Two hundred fish caught in Cayuga Lake had a mean length of 14.1 inches. The population standard deviation is 2.9 inches. (Give your answer correct to two decimal places.)
(a) Find the 90% confidence interval for the population mean length.
Lower Limit
Upper Limit
(b) Find the 98% confidence interval for the population mean length.
Lower Limit
Upper Limit
a) 90% = mean ± 1.645 SEm
SEm = SD/√n
b) 98% = mean ± 2.33 SEm
I have honestly tried to work this out several times and I just do not see where they are getting the information to use in the formula. Can someone help me.
To find the confidence intervals for the population mean length, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard deviation / √(sample size))
The critical value is determined based on the confidence level, which represents the level of confidence we have in our estimate.
(a) For a 90% confidence interval, the critical value can be found using a t-distribution table or a statistical software. Since we have a large sample size (n = 200), we can use the Z-distribution instead of the t-distribution.
The critical value for a 90% confidence level can be found using the following formula:
Critical value = Z(α/2), where α = 1 - confidence level
For a 90% confidence level, α = 1 - 0.90 = 0.10
The critical value for α/2 = 0.10/2 = 0.05 is 1.645 (from a standard normal distribution table).
Substituting the values into the confidence interval formula:
Confidence Interval = 14.1 ± (1.645 * 2.9 / √200)
Calculating the values:
Confidence Interval = 14.1 ± (1.645 * 2.9 / 14.14) ≈ 13.95, 14.25
Therefore, the 90% confidence interval for the population mean length is approximately:
Lower Limit = 13.95 inches
Upper Limit = 14.25 inches
(b) For a 98% confidence interval, we need to find the critical value using the Z-distribution table.
The critical value for a 98% confidence level can be found using the same formula as before:
Critical value = Z(α/2), where α = 1 - confidence level
For a 98% confidence level, α = 1 - 0.98 = 0.02
The critical value for α/2 = 0.02/2 = 0.01 is 2.33 (from a standard normal distribution table).
Substituting the values into the confidence interval formula:
Confidence Interval = 14.1 ± (2.33 * 2.9 / √200)
Calculating the values:
Confidence Interval = 14.1 ± (2.33 * 2.9 / 14.14) ≈ 13.79, 14.41
Therefore, the 98% confidence interval for the population mean length is approximately:
Lower Limit = 13.79 inches
Upper Limit = 14.41 inches