SOLUTION: Acid rain from the burning of fossil fuels has caused many of the lakes around the world to become acidic. The biology in these lakes often collapses because of the rapid and unfav

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Question 1205340: Acid rain from the burning of fossil fuels has caused many of the lakes around the world to become acidic. The biology in these lakes often collapses because of the rapid and unfavorable changes in water chemistry. A lake is classified as nonacidic if it has a pH greater than 6. A. Marchetto and A. Lami measured the pH of high mountain lakes in the Southern Alps and reported their findings in the paper “Reconstruction of pH by Chrysophycean Scales in Some Lakes of the Southern Alps” (Hydrobiologia, Vol. 274, pp. 83-90). The pH levels obtained by the researchers for 15 lakes are 7.2, 7.3, 6.1, 6.9 6.6, 7.3, 6.3, 5.5, 6.3, 5.7, 6.9, 6.7, 7.9, and 5.8. At the 5% significance level, do the data provide sufficient evidence to conclude that, on average, high mountain lakes in the Southern Alps are nonacidic? A. State null hypothesis
B. State alternative hypothesis
C. Level of significance
D. Test statistic E. Decision rule based on p-value
F. Statistical decision
G. Conclusion

Answer by Theo(13342) About Me  (Show Source):
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this was analyzed using the statistic calculator at https://www.statskingdom.com/130MeanT1.html

excerpts from the results are shown bwlow:











this was a one sample t-test with a right tailed confidence interval.
the p value was .003098.
the t-value was 3.2609
the mean of the sample was 6.61
the standard deviation of the sample was .6967
the sample size was 14
the degrees of freedom was 13
the standard error was .6967/sqrt(14) = .1862

the critical t-value at .05 level of significance at the right tail with 13 degrees of freedom is equal to 1.7709.
the test t-value at 3.2609 was greater than this.
the test p-value at .003098 was less than .05.

the results were considered significant.
the conclusion was that the lakes were nonacidic since they contained a pH greaetr than 6.
the test values, being significant, indicated the difference in the maen of the sample and a pH of 6 was not due to random variation in the mean of sample of size 14.

note:
you said 15 elements but i counted only 14.
sample size was 14.
degrees of freedom were sample size minus 1 = 13.

the standard error = standard deviation / sqrt(sample size) = .6967/sqrt(14) = .1862.