One sample Z test
One sample Z test is used to
test the claim about population mean (µ). Assumptions
for the one-sample Z test are as follows.
- The data should be continuous.
- The observations in the data should be independent.
- The data should come from the normal distribution.
- The population standard deviation is known.
Consider the following example
to understand how does the one-sample Z test works.
Question
Last week, your manager asked
you to analyze staffing needs for the Foreclosure Department. She was so
impressed, and she wants you to create another report for her. Her intention is
to decrease the processing time per document.
Based on last week's report,
the average number of processed documents per hour was 15.11, with a standard
deviation of 2.666. That is, one document was reviewed in 238.25 seconds. To be
objective as much as possible, the manager spoke with an employee whose average
was exactly 15 documents per hour. The employee claimed that if she was given a
larger monitor, the processing time would be shorter.
They conducted an experiment
with a large monitor and measured processing time. After reviewing 20
documents, the calculated average processing time per document was 190.58
seconds. The manager believes that a bigger monitor helped reduce the
processing time for reviewing foreclosure documents. Conduct a hypothesis test
using a 95% confidence level, which means that significance level α = 0.05.
Use the 5-step process, and
explain each term or concept mentioned in each section in the following.
Solution
Step 1: Write down the given information
Mean and standard deviation are given in terms of “number of documents”.
Average number of processed documents per hour (µ0) = 15.11 and
standard deviation (σ) = 2.666.
The sample mean is given in time (seconds). Using calculated
average processing document per document, convert sample mean (X.bar) number of
documents per hour.
Calculated average processing document per document = 190.58
seconds.
Thus, sample mean (X.bar) = 3600/190.58 = 18.89 documents per
hour.
Step 2: Identify the claim and write the null and alternative hypothesis
The employee claimed that if she was given a larger monitor, the processing time would be shorter. As the processing time gets shorter the number of documents per hour would increase.
H0: µ = 15.11
Vs
H1: µ > 15.11
Step 3: To determine which test would be appropriate
If population standard deviation (σ) is known, use the Z test while if population standard deviation (σ) is unknown then use the T-test.
In the above example, the population standard deviation (σ) is
known so the one-sample Z test is more appropriate.
Step 4: Calculate the value of the Z statistic
Where µ0 = population mean under null hypothesis, x.bar = sample mean, σ = population standard deviation, n = size of the sample.
In the above example, X.bar = 18.89 documents per hour, µ0
= 15.11 documents per hour and σ = 2.666 documents per hour, sample size
(n) = 20.
Plugging these values in the above formula we get, Z0 = 6.3408.
Step 5: calculate the p-value
To find the p-value you can use the Z table or technology like excel, Ti-84 calculator on StatCrunch. Identify the tail of the test and calculate the p-value accordingly.
In the above example, the alternative hypothesis contains the
">” sign so the given test is a right tailed test. Therefore, the
p-value is given by; p-value = P(Z > Z0) = P(Z > 6.3408) = 1.14*10-10
= 0 (approximately).
Step 6: Decision rule
If p-value ≤ α then reject the null hypothesis. If p-value > α then fails to reject the null hypothesis.
In the above example,
p-value = 0 < α = 0.05. Therefore, we reject the null hypothesis.
Conclusion
There is sufficient evidence to support the employee’s claim that if she was given a larger monitor, the processing time would be shorter.
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