# Isom2500-Krisssassignment3

Submitted By AYChester
Words 553
Pages 3
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

3)
H0: μ≥6
H1: μ<6 α=5%=0.05 x= 5.6875
S= 1.51308804/12
d.f : 12-1 =11

For α=0.05 with d.f = 11
-tα= -t0.05= -1.796

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)

Cell (i)
1 65 129(104)/216 = 62.11 2.89 .13
2 39 87(104)/216 = 41.89 -2.89 .20
3 64 129(112)/216 =66.89...