Math Problem Statement

A state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. Six water specimens taken from a river at a specific location during the low-water season (July) gave readings of 5.0, 5.2, 4.8, 4.9, 5.1, and 4.8 ppm of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using 𝛼 = 0.05. State the test statistic. t = ?

Solution

To determine whether the dissolved oxygen content is significantly less than 5 ppm, we conduct a one-sample tt-test for the mean. Here are the steps:

Step 1: State the Hypotheses

  • Null hypothesis (H0H_0): μ=5\mu = 5 ppm (mean dissolved oxygen content is 5 ppm).
  • Alternative hypothesis (HaH_a): μ<5\mu < 5 ppm (mean dissolved oxygen content is less than 5 ppm).

Step 2: Calculate the Test Statistic

The formula for the test statistic is:

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

where:

  • xˉ\bar{x}: Sample mean
  • μ0\mu_0: Population mean under the null hypothesis (5 ppm)
  • ss: Sample standard deviation
  • nn: Sample size

Step 3: Calculate Sample Mean (xˉ\bar{x}) and Standard Deviation (ss)

The data are: 5.0,5.2,4.8,4.9,5.1,4.85.0, 5.2, 4.8, 4.9, 5.1, 4.8.

  1. Compute the mean: xˉ=xin=5.0+5.2+4.8+4.9+5.1+4.86\bar{x} = \frac{\sum x_i}{n} = \frac{5.0 + 5.2 + 4.8 + 4.9 + 5.1 + 4.8}{6}

  2. Compute the standard deviation: s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

Let me calculate these values.import numpy as np

Data

data = np.array([5.0, 5.2, 4.8, 4.9, 5.1, 4.8])

Calculations

mean = np.mean(data) std_dev = np.std(data, ddof=1) # Sample standard deviation n = len(data)

mean, std_dev(4.966666666666667, 0.16329931618554525)(4.966666666666667, 0.16329931618554525)successThe sample mean (xˉ\bar{x}) is approximately 4.9674.967, and the sample standard deviation (ss) is approximately 0.1630.163.

Step 4: Compute the tt-Statistic

Now, substitute these values into the formula:

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

Let me compute the tt-statistic.# Population mean under H0 mu_0 = 5

Calculate the t-statistic

t_statistic = (mean - mu_0) / (std_dev / np.sqrt(n)) t_statistic-0.499999999999998-0.499999999999998successThe tt-statistic is approximately 0.50-0.50.

Would you like to proceed with determining the critical value and making a conclusion, or do you need clarification on any steps so far?

Here are some follow-up questions:

  1. What is the critical value for a one-tailed test at α=0.05\alpha = 0.05 for this sample size?
  2. How is the tt-distribution used to assess the p-value in hypothesis testing?
  3. What assumptions are made in a one-sample tt-test?
  4. How would the test outcome change if the sample size were larger?
  5. How does the standard deviation affect the tt-statistic?

Tip: Always check if the assumptions of normality or sample size adequacy are met when performing tt-tests!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
One-Sample t-Test

Formulas

t = (x̄ - μ₀) / (s / √n)
Sample mean: x̄ = Σxᵢ / n
Sample standard deviation: s = √(Σ(xᵢ - x̄)² / (n-1))

Theorems

t-Distribution
Central Limit Theorem

Suitable Grade Level

Grades 11-12 or College Level