You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:
29.1 |
58.6 |
87.6 |
45.5 |
22.1 |
57.8 |
46.7 |
47.8 |
47.8 |
24.3 |
63.1 |
Find the 98% confidence interval. Enter your answer as an
open-interval (i.e., parentheses)
accurate to two decimal places (because the sample data are
reported accurate to one decimal place).
98% C.I. =
Answer should be obtained without any preliminary rounding.
The data is provided as:
29.1 | |
58.6 | |
87.6 | |
45.5 | |
22.1 | |
57.8 | |
46.7 | |
47.8 | |
47.8 | |
24.3 | |
63.1 | |
Count | 11 |
Mean | 48.2182 |
St. Dev | 18.9783 |
The number of degrees of freedom are df=11−1=10, and the significance level is α=0.02.
Based on the provided information, the critical t-value for α=0.02 and df=10 degrees of freedom is tc=2.764.
The 98% confidence for the population mean μ is computed using the following expression
Therefore, based on the information provided, the 98 % confidence for the population mean μ is
Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!
Get Answers For Free
Most questions answered within 1 hours.