Let's solve each part of the hypothesis test step-by-step:
### Given Data:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu \geq 20\)[/tex] (or equivalently [tex]\( \mu = 20 \)[/tex] when testing)
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu < 20\)[/tex]
- Sample Size ([tex]\(n\)[/tex]): 50
- Sample Mean ([tex]\(\bar{x}\)[/tex]): 19.4
- Population Standard Deviation ([tex]\(\sigma\)[/tex]): 2
- Significance Level ([tex]\(\alpha\)[/tex]): 0.05
### a) Compute the critical value (location of end point(s) of rejection region):
For a one-tailed test with [tex]\(\alpha = 0.05\)[/tex], the critical value is the z-value such that the cumulative probability to the left of this z-value is 0.05. Using the standard normal distribution table, the critical value corresponding to [tex]\(\alpha = 0.05\)[/tex] is:
[tex]\[ z_{\text{critical}} = -1.64485 \][/tex]
### b) Formulate the rejection rule:
To reject the null hypothesis, the test statistic must be in the rejection region. For a one-tailed test where [tex]\(H_a: \mu < 20\)[/tex]:
[tex]\[ \text{Reject } H_0 \text{ if the test statistic } z \text{ is less than } z_{\text{critical}} \][/tex]
[tex]\[ \text{Reject } H_0 \text{ if } z < -1.64485 \][/tex]
### c) What is the value of the test statistic?
The test statistic for a z-test is calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ z = \frac{19.4 - 20}{2/\sqrt{50}} \][/tex]
[tex]\[ z = \frac{-0.6}{2/7.071} \][/tex]
[tex]\[ z \approx -2.12132 \][/tex]
### d) What is the conclusion?
Compare the test statistic to the critical value:
[tex]\[ z = -2.12132 \][/tex]
[tex]\[ z_{\text{critical}} = -1.64485 \][/tex]
Since [tex]\(-2.12132 < -1.64485\)[/tex], we reject the null hypothesis [tex]\(H_0\)[/tex].
### e) Calculate the p-value and formulate the rejection rule:
The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed statistic under the null hypothesis. For a z-test, the p-value can be found using the cumulative distribution function (CDF) of the standard normal distribution.
[tex]\[ \text{p-value} = P(Z < -2.12132) \approx 0.01695 \][/tex]
Formulate the rejection rule in terms of the p-value:
[tex]\[ \text{Reject } H_0 \text{ if } \text{p-value} < \alpha \][/tex]
Compare the p-value to [tex]\(\alpha\)[/tex]:
[tex]\[ \text{p-value} = 0.01695 \][/tex]
[tex]\[ \alpha = 0.05 \][/tex]
Since [tex]\(0.01695 < 0.05\)[/tex], we reject the null hypothesis [tex]\(H_0\)[/tex].
### Summary of the results:
- Critical value: [tex]\(-1.64485\)[/tex]
- Test statistic: [tex]\(-2.12132\)[/tex]
- Conclusion: Reject [tex]\(H_0\)[/tex]
- p-value: [tex]\(0.01695\)[/tex]
- Rejection by p-value: True
Thus, we conclude that there is sufficient evidence to reject the null hypothesis [tex]\(\mu = 20\)[/tex] in favor of the alternative hypothesis [tex]\(\mu < 20\)[/tex] at the 0.05 significance level.
