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:
We conclude that her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company states.
Step-by-step explanation:
We are given;
Population mean; μ = 30 minutes
Population standard deviation; σ = 3 minutes
Sample size; n = 10
Sample mean; x¯ = 25
Significance level = 5% = 0.05
Let's define the hypotheses;
Null hypothesis; H0: μ = 30
Alternative hypothesis: Ha: μ ≠ 30
Let's find the test statistic;
z = (x¯ - μ)/(σ/√n)
z = (25 - 30)/(3/√10)
z = -5.27
From online p-value from z-score calculator attached using, z = -5.27, significance level = 0.05, two tailed hypothesis, we have;
p < 0.00001
This is less than the significance level, and so we will reject the null hypothesis and conclude that Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.
