Unformatted text preview: the value of the test statistic z.
2. If z is in the upper tail (z > 0), find the area under the standard normal curve to the right of z. If z is in the lower tail (z < 0), find the area under the standard normal curve to the left of z.
3. Double the tail area obtained in step 2 to obtain the p –value. The rejection rule: Reject H0 if the pvalue < α . Critical Value Approach to Critical Value Approach to TwoTailed Hypothesis Testing
The critical values will occur in both the lower and
upper tails of the standard normal curve. Use the standard normal probability distribution table to find zα/2 (the zvalue with an area of α /2 in the upper tail of the distribution).
s The rejection rule is: Reject H0 if z < zα/2 or z > zα/2. Example: Glow Toothpaste
• TwoTailed Test About a Population Mean: σ Known The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
will be selected in order to check the
Glow Quality assurance procedures call for
the continuation of the filling process if the
sample results are consistent with the assumption that
the mean filling weight for the population of toothpaste
tubes is 6 oz.; otherwise the process will be adjusted. Example: Glow Toothpaste
Example: Glow Toothpaste
s TwoTailed Test About a Population Mean: σ Known Assume that a sample of 30 toothpaste
tubes provides a sample mean of 6.1 oz.
The population standard deviation is believed to be 0.2 oz. Perform a hypothesis test, at the .03
level of significance, to help determine
whether the filling process should continue
operating or be stopped and corrected. oz. Glow TwoTailed Tests About a Population Mean:
σ Known p –Value and Critical Value Approaches
1. Determine the hypotheses.
1. Determine the hypotheses. H0: µ = 6
Ha: µ ≠ 6 2. Specify the level of significance. α = .03 3. Compute the value of the test statistic. x − µ0
6.1 − 6
σ / n .2 / 30 Glow TwoTailed Tests About a Population Mean:
σ Known p –Value Approach
4. Compute the p –value.
For z = 2.74, cumulative probability = .9969
p–value = 2(1 − .9969) = .0062
5. Determine whether to reject H0.
Because p–value = .0062 < α = .03, we reject H0.
There is sufficient statistical evidence to
infer that the alternative hypothesis is true (i.e. the mean filling weight is not 6 ounces). Glow TwoTailed Tests About a Population Mean:
σ Known Glow pValue Approach 1/2
= .0031 1/2
= .0031 α /2 = .015 α /2 = .015
z = 2.74 z zα/2 = 2.17 0 zα/2 = 2.17 z = 2.74 TwoTailed Tests About a Population Mean:
s Excel Formula Worksheet
Sample Size =COUNT(A2:A31)
Sample Mean =AVERAGE(A2:A31)
Population Std. Dev. 0.2
Hypothesized Value 6
Standard Error =C4/SQRT(C1)
Test Statistic z =(C2-C5)/C7
p -Value (lower tail) =NORMSDIST(C8)
p -Value (upper tail) =1-C10
p -Value (two tail) =2*(MIN(C10,C11)) Note: Rows...
View Full Document
This note was uploaded on 08/28/2011 for the course BUS 300 taught by Professor White during the Spring '09 term at Rutgers.
- Spring '09