Question

It is believed that the mean population age in an area is 18.9. An investigator took a sample of 200 people and found a sample mean age of 21 with a sample standard deviation (σ) of 5. Test the hypothesis that the population mean is 18.9 at an alpha level of 0.05.

a) State the null hypothesis and the alternative hypothesis.

            b) State the test that should be used to test this hypothesis?

            c) Compute the appropriate test statistic for this test.

            d) Determine the critical value(s) for the significance level.

            e) Determine the p-value for this test.

            f) Based on the p-value or critical value(s) you found above, what is your                               conclusion. Explain.

            g) Find a 95% confidence interval and explain how this supports your conclusion.

Answer 1

a) $H_{0}: \mu = 18.9$

    $H_{1}: \mu \neq 18.9$

b) We should use one sample z-test for mean.

c) The test statistic is

$z = \frac{\bar x - \mu}{\sigma/\sqrt n}$

    $= \frac{21 - 18.9}{5/\sqrt {200}}$

   

e) P-value = 2 * P(Z > 5.94)

                 = 2 * (1 - P(Z < 5.94))

                 = 2 * (1 - 1)

                 = 2 * 0 = 0

f) Since the P-value is less than the significance level (0 < 0.05), so we should reject the null hypothesis.

At 0.05 significance level, there is not sufficient evidence to conclude that the population mean is 18.9.

g) For 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

$\bar x \pm z_{0.025} * \frac{\sigma}{\sqrt n}$

$= 21 \pm 1.96 * \frac{5}{\sqrt {200}}$

$= 21 \pm 0.69$

Since the confidence interval does not contain the hypothesized population mean value 18.9, so we should reject the null hypothesis.