Question

t est to gather oation. The outut of this What is the p-value of this test? What is the conclusion? (3 credins) Assume that the distribution of the pH Sevel is normal with ox1.20. Test 7 uing a twotaled test (significance level 001). (4 creds) n.:μ " 7 verm / μ c) Assume that the distribution of the pH level is normal with σ120. Test p7 versus H:p <7 using a one-tailed test (significance level 0.01). (3 credits) As in problem 4, Jim Lee collected a random sample with 53 observed values. The mean of the sample is F 6.59. Assume that the distribution of the pH is normal with o-1.20. Please construct a 95% confidence interval for the population mean and interpret the meaning of the 95% confidence interval" (10 credits) 5.

Answer 1

a.

p-value of the test = 0.61

Since p-value (=0.61) > 0.01 assumed significance level, we do not reject the null hypothesis that the sample is from normal distribution

b.

From the data

Given

$\overline{x} = 6.59$ ; $\sigma = 1.20$ ;

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{6.59 - 7}{\frac{1.20}{\sqrt{53}}} = -2.4874$

At 0.01 significance level, for a two-tail test z-critical value is 2.58

Since the calculated absolute value of z i.e., $\left | z \right | = 2.4874 < z_{critical} = 2.58$ we do not reject null hypothesis and infer that the population mean pH is 7

c.

From the data

Given

$\overline{x} = 6.59$ ; $\sigma = 1.20$ ;

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{6.59 - 7}{\frac{1.20}{\sqrt{53}}} = -2.4874$

At 0.01 significance level, for a left-tail test z-critical value is -2.33

Since the calculated absolute value of z i.e., we reject null hypothesis and infer that the population mean pH is less than 7

5.

95% confidence interval for the population mean $\mu$ is

$(\overline{x} - z_{critical} \ast \frac{\sigma}{\sqrt{n}}{ , }\overline{x} + z_{critical} \ast \frac{\sigma}{\sqrt{n}})$

At 95% confidence z-critical value is 1.96

Given

$\overline{x} = 6.59$ ; $\sigma = 1.20$ ;

95% confidence interval for the population mean $\mu$ of pH is

$= (6.59 - 1.96 \ast \frac{1.20}{\sqrt{53}}{ , }6.59 + 1.96 \ast \frac{1.20}{\sqrt{53}})$

Interpretation:

We are 95% confident that the population mean pH is between 6.2669 and 6.9131