Answer :
The student should perform a one-sample t-test to compare the average number of weather delays at KXYZ Airport to the national average. The null hypothesis states no difference in the number of delays, while the alternative hypothesis states a significant difference exists. Calculations for the mean, standard deviation, degrees of freedom, and t statistic will depend on specific sample data provided by the student.
To address the research question, the student should use a one-sample t-test because it's designed to determine if a sample mean significantly differs from a known population mean and the population standard deviation is known.
The null hypothesis (H-o) is that there is no significant difference in the number of weather delays at KXYZ Airport compared to the national average, which is 59 flights per day.
The alternative hypothesis (Ha) is that there is a significant difference in the number of weather delays at KXYZ Airport compared to the national average.
The mean (\\(ar{x}\\)) and standard deviation (s) for the population are given as 59 flights and 34.83, respectively.
The mean and standard deviation for the sample needs to be calculated based on the sample data provided by the student.
Degrees of freedom (df) can be computed as n-1, where n is the sample size. For example, if the sample size is 30, then df = 30 - 1 = 29.
Cohen's d effect size can be calculated if the sample mean and standard deviation are known.
The test value would be the national average of delays, which is 59.
The t statistic is calculated using the sample mean, the population mean (test value), the sample standard deviation, and the square root of the sample size.
Answer:
a.) one sample t test
b.) H0 : μ = 59.3
c.) H1 : μ > 59.3
d.) μ = 59.3 ; σ = 39.84
e.) xbar = 79.4 ; s = 61.36
Test statistic = 3.16
Step-by-step explanation:
Given the sample data:
49.00 49.00 49.00 49.00 49.00 63.00 63.00 63.00 63.00 63.00 199.00 199.00 199.00 199.00 199.00 38.00 38.00 38.00 38.00 38.00 48.00 48.00 48.00 48.00 48.00 49.00 63.00 199.00 38.00 48.00
Sample size, n = 30
Using calculator :
xbar from the data above = 79.4
Standard deviation = 61.359
H0 : μ = 59.3
H1 : μ > 59.3
Test statistic :
(Xbar - μ) ÷ (σ/sqrt(n)
σ = 34.83
(79.4 - 59.3) ÷ (34.83/sqrt(30))
20.1 ÷ 6.359
Test statistic = 3.16