Since the population standard deviation is unknown, you need to use a t-test for the hypothesis. The formula for the t-test statistic is:

$t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}$

Where:

• $\bar{X} = 90.8$ (sample mean)
• $\mu_0$ = hypothesized population mean (typically 0 or whatever value is relevant)
• $s = 10.6$ (sample standard deviation)
• $n = 14$ (sample size)

If we assume $\mu_0 = 0$ (as is common when testing), the t-statistic becomes:

$t = \frac{90.8 - 0}{10.6 / \sqrt{14}}$

First, calculate the denominator:

$\frac{10.6}{\sqrt{14}} = \frac{10.6}{3.7417} \approx 2.833$

Now, calculate the t-statistic:

$t = \frac{90.8}{2.833} \approx 32.056$

So the t-statistic is:

$t = 32.056$

Step 2: Calculate the p-value (right-tailed)

Since this is a right-tailed test, we need the p-value corresponding to the computed t-statistic with n - 1 = 13 degrees of freedom. You can find this using a t-distribution table or statistical software.

Using a t-table:
For t = 32.056 with 13 degrees of freedom, the p-value will be extremely small (near 0). In fact, for such a high t-value, the p-value is effectively:

$\text{p-value} \approx 0.000$

This means the result is highly significant.

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Would you like me to go into further details or check your specific hypothesis value?

Here are 5 related questions you may explore:

1. How do you determine whether to use a t-test or z-test?
2. What is the difference between one-tailed and two-tailed tests?
3. How do degrees of freedom affect the shape of the t-distribution?
4. What are Type I and Type II errors in hypothesis testing?
5. How would the p-value change if it were a two-tailed test?

Tip: Always report your hypothesis $H_0$ and $H_1$ clearly before conducting tests for clarity and reproducibility.