Hypothesis testing for population mean

Hypothesis Tests for Population Mean (σ Unknown) with P-value Method on TI-84 Plus

Introduction

  • When testing a hypothesis for a population mean ($\mu$) and the population standard deviation ($\sigma$) is unknown, we use the t-test.
  • This involves replacing $\sigma$ with the sample standard deviation (s) and using the t-statistic.

T-Test Statistic

  • The t-test statistic is calculated as: t=xˉ−μs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}
    • $\bar{x}$ = sample mean
    • $\mu$ = population mean (from the null hypothesis)
    • s = sample standard deviation
    • n = sample size
  • When the null hypothesis is true, the t-statistic follows a Student's t-distribution with $n-1$ degrees of freedom.

Assumptions for T-Test

  • Simple random sample.
  • Either:
    • The sample size is large ($n > 30$).
    • The population is approximately normal.

Example 1: Weight Loss Study

  • Scenario: A medical study of 76 subjects on a low-fat diet showed a sample mean weight loss of $\bar{x} = 2.2$ kg with a sample standard deviation of $s = 6.1$ kg. Test if the mean weight loss is greater than zero.
  • Null Hypothesis ($H_0$): $\mu = 0$
  • Alternate Hypothesis ($H_1$): $\mu > 0$ (right-tailed test)
  • Test Statistic Calculation:
    t=2.2−06.1/76=3.144t = \frac{2.2 - 0}{6.1 / \sqrt{76}} = 3.144
  • Degrees of Freedom: $n - 1 = 76 - 1 = 75$
  • P-value: Using technology, the p-value is found to be $p = 0.0012$.
  • Decision: Since $p < 0.05$, reject the null hypothesis at the 0.05 level.
  • Conclusion: The mean weight loss is greater than zero.
Using TI-84 Plus Calculator
  1. Press STAT, highlight TESTS, and select T-Test.
  2. Choose Stats as the input option.
  3. Enter the following values:
    • $\mu_0$: 0
    • $\bar{x}$: 2.2
    • s: 6.1
    • n: 76
  4. Select > \mu_0 for the right-tailed test.
  5. Select Calculate.

Example 2: Antifungal Ointment Testing

  • Scenario: The FDA is testing a generic antifungal ointment. The brand name ointment delivers a mean of $\mu = 3.5$ micrograms of active ingredient per square centimeter. Seven subjects apply the generic ointment, and the absorbed amounts are measured.
  • Data: Absorbed amounts in micrograms.
  • Significance Level: $\alpha = 0.01$
Assumptions Check

  • Small sample size requires checking for approximate normality.

  • Dot plot of the data shows no strong skewness or outliers, so the normality assumption is reasonable.

  • Null Hypothesis ($H_0$): $\mu = 3.5$

  • Alternate Hypothesis ($H_1$): $\mu \neq 3.5$ (two-tailed test)

Using TI-84 Plus Calculator
  1. Enter the data into list L1.
  2. Press STAT, highlight TESTS, and select T-Test.
  3. Choose Data as the input option.
  4. Enter the following values:
    • $\mu_0$: 3.5
    • List: L1
    • Freq: 1
  5. Select $\neq \mu_0$ for the two-tailed test.
  6. Select Calculate.
  • P-value: The p-value is found to be $p = 0.0256$.
  • Decision: Since $p > 0.01$, we do not reject the null hypothesis at the 0.01 level.
  • Conclusion: There is not enough evidence to conclude that the mean amount of drug absorbed differs from 3.5 micrograms.

Statistical Significance

  • If $p \le \alpha$, we reject the null hypothesis and the result is statistically significant at the level $\alpha$.
  • If $p > \alpha$, we do not reject the null hypothesis and the result is not statistically significant.
Example: Given $p = 0.03$
  • At $\alpha = 0.05$: Since $0.03 < 0.05$, the result is statistically significant.
  • At $\alpha = 0.01$: Since $0.03 > 0.01$, the result is not statistically significant.

Example 3: Restaurant Dessert Offer

  • Scenario: A restaurant offers a free dessert on Monday nights to increase business. Before the offer, the mean number of dinner customers was 150. We have a random sample of 12 days with the free dessert offer.
  • Data: Number of diners on 12 random Mondays while the offer was in effect.
  • Significance Level: $\alpha = 0.01$
  • Null Hypothesis ($H_0$): $\mu = 150$
  • Alternate Hypothesis ($H_1$): $\mu > 150$
Assumptions Check
  • Assume simple random sample.
  • Check for approximate normality via a box plot. If reasonable, proceed with the t-test.
Steps
  1. Enter the data into list L1 in the calculator.
  2. Press STAT, highlight TESTS, and select T-Test.
  3. Choose Data as the input option.
  4. Enter the following values:
    • $\mu_0$: 150
    • List: The list where data was entered, e.g., L1
    • Freq: 1
  5. Select > \mu_0 for the right-tailed test.
  6. Select Calculate.
  • P-value: Approximately $p = 0.005$
  • Decision: Since $p < 0.01$, we reject $H_0$.
  • Conclusion: The evidence is very strong that the mean number of diners increased with the free dessert offer.