9.14 The mean potassium content of a popular sports drink is listed as 140 mg in a 32 oz bottle. Analysis of 20 bottles indicates a sample mean of 139.4 mg.

(a) Write the hypotheses for a two-tailed test of the claimed potassium content.

(b) Assuming a known standard deviation of 2.00 mg, calculate the z test statistic to test the manufacturer's claim.

(c) At the 10 percent level of significance ($\alpha = .10$) does the sample contradict the manufacturer's claim?

(d) Find the p-value.

#### Solution

Given that $n=20$, $\overline{x} =139.4$ mg, $\sigma=2.00$ mg.

(a) The null hypothesis is $H_0 : \mu = 140$ mg against $H_a : \mu \neq 140$ mg.

(b) The test statistic is

` $$ \begin{aligned} Z &= \frac{\overline{x}-\mu_0}{\sigma / \sqrt{n}}\\ &= \frac{139.4-140}{2 / \sqrt{20}}\\ & = -1.342. \end{aligned} $$ `

(c) The critical values of $Z$ at $0.1$ level of significane are `$Z_{crit}=-1.64$`

and `$Z_{crit}=1.64$`

.

The critical value based decision criteria is if $|Z| \geq Z_{\alpha/2}$, we reject the null hypothesis, otherwise fail to reject the null hypothesis.

Since $|Z| < Z_{crit}$, i.e., $|-1.342| < 1.64$ we fail to reject $H_0$ at $0.1$ level of significance.

We accept the null hypothesis that the mean potassium content of a particular sports drink is 140 mg.

(d) The $p$-value is $2*P(Z\geq 1.342) = 0.18$.

The $p$-value is not less than $0.1$ level of significance, we fail to reject $H_0$ at $0.1$ 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