9.13 GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury.

(a) Write the hypotheses for a right tailed test, using GreenBeams's claim as the null hypothesis about the mean.

(b) Assuming a known standard deviation of 0.18 mg, calculate the z test statistic to test the manufacturer's claim.

(c) At the 1 percent level of significance ($\alpha = .01$) does the sample exceed the manufacturer's claim?

(d) Find the p-value.

#### Solution

Given that $n=25$, $\overline{x} =3.59$ mg, $\sigma=0.18$ mg.

(a) The null hypothesis is $H_0 : \mu \leq 3.50$ mg against $H_a : \mu > 3.50$ mg.

(b) The test statistic is

` $$ \begin{aligned} Z &= \frac{\overline{x}-\mu_0}{\sigma / \sqrt{n}}\\ &= \frac{3.59-3.5}{0.18 / \sqrt{25}}\\ & = 2.5. \end{aligned} $$ `

(c) The critical value of $Z$ at $0.01$ level of significane is $Z_{crit}=2.33$.

Since $2.5 >2.33$ we reject $H_0$ at $0.01$ level of significance.

The manufacturer's claim that its compact fluorescent bulbs average no more than 3.5 mg in mercury is not supported by evidence. Instead we conclude that $H_a$ is true at $0.01$ level of significance.

(d) The $p$-value is $P(Z>2.5) = 0.006$.

The $p$-value is less than $0.01$ level of significance, we reject $H_0$ at $0.01$ level of significance.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators