While developing x-bar and R charts, the manager took 10 samples each of size 6. The average of sample means was 3.2. The averag
1 answer:
Answer:
Option a. 1.2 is the correct answer
Step-by-step explanation:
Given the data in the question;
sample size n = 10
sub group = 6
The average of sample means x" = 3.2
Average range value R' = 0.6
the value of UCL for the R-chart = ?
To get the value of UCL for the R-chart, we say;
= D₄ × R'
from table of constants "Tabular values for x-bar and range charts
for size 6, the factor for control limits for range D₄ = 2.004
so we substitute
= 2.004 × 0.6
= 1.2024 ≈ 1.2
Hence, Option a. 1.2 is the correct answer.
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Answer:
1.7 × 10⁻⁴
Step-by-step explanation:
The question relates to a two sample z-test for the comparison between the means of the two samples
The null hypothesis is H₀: μ₁ ≤ μ₂
The alternative hypothesis is Hₐ: μ₁ > μ₂
Where;
= 13.5
= 12
σ₁ = 2.5
σ₂ = 1.5
We set our α level at 0.05
Therefore, our critical z = ± 1.96
For n₁ = n₂ = 23, we have;
We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96
Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983
Therefore;
The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.