1)
a) H0: μ≥6 H1: μ<6
b) Making a Type I error if the manager believe that the new system will reduce waiting times from the current 9 to 10 minutes to less than 6 minutes. However, the actual waiting time is still between 9 to 10 minutes.
Making a Type II error if the manager failed to realize that the new system would not reduce waiting times from the current 9 to 10 minutes to less than 6 minutes.
2)
H0: μ≤ 100
H1: μ> 100
α=1%=0.01 n= 40 x= 105.7 (calculate from the excel data) σ=16 Z0.01= 2.325
Z test statistics = 105.7-10016/40
=2.253
<Z0.01= 2.325
Therefore, we cannot conclude that at 1% significance, the site is acceptable based on the data provided
T test statistics: 105.75.6875-61.51308804/12 = - 0.31250.436790983 = -0.715445318 Because t=-0.71545> - 1.796 Therefore, at 5% of significance level, we cannot conclude that the mean time of delivery is less than 6 hours in advertisement.
4)
The below is our assumption made:
Null Hypothesis H0: p = .2155,
Alternative Hypothesis H1: p ≠ .2155
By using the P-value Approach:
.
Given: Sample Mean: 0.2442, Hypothesis Mean: 0.2155, Sample Size: 1040
=
= 2.25
P-value = 2P(Z > 2.25) = 2(1– .9878) =.0244.
There is enough evidence to prove that the proportion of 4-4-3-2 hands differs from the theoretical probability. The reason of the differences is not enough shuffling
5)
The hypothesis we assumed:
a. The two variables (predicted change & actual change), both are independent The two variables are not independent (dependent)