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To solve Problem 5, let's break it down step by step. Here's what the problem states:

We have a **normal population** from which a sample of **10 elements** is drawn. The sample has the following values:
- Sample mean \(\bar{x} = 10.9\)
- Sample standard deviation \(s = 0.58\)

We're tasked with testing the null hypothesis \( H_0 : \mu = 10 \) against the alternative hypothesis \( H_1 : \mu \neq 10 \) at a **significance level of 5%** (\(\alpha = 0.05\)).

### Step 1: Formulate the Hypotheses
The null and alternative hypotheses are:
\[
H_0 : \mu = 10
\]
\[
H_1 : \mu \neq 10
\]

This is a two-tailed test since the alternative hypothesis is that \(\mu\) is not equal to 10.

### Step 2: Compute the Test Statistic
The test statistic for a one-sample \( t \)-test is given by:
\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]
where:
- \(\bar{x}\) = sample mean = 10.9
- \(\mu_0\) = population mean under the null hypothesis = 10
- \(s\) = sample standard deviation = 0.58
- \(n\) = sample size = 10

Let's plug in the values:

\[
t = \frac{10.9 - 10}{\frac{0.58}{\sqrt{10}}}
\]

Now, I will calculate the value of the \(t\)-statistic.The calculated \(t\)-statistic is approximately **4.91**.

### Step 3: Determine the Critical Value or P-value
Since this is a two-tailed test with \(\alpha = 0.05\), we need to find the critical \(t\)-values for a two-tailed test with \(n - 1 = 9\) degrees of freedom. Alternatively, we can find the p-value for the computed \(t\)-statistic.

Next, I'll calculate the p-value associated with the test statistic.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you on how to find the p-value:

1. You can use statistical software or a calculator that supports the Student's \( t \)-distribution.
2. For a \( t \)-statistic of approximately 4.91 and 9 degrees of freedom, input this into a two-tailed \( t \)-distribution calculator to find the p-value.
3. The p-value will be very small, likely less than 0.01, suggesting that you reject the null hypothesis at the 5% significance level.

If you have any questions about the process or want further details, feel free to ask!

### Here are 5 follow-up questions you might find useful:
1. How is the \( t \)-distribution different from the normal distribution?
2. What is the interpretation of a p-value in hypothesis testing?
3. How do degrees of freedom affect the shape of the \( t \)-distribution?
4. What does it mean to reject the null hypothesis?
5. How can you calculate confidence intervals in this context?

**Tip:** The smaller the p-value, the stronger the evidence against the null hypothesis.

Solve problem 5. We have a normal population where a sample of 10 elements was drawn. The sample has a mean of 10.9 and a standard deviation of 0.58. Test the null hypothesis H₀: μ = 10 against the alternative H₁: μ ≠ 10 at a 5% significance level.

Perform hypothesis testing on a sample with mean 10.9 and standard deviation 0.58, testing the hypothesis H₀: μ = 10 against H₁: μ ≠ 10. This example applies Student's t-test at a 5% significance level.

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