Chapter 5 Confidence Interval and Hypothesis Testing (Section on Feb 24th)

Sampling Distribution

Exercise 5.1 The ages for a random sample of 25 individuals are recorded. If the age of each individual follows a normal distribution with \(\mu = 25.36, \sigma = 2.6\). Find the following:

  1. Find \(P(\bar{X}\leq 26.3)\)

  2. Find \(P(\bar{X}\geq 22.3)\)

  3. Find \(P(21.2 \leq \bar{X} \leq 26.3)\)

Proof. Transfer to Standard Normal Random Variable \(\rightarrow\) Refer to a distribution Table or Calculator

In gereral, if the samples are from a normal distribution with mean \(\mu\) and variance \(\sigma^2\), then the sample mean \(\bar{X}\) follows a noraml distribution with mean \(\mu\) and variance \(\frac{\sigma^2}{n}\).

In this case, \(\bar{X}\sim N(25.36,\frac{2.6^2}{25})\). The satndard deviation is then \(\sqrt{\frac{2.6^2}{25}}=\frac{2.6}{5}\) (a) \[\begin{equation} P(\bar{X}\leq 26.3) = P(Z\leq \frac{26.3-25.36}{2.6/5}) = P(Z\leq 1.81) \approx 0.9649 \tag{5.1} \end{equation}\]

  1. \[\begin{equation} \begin{split} P(\bar{X}\geq 22.3) &= P(Z\geq \frac{22.3-25.36}{2.6/5}) = P(Z\geq -5.8846)\\ &=1-P(Z\leq -5.8846)\approx 0.9999 \end{split} (\#eq:05002) \end{equation}\]

  2. \[\begin{equation} \begin{split} P(21.2 \leq \bar{X} \leq 26.3) &= P(\bar{X}\leq 26.3) - P(\bar{X}\leq 21.2)\\ &=P(Z\leq 1.81) - P(Z\leq \frac{21.2-25.36}{2.6/5})\\ &=P(Z\leq 1.81) - P(Z\leq -8)\\ &=0.9649 - 0.0001\approx 0.9648 \end{split} (\#eq:05003) \end{equation}\]

Reference: Standard Normal Distribution Table

Confidence Interval

    1. What is a confidence interval?

    An estimate of plausible values for the population parameter based on data.

    1. Will the true parameter always be within the interval?

    Not necessarily.

    1. What is the point estimate?

    Sample Statistics.

    1. How do you calculate the margin of error?

    Critical Value multiplied by the Standard Error.

Critical value is usually Z-score (when knowing variance or sample size is larger) or T-score (when variance is unknown). More information is available here.

Hypothesis Testing

State the 4 steps within hypothesis testing discussed in class.

    1. State Null and Alternative Hypotheses.
    1. State significance level and Determine Critical Value.
    1. Perform Statistical Test.
    1. State Conclusion.
Exercise 5.2 A simple random sample of 50 adults is obtained, and each person’s red blood cell count (in cells per micro- liter) is measured. The sample mean is 5.23. The population standard deviation for red blood cell counts is 0.54. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 5.4, which is a value often used for the upper limit of the range of normal values. What type of error could you be making and why? (ASSUME Normal)

Proof. 1. State Null and Alternative Hypotheses: \(H_0:\mu=5.4\) vs \(H_1: \mu<5.4\).

  1. State significance level and Determine critical value: \(\alpha=0.01\), \(CV=|-2.33|=2.33\).

  2. Perform Statistical Test: \(TS=|\frac{5.23-5.4}{0.54/\sqrt{50}}|=|-2.226|=2.226\). \(|TS|<CV\), therefore we fail to reject \(H_0\).

  3. State Conclusion

    • Sample Size \(n = 50\)

    • The population mean is equal to 5.4

    • One Sample Hypothesis Test (Mean)

    • Fail to reject that the population mean is equal to 5.4

There maybe Type-II error.

Error Types

  • Type-I error: rejection of a true null hypothesis

  • Type-II error: failure to reject a false null hypothesis

Exercise 5.3 Reporting Income In a Pew Research Center poll of 745 randomly selected adults, 589 said that it is morally wrong to not report all income on tax returns. Use a 0.01 significance level to test the claim that 75% of adults say that it is morally wrong to not report all income on tax returns. What type of error could you be making and why?

Proof. 1. State Null and Alternative Hypotheses: \(H_0:p=0.75\) vs \(H_1: p\neq 0.75\).

  1. State significance level and Determine critical value: \(\alpha=0.01\), \(CV=|-2.57|=2.57\).

  2. Perform Statistical Test: \(TS=|\frac{0.7906-0.75}{\sqrt{(0.75\times 0.25)/745}}|=|2.5591|=2.5591\). \(|TS|<CV\), therefore we fail to reject \(H_0\).

Here sample proportion \(\hat{p}=\frac{589}{745}=0.7906\) and variance \(\frac{0.75\times 0.25}{745}\).

  1. State Conclusion

    • Sample Size \(n = 745\)

    • \(75\%\) of adults say that it is morally wrong

    • One Sample Hypothesis Test (Proportion)

    • Fail to reject that \(75\%\) of adults say that it is morally wrong

There maybe Type-II error.