Question

Suppose a geologist believes that the average chloride concentration in her local water supply is above the national average of 75 ppm. The geologist defines a null hypothesis, Ho:u = 75 ppm, to reflect the mean chloride concentration in the national water supply. She formulates an alternative hypothesis, H1: u > 75 ppm, to indicate that her locality has a higher concentration of chloride in its water supply, on average. In these hypotheses, y represents the true mean concentration of chloride in the local water supply The geologist samples groundwater from across her locality and determines that the average concentration of chloride in her sample is 78 ppm, with a standard deviation of 17 ppm. Experience has shown that these data can be treated as random samples from a normal population. The geologist uses a one-sample, right-tailed t-test for a mean to test her hypotheses. She computes a t-statistic with a value of 0.558 and 9 degrees of freedom. Decide the outcome of the geologist's test using a significance level of a = 0.05. O Reject the null hypothesis because the P-value is less than a. Fail to reject the null hypothesis because the P-value is less than a. Reject the null hypothesis because the P-value is greater than a. Fail to reject the null hypothesis because the P-value is greater than a. Unable to make this decision because the P-value cannot be found in the t-distribution table.

Answer 1

1). t = 0.558

df = 9

α = 0.05

Right tailed p-value = T.DIST.RT(0.558, 9) = 0.2952

p-value = 0.2952 > 0.05, we fail to reject the null hypothesis.

Answer: Fail to reject the null hypothesis because P-value is greater than α.

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2). Answer: A matched-pairs t-test is not valid because the sample is not random.