Question

Consider the following hypothesis test \( H_0: \mu = 18 \), \( H_a: \mu \neq 18 \). A sample of 81 provided a sample mean \( \bar{x} = 17 \) and a population standard deviation \( \sigma = 4.5 \). The value of test statistic and degree of freedom are:

• A. t = -1.7, degree of freedom = 16
• B. t = -4.5, degree of freedom = 17
• C. t = -2, degree of freedom = 80
• D. t = -1.54, degree of freedom = 48

Hint:

For samples \( n > 30 \), the t-distribution approaches the normal (z) distribution, but the degrees of freedom remain \( n-1 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a hypothesis test with a large sample, we use the z-test or t-test. When the population standard deviation is given (or sample size is large), the test statistic is calculated as \( z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \).

Step 2: Key Formula or Approach:
\( \bar{x} = 17 \), \( \mu = 18 \), \( \sigma = 4.5 \), \( n = 81 \).
Degree of freedom for t-distribution is \( n - 1 \).

Step 3: Detailed Explanation:
\[ t = \frac{17 - 18}{4.5 / \sqrt{81}} = \frac{-1}{4.5 / 9} = \frac{-1}{0.5} = -2 \]
Degrees of freedom \( = 81 - 1 = 80 \).

Step 4: Final Answer:
The test statistic is \( -2 \) and the degrees of freedom is \( 80 \).