In: Statistics and Probability
GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.41 mg of mercury. A sample of 45 bulbs shows a mean of 3.46 mg of mercury.
(a) State the hypotheses for a right-tailed test, using GreenBeam’s claim as the null hypothesis about the mean.
a. H0: μ ≥ 3.41 mg vs. H1: μ < 3.41 mg
b. H0: μ ≤ 3.41 mg vs. H1: μ > 3.41 mg
c. H0: μ = 3.41 mg vs. H1: μ ≠ 3.41 mg
(b) Assuming a known standard deviation of 0.21 mg, calculate the z test statistic to test the manufacturer’s claim. (Round your answer to 2 decimal places.)
Test statistic___
(c) At the 5 percent level of significance (α = 0.05) does the sample exceed the manufacturer’s claim?
No
Yes
(d) Find the p-value. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.)
p-value____
(A)
Here we have given that,
Claim: To check whether the compact fluorescent bulbs average no more than 3.41 mg of mercury.
The null and alternative hypothesis is as follows
v/s
This is the right one-tailed test as our interest is to see the average no more than 3.14 mg of mercury.
Option B is correct.
(B)
We have given that,
n= Number of bulbs=45
= sample mean of mercury=3.46 mg
= population standard deviation =0.21 mg
Here, Population standard deviation is known and we are using the standard normal z test.
Now, we can find the test statistic
=
= 1.60
we get,
the Test statistic is 1.60
(C) and (D)
Now we find the P-value
= level of significance=0.05
This is the right one-tailed test
P-value =(P(Z > z)
=[1- P( Z < 1.60)]
=[ 1 - 0.94520] using standard normal z probability table
=0.0548
we get the P-value is 0.0548
Decision:
P-value (0.0548) > 0.05 ()
Conclusion
we fail to reject Ho (Null Hypothesis)
There is not sufficient evidence to support the claim the compact fluorescent bulbs average no more than 3.41 mg of mercury.
i.e. Yes, The sample exceed the manufacturer's claim.