Question

An industrial company claims that the mean pH level of the water in a nearby river is 6.7. You randomly select 21 water samples and measure the pH of each. The sample has a mean of 6.82 and a standard deviation of 0.26.  Assume the population is normally distributed.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 6.76.7

Ha: μ ≠ 6.76.7

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc​=2.086.

The rejection region for this two-tailed test is R={t:∣t∣>2.086}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=2.115>tc​=2.086, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0472, and since p=0.0472<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 6.7, at the 0.05 significance level.