ME = 4.032 * 2.734 ≈ 10.99.
7. Calculate the lower and upper bounds of the confidence interval.
- Lower bound = x̄ - ME
- Upper bound = x̄ + ME
- In this case, lower bound = 51.2 - 10.99 ≈ 40.21, and upper bound = 51.2 + 10.99 ≈ 62.19.
Therefore, using the information from the 6 tracked turtles, we can estimate with 99% confidence that the true mean shell length of all green sea turtles in the area is between 40.21 and 62.19.