Questions: For a hypothesis test: Null hypothesis: μ=300 Alternative hypothesis: μ ≠ 300 The sample standard deviation, 5=25. Which type of distribution would you use to get the test statistic and p-value? Uniform Distribution Student t-distribution Binomial Distribution Normal Distribution

For a hypothesis test:
Null hypothesis: μ=300
Alternative hypothesis: μ ≠ 300
The sample standard deviation, 5=25.
Which type of distribution would you use to get the test statistic and p-value?
Uniform Distribution
Student t-distribution
Binomial Distribution
Normal Distribution

Transcript text: For a hypothesis test: Null hypothesis: $\mu=300$ Alternative hypothesis: $\mu \neq 300$ The sample standard deviation, $5=25$. Which type of distribution would you use to get the test statistic and p-value? Uniform Distribution Student t-distribution Binomial Distribution Normal Distribution

Solution

Solution Steps

Step 1: Standard Error Calculation

The standard error \( SE \) is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{30}} \approx 4.5644 \]

Step 2: Test Statistic Calculation

The test statistic \( Z_{test} \) is calculated as follows:

\[ Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{300 - 300}{4.5644} = 0.0 \]

Step 3: P-value Calculation

For a two-tailed test, the p-value \( P \) is calculated using the formula:

\[ P = 2 \times (1 - T(|z|)) = 1.0 \]

Final Answer

The test statistic is \( 0.0 \) and the p-value is \( 1.0 \).

Thus, the answer is: \[ \boxed{Z_{test} = 0.0, \, P = 1.0} \]

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