Math Problem Statement

Use a t-test to test the claim about the population mean u at the given level of significance a using the given sample statistics. Assume the population is normally distributed. Claim: + = 51,500; a = 0.10 Sample statistics: x= 50,635, s = 2800, n = 19 Click the icon to view the t-distribution table. ¼ Question 1 • Question 2 • Question 3 • Question 4 What are the null and alternative hypotheses? Choose the correct answer below. • А. Но: н # 51,500 Ha: M = 51,500 У С. Но: M = 51,500 Ha: M¥51,500 O B. Ho: H≤ 51,500 Ha: M > 51,500 • D. Ho: #≥51,500 Ha: M < 51,500 What is the value of the standardized test statistic? The standardized test statistic is • (Round to two decimal places as needed.)

Solution

Step 1: Identify the Hypotheses

The claim in this case is that the population mean (μ\mu) is equal to 51,500. This suggests we are testing whether the population mean is equal to 51,500 or not.

The null hypothesis (H0H_0) is that the population mean is equal to 51,500, and the alternative hypothesis (HaH_a) is that the population mean is not equal to 51,500. This represents a two-tailed test.

Thus, the correct hypotheses are:

  • H0:μ=51,500H_0: \mu = 51,500 (null hypothesis)
  • Ha:μ51,500H_a: \mu \neq 51,500 (alternative hypothesis)

The correct answer is: C. H0:μ=51,500H_0: \mu = 51,500, Ha:μ51,500H_a: \mu \neq 51,500.

Step 2: Calculate the Test Statistic

We use the t-test statistic formula for a sample mean:

t=xˉμ0snt = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}

Where:

  • xˉ=50,635\bar{x} = 50,635 (sample mean)
  • μ0=51,500\mu_0 = 51,500 (claimed population mean)
  • s=2,800s = 2,800 (sample standard deviation)
  • n=19n = 19 (sample size)

Substitute the values into the formula:

t=50,63551,5002,80019t = \frac{50,635 - 51,500}{\frac{2,800}{\sqrt{19}}}

Step 3: Compute the Denominator

2,800192,8004.3589642.35\frac{2,800}{\sqrt{19}} \approx \frac{2,800}{4.3589} \approx 642.35

Step 4: Compute the Test Statistic

Now, calculate the numerator:

50,63551,500=86550,635 - 51,500 = -865

Now, calculate the entire expression:

t=865642.351.35t = \frac{-865}{642.35} \approx -1.35

Step 5: Conclusion

The value of the standardized test statistic is approximately -1.35.


Would you like me to help you with the next steps, such as finding the critical value or determining whether to reject the null hypothesis based on the t-distribution table?

Additional Questions:

  1. How do you interpret a two-tailed test result?
  2. What is the meaning of the significance level (α\alpha) in hypothesis testing?
  3. How can you calculate the p-value from the t-statistic?
  4. What is the difference between a one-tailed and two-tailed t-test?
  5. How do you determine the degrees of freedom for a t-test?

Tip: Always ensure your sample size is large enough when performing t-tests to guarantee reliable results. For smaller sample sizes, ensure that the population is normally distributed.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
t-Test
Significance Level
Normal Distribution

Formulas

t = (x̄ - μ₀) / (s / √n)

Theorems

Central Limit Theorem
t-distribution

Suitable Grade Level

Grades 11-12