Question

Benzene is a toxic chemical used in the manufacturing of medicinal chemicals, dyes, artificial leather, and linoleum. A manufacturer claims that its exit water meets the federal regulation with a mean of less than 7980 ppm of benzene. To assess the benzene content of the exit water, 10 independent water samples were collected and found to have an average of 7900 ppm of benzene. Assume a known population standard deviation of 80 ppm and use a significance level of 0.01. Test the manufacturer’s claim.(a) State the Null and Alternative hypothesis. (b) calculate the test statistics. (c) Find the P-value and draw your conclusion. Round your answer to 5 decimal places. (d) Find a 99% two-sided confidence bound on the true mean. Round your answer to 2 decimal places.

Answer 1

The Null hypothesis states that the mean of the benzene content in the exit water is greater than or equal to 7980 ppm. The Alternative hypothesis states that the mean of the benzene content in the exit water is less than 7980 ppm. The test statistic is calculated using the given values and compared to the critical value. The P-value is determined based on the comparison, and a confidence interval is calculated to evaluate the true mean.

Step-by-step explanation:

a. The Null hypothesis states that the mean of the benzene content in the exit water is greater than or equal to 7980 ppm. The Alternative hypothesis states that the mean of the benzene content in the exit water is less than 7980 ppm.

b. To calculate the test statistic, we can use the formula:

t = (sample mean - hypothesized mean) / (population standard deviation / √sample size)

Substituting the given values, we get:

t = (7900 - 7980) / (80 / √10) = -1.772

c. Using a significance level of 0.01, we can find the P-value by comparing the test statistic to the critical value for a one-tailed test. Looking up the critical value for a one-tailed test with a significance level of 0.01 and degrees of freedom equal to the sample size minus 1 (10-1=9), we find a critical value of -2.821. Since the test statistic (-1.772) is greater than the critical value (-2.821), the P-value is greater than 0.01. Therefore, we fail to reject the Null hypothesis.

d. To find a 99% two-sided confidence bound on the true mean, we can use the formula:

Confidence Interval = sample mean ± (critical value * (population standard deviation / √sample size))

Substituting the given values, we get:

Confidence Interval = 7900 ± (2.821 * (80 / √10)) = 7900 ± 84.67 = (7815.33, 7984.67)

Answer 2

a: H0: µ = 7980, Ha: µ < 7980 (claim)

b: -3.162

c: P-value is 0.00576

d: 7821.61 < µ < 7978.39

Step-by-step explanation:

The given data for the situation is:

µ = 7980, σ = 80, x = 7900, n = 10

The hypothesis are

H0: µ = 7980

Ha: µ < 7980 (claim)

This is a left tailed test, with a significance level of 0.01, so the critical value for the situation is: t < -2.821

See the attached photo 1 for the calculation of the test statistic...

Use a chart, software or online calculator to find the p-value of the test statistic

See attached photo 2 for the construction of the confidence interval