Question

In: Statistics and Probability

The urinary fluoride concentration (parts per million) was measured both for a sample of livestock grazing...

The urinary fluoride concentration (parts per million) was measured both for a sample of livestock grazing in an area previously exposed to fluoride pollution and for a similar sample grazing in an unpolluted region:

Polluted. 21.3. 18.7 23.0. 17.1. 16.8. 20.9. 19.7

Unpolluted. 14.2. 18.3. 17.2. 18.4. 20.0

Does the data indicate strongly that true average fluoride concentration for livestock grazing in the polluted region is larger than for the unpolluted region? Use the appropriate test at level α = 0.01.

Solutions

Expert Solution

1 represents the polluted region and 2 represents the unpolluted region.

The sample means are shown below:

Also, the sample standard deviations are:

s1​=2.274392, s2​=2.15685

and the sample sizes are n1​=7 and n2​=5.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​.

True average fluoride concentration for livestock grazing in the polluted region is equal to the of unpolluted region

Ha: μ1​ > μ2​.

True average fluoride concentration for livestock grazing in the polluted region is larger than for the unpolluted region

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.083 and FU​=21.975, and since F = 1.112, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

The significance level is α = 0.01, and the degrees of freedom are df = 10. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this right-tailed test is tc ​= 2.764, for α=0.01 and df = 10.

The rejection region for this right-tailed test is R = { t : t > 2.764}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

​​

(4) Decision about the null hypothesis

Since it is observed that t = 0.696 ≤ tc ​= 2.764, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2512, and since p = 0.2512 ≥ 0.01, 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 μ1​ is greater than μ2​, at the 0.01 significance level.

Therefore, there is not enough evidence to claim that the true average fluoride concentration for livestock grazing in the polluted region is larger than for the unpolluted region at the 0.01 significance level.

Confidence Interval

The 99% confidence interval is −7.19<μ1​−μ2​<11.236.


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