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P- VALUE APPROACH

The p-value is the probability of obtaining a test statistic at least as extreme as the test statistic we calculated from the sample.
The p-value is also known as the observed significance level.
It adds a degree of significance to the result of the hypothesis.
We can now determine how strongly we “reject” or “fail to reject” the null hypothesis.
If p-value < α, we reject the null hypothesis. If the p-value ≥ α, we fail to reject the null hypothesis.
The farther the p-value is from α, the stronger the decision.

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Example

Company packages salted and unsalted peanuts in 16 – ounce sacks. The company’s filling strives for an average fill amount equal to 16 ounces.

Suppose, z = 3.32 and α = 0.10

then the p- value = 2(0.0005) = 0.0010.

Decision Rule:

Since the p-value = 0.0010 < α = 0.010, we reject H₀

and conclude that the average filling amount is not equal to 16 ounces

Note:

When the alternative hypothesis has the not equal to sign then it is a two tailed test.
Example

H₀ : µ ≤ 16

H₁ : µ ≠ 16

HYPOTHESIS TEST FOR µ , σ

Decision Rule

When the t calculated is less than the α level we fail to reject H₀ and conclude using the null hypothesis.
When the t calculated is greater than the α level we reject H₀ and conclude using the alternative hypothesis

Controlling α and β

Hypothesis testing is subject to two potential errors α and β these can have adverse conditions on hypothesis testing.
Power of Test
The power of a test is the probability that the hypothesis test will correctly reject the null hypothesis when the null hypothesis is false .
The power test for any value is the ability of the test to reject the null hypothesis with certainty.
Its value is between 0 and 1 inclusive.

power = 1 - β

Two Sample Z-Test: Definition, Formula,�

where:
x₁, x₂: sample means
σ₁, σ₂: population standard deviations
n₁, n₂: sample sizes
If the p-value that corresponds to the z test statistic is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Example

Suppose the IQ levels among individuals in two different cities are known to be normally distributed each with population standard deviations of 15.A scientist wants to know if the mean IQ level between individuals in city A and city B are different, so she selects a simple random sample of 20 individuals from each city and records their IQ levels.
To test this, she will perform a two sample z-test at significance level α = 0.05.
Suppose she collects two simple random samples with the following information:
x₁ (sample 1 mean IQ) = 100.65
n₁ (sample 1 size) = 20
x₂ (sample 2 mean IQ) = 108.8
n₂ (sample 2 size) = 20
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Examples

The average wait time to see an E.R. doctor is said to be 150 minutes. You think the wait time is actually less. You take a random sample of 30 people and find their average wait is 148 minutes with a standard deviation of 5 minutes. Assume the distribution is normal find the p-value for this test.
A new stockbroker named XYZ, claims that their brokerage fess are lower than that of your current broker ABC. Data available from the independent research firm indicates that the mean and standard deviation of all ABC broker clients are $18 and $6 respectively. A sample of 100 clients of ABC is taken and brokerage charges are calculated with new rates of XYZ broker. If the mean of the sample is $18.75 and the sample standard deviation is $6, can any inference be made about the difference in the average brokerage bill between ABC and XYZ broker?

Examples

Suppose a standard fertilizer has been shown to cause a species of plants to grow by an average of 10 inches. However, one botanist believes a new fertilizer can cause this species of plants to grow by an average of greater than 10 inches. To test this claim, she applies the new fertilizer to a simple random sample of 15 plants and obtains the following information:
n = 15 plants
x = 11.4 inches
s = 2.5 inches

Find the p-value and draw your conclusion

A teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. If a sample of 81 students was selected with a mean score of 90 then check if there is enough evidence to support this claim at a 0.05 significance level.
An online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes. Is there enough evidence to support this claim at a 0.05 significance level if 49 orders were examined with a mean of 100 minutes?