Rachel is a lunchroom supervisor at West School. The children eat lunch at 15 long tables. When all tables are used, 240 childre
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Answer:
The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;
The change in the call waiting time is not statistically significant
Step-by-step explanation:
The given call waiting times are;
24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77
19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27
From the data we have;
The mean waiting time before the downsize, = 18.7895
The mean waiting time before the downsize, s₁ = 2.705152
The sample size for the before the downsize, n₁ = 20
The mean waiting time after the downsize, = 19.5125
The mean waiting time after the downsize, s₂ = 3.155945
The sample size for the after the downsize, n₂ = 20
The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20 - 2 = 38
df = 38
At 95% significance level, using a graphing calculator, we have; = ±2.026192
The t-confidence interval is given as follows;
Therefore;
(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)
The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668
By approximation, we have;
The 95% CI = -2.61 < μ₂ - μ₁ < 1.16
Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.