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9. Conduct a hypothesis test showing all of your steps. A community center claims that the chlorine level in its pool has a standard deviation of 0.48 parts per million (ppm). Samples were taken 27 times over the past month that yielded a standard deviation of 0.61 ppm. Test the center's claim with a 5% level of significance. a) State the hypotheses and label the claim. b) Is it left-tailed, right-tailed, or two-tailed (circle one)? c) What are the critical values? Draw a picture. d) What is the test-statistic? e) Give one reason for your decision to reject/not reject Ho based on: 1) Test statistic compared to critical value(s). State the conclusion, relating it back to the original claim.

Answer 1

The provided sample variance is s2 = 0.3721 and the sample size is given by = 27. n = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:σ = 0.48 Ha : 0 + 0.48 This corresponds to a two-tailed test for which a Chi-Square test for one population variance will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the the rejection region for this two-tailed test is R= {x? : x2 < 13.844 or x? > 41.923} (3) Test Statistics The Chi-Squared statistic is computed as follows: (n − 1)s2 x (27 – 1) · 0.3721 0.2304 41.99 02 (4) Decision about the null hypothesis = 41.99 > Xỉ = 41.923, it is then concluded that the Since it is observed that x2 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 * s.d. o is different than 0:48 at the 0.05 significance level.

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