Use the above table to find the missing values , beta, alpha, lambda and theta

Here are the solutions for the problems.

: ANOVA Table

We need to find the missing values $\eta, \beta, \alpha, \lambda, \theta$ in the ANOVA table.

Step 1: Calculate $\eta$ (Sum of squares for Within group). The total sum of squares is the sum of the sum of squares for "Source of starch" and "Within group". $SS_{Total} = SS_{Between} + SS_{Within}$ $58229.750 = 12112.250 + \eta$ $\eta = 58229.750 - 12112.250$ $\eta = \boxed{46117.500}$

Step 2: Calculate $\beta$ (df for Within group). Mean Square (MS) is calculated as Sum of Squares (SS) divided by df. $MS_{Within} = \frac{SS_{Within}}{df_{Within}}$ $3843.125 = \frac{\eta}{\beta}$ $3843.125 = \frac{46117.500}{\beta}$ $\beta = \frac{46117.500}{3843.125}$ $\beta = \boxed{12}$

Step 3: Calculate $\alpha$ (df for Total). The total degrees of freedom is the sum of the degrees of freedom for "Source of starch" and "Within group". $df_{Total} = df_{Between} + df_{Within}$ $\alpha = 3 + \beta$ $\alpha = 3 + 12$ $\alpha = \boxed{15}$

Step 4: Calculate $\lambda$ (Mean square for Source of starch). $MS_{Between} = \frac{SS_{Between}}{df_{Between}}$ $\lambda = \frac{12112.250}{3}$ $\lambda = \boxed{4037.417}$ (rounded to three decimal places)

Step 5: Calculate $\theta$ (F-statistic). The F-statistic is the ratio of the Mean Square for "Source of starch" to the Mean Square for "Within group". $F = \frac{MS_{Between}}{MS_{Within}}$ $\theta = \frac{\lambda}{3843.125}$ $\theta = \frac{4037.417}{3843.125}$ $\theta = \boxed{1.051}$ (rounded to three decimal places)

: Hypothesis Test for Mean Breaking Strength

We are testing the hypothesis that the mean breaking strength of the fishing line is 9 kg.

Step 1: State the null and alternative hypotheses. The null hypothesis ($H_0$) is that the mean breaking strength is 9 kg. The alternative hypothesis ($H_1$) is that the mean breaking strength is not 9 kg. $H_0: \mu = 9 \text{ kg}$ $H_1: \mu \neq 9 \text{ kg}$

Step 2: Determine the level of significance and critical values. The level of significance is $\alpha = 0.05$. Since this is a two-tailed test, we divide $\alpha$ by 2: $\alpha/2 = 0.05/2 = 0.025$. The critical z-values for a 95% confidence level (or $\alpha=0.05$) in a two-tailed test are $\pm 1.96$.

Step 3: Calculate the test statistic. We use the z-test statistic because the population standard deviation ($\sigma$) is known and the sample size ($n$) is large. The formula is: $z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$ Given: Sample mean ($\bar{x}$) = 10 kg Hypothesized population mean ($\mu_0$) = 9 kg Population standard deviation ($\sigma$) = 0.5 kg Sample size ($n$) = 50 $z = \frac{10 - 9}{0.5 / \sqrt{50}}$ $z = \frac{1}{0.5 / 7.0710678}$ $z = \frac{1}{0.070710678}$ $z \approx \boxed{14.142}$

Step 4: Make a decision. Compare the calculated z-statistic to the critical values. The calculated z-statistic is $14.142$. The critical z-values are $\pm 1.96$. Since $14.142 > 1.96$, the calculated z-statistic falls into the rejection region.

Step 5: State the conclusion. Since the test statistic ($z = 14.142$) is greater than the critical value ($1.96$), we reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance to conclude that the mean breaking strength of the new synthetic fishing line is significantly different from 9 kg.

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