Question

An industrial company claims that the mean pH level of the water in a nearby river is 6.8 you randomly select 19 water sample mean and standard deviation are 6.7 and 0.24, respectively. is there enough evidence to reject the company claim at a=0.05? Assume the population normally distributed.

Answer 1

The provided sample mean is 6.7 and the sample standard deviation is s = 0.24 , and the sample size is n = 19 .

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 6.8

Ha: μ ≠ 6.8

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 t_c = 2.101

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that |t| = 1.816 < t_c = 2.101 , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.086 , and since p = 0.086 > 0.05 , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 6.8, at the 0.05 significance level.