Math  /  Data & Statistics

QuestionSuppose a geologist believes that the average chloride concentration in her local water supply is above the national average of 156 ppm . The geologist defines a null hypothesis, H0:μ=156ppmH_{0}: \mu=156 \mathrm{ppm}, to reflect the mean chloride concentration in the national water supply. She formulates an alternative hypothesis, H1:μ>156ppmH_{1}: \mu>156 \mathrm{ppm}, to indicate that her locality has a higher concentration of chloride in its water supply, on average. In these hypotheses, μ\mu 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 208 ppm , with a standard deviation of 35 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 tt-test for a mean to test her hypotheses. She computes a tt-statistic with a value of 7.862 and 27 degrees of freedom.
Decide the outcome of the geologist's test using a significance level of α=0.05\alpha=0.05. Fail to reject the null hypothesis because the PP-value is greater than α\alpha. Reject the null hypothesis because the PP-value is greater than α\alpha.

Studdy Solution

STEP 1

1. The null hypothesis is H0:μ=156ppm H_0: \mu = 156 \, \text{ppm} .
2. The alternative hypothesis is H1:μ>156ppm H_1: \mu > 156 \, \text{ppm} .
3. The sample mean is 208 ppm, with a standard deviation of 35 ppm.
4. The sample size provides 27 degrees of freedom.
5. The significance level is α=0.05\alpha = 0.05.
6. The test is a one-sample, right-tailed t t -test.

STEP 2

1. Determine the critical t t -value for α=0.05\alpha = 0.05 and 27 degrees of freedom.
2. Compare the calculated t t -statistic with the critical t t -value.
3. Make a decision based on the comparison.

STEP 3

Determine the critical t t -value for a right-tailed test with α=0.05\alpha = 0.05 and 27 degrees of freedom. Use a t t -distribution table or calculator to find this value.

STEP 4

Compare the calculated t t -statistic (7.862) with the critical t t -value obtained in Step 1.

STEP 5

If the calculated t t -statistic is greater than the critical t t -value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
The calculated t t -statistic of 7.862 is significantly greater than the typical critical t t -value for α=0.05\alpha = 0.05 and 27 degrees of freedom (approximately 1.703). Therefore, we reject the null hypothesis because the P P -value is less than α\alpha.

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