Question
In: Statistics and Probability
The following data comparing wait times at two rides at Disney are listed below: Position Pirates...
The following data comparing wait times at two rides at Disney are listed below:
|
Position |
Pirates |
Splash Mountain |
|
Sample Size |
32 |
30 |
|
Average Wait Time (In Minutes) |
14.68 |
18.77 |
|
Population Standard Deviation |
11.87 |
16.79 |
- What is the 98% confidence interval for the difference in wait times between pirates and splash mountain?
- What is the test statistic for testing to see if there is a significant difference in wait times between pirates and splash mountain?
Solutions
Expert Solution
a)
Given CI level is 0.98, hence α = 1 - 0.98 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.33
Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(140.8969/32 + 281.9041/30)
sp = 3.7148
Margin of Error
ME = zc * sp
ME = 2.33 * 3.7148
ME = 8.655
CI = (x1bar - x2bar -zc * sp , x1bar - x2bar +zc * sp)
CI = (14.68 - 18.77 - 2.33 * 3.7148 , 14.68 - 18.77 - 2.33 *
3.7148
CI = (-12.7455 , 4.5655)
b)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 ≠ μ2
Rejection Region
This is two tailed test, for α = 0.02
Critical value of z are -2.326 and 2.326.
Hence reject H0 if z < -2.326 or z > 2.326
Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(140.8969/32 + 281.9041/30)
sp = 3.7148
Test statistic,
z = (x1bar - x2bar)/sp
z = (14.68 - 18.77)/3.7148
z = -1.10
P-value Approach
P-value = 0.2713
As P-value >= 0.02, fail to reject null hypothesis.
orchestra answered 2 years agoRelated Solutions
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