Homser Lake, Oregon, has an Atlantic salmon catch-and-release program that has been very successful. The average fisherman's catch has been \( \mu=8.4 \). Atlantic salmon per day. Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of fishermen gave the following catches per day: \( 11,6,11,11,5,2,1,7,7,8,6,2,12,13 \) Assuming the catch per day has an approximately normal distribution, does the sample indicate at a 0.1 level of significance that the population average catch per day is now different from 8.4 ? A) Use a calculator with the "Stats" feature to find \( \overline{\bar{x}} \) and \( s \). (Round your answer to 2 decimal places). \[ \bar{x}=7.29 \] \[ s=3.93 \] B) State the Null and Alternate hypotheses. \( H_{0}: \mu=8.4 ; H_{1}: \mu>8.4 \) \( H_{0}: \mu>8.4 ; H_{1}: \mu=8.4 \) \( H_{0}{ }^{-} \mu \neq 8.4: H_{1}: \mu=8.4 \) \( H_{0}^{*}{ }^{*} \mu=8.4: H_{1}: \mu<8.4 \) ० \( H_{0}: \mu=8.4 ; H_{1}: \mu \neq 8.4 \) C) Is this a (left-tail, right-tail, one-tail, two-tail) test? one-tail two-tail left-tail right-tail D) Accurate to 3 decimal places, find the \( t^{*} \) value of the sample. \[ -1.018 \times \] E) Find the degree of freedom of the sample. \[ 13 \] F) Accurate to 4 decimal places, what's the p-value of the sample? \( \square \)

**A) Calculation of the Sample Mean and Standard Deviation** The sample mean and standard deviation are given by \[ \bar{x}=7.29 \quad \text{and} \quad s=3.93. \] **B) Statement of Null and Alternative Hypotheses** Since we are testing whether the population mean has changed from \(8.4\) (i.e. “different than” \(8.4\)), the hypotheses are \[ H_0: \mu = 8.4 \quad \text{and} \quad H_1: \mu \neq 8.4. \] **C) Type of Test** Because the alternative hypothesis is “\(\mu \neq 8.4\),” this is a **two-tail** test. **D) Computation of the Test Statistic \(t^*\)** The test statistic is given by \[ t^* = \frac{\bar{x} - \mu_0}{\dfrac{s}{\sqrt{n}}}, \] where \(\bar{x}=7.29\), \(\mu_0 = 8.4\), \( s=3.93 \), and \( n = 14 \). 1. Compute the numerator: \[ \bar{x} - \mu_0 = 7.29 - 8.4 = -1.11. \] 2. Compute the denominator: \[ \frac{s}{\sqrt{n}} = \frac{3.93}{\sqrt{14}} \approx \frac{3.93}{3.7417} \approx 1.05. \] 3. Therefore, \[ t^* \approx \frac{-1.11}{1.05} \approx -1.057. \] Rounded to three decimal places, \[ t^* \approx -1.057. \] **E) Degrees of Freedom** Degrees of freedom is given by \[ df = n - 1 = 14 - 1 = 13. \] **F) Calculation of the p-value** Since this is a two-tail test, the p-value is \[ \text{p-value} = 2\;P\left(T \le t^*\right) \] with \(df = 13\) and \(t^* \approx -1.057\). Using a \(t\)-distribution calculator or table, we find that the one-tail probability corresponding to \(t^*=-1.057\) (or equivalently, \(|t^*|=1.057\)) for \(13\) degrees of freedom is approximately \(0.1555\). Thus, \[ \text{p-value} \approx 2\times 0.1555 = 0.3110. \] Rounded to four decimal places, \[ \text{p-value} \approx 0.3110. \]