A caviar producing company packs its product in containers. They claim that the average weight of the caviar in the containers is 1 kilogram. They sell each container for 8,000 CAD. We had a budget of approximately 90,000 CAD. So, we could buy 11 of the containers and weighted the caviar in each. Here are the results in Kilograms:
0.89, 1.01, 1.00, 0.90, 0.90, 0.91, 0.91, 0.89, 0.95, 1.00, 0.96.
If we assume that the weights are normally distributed, is there sufficient evidence for us to believe that the actual average weight of caviar in the containers is less than what the producer claims it is. Use the critical value approach with α = 0.05.
My work:
H0: μ = 1 and Ha: μ < 1. I then found the mean of the results, x̄ = 10.32/11 = 0.9382, along with the standard deviation: 0.0475.
Using this, I found the test statistic: t=(0.9382-1)/[0.0475/sqrt(11)] = -0.0618/0.0143 = -4.3217 and then the critical value: t(n-1),α = t10,0.05 = 1.812.
-4.3217 < -1.812 is true, so therefore H0 should not be rejected, there is enough evidence for us to believe that the actual average weight of caviar in the containers is less than what the producer claims it is.
I'm not sure if I completed the problem properly, any help/guidance is great, thanks!
The test is one-sided: $$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_a : \mu < \mu_0$$ where $\mu_0 = 1$ is the hypothesized mean. The test statistic is $$T = \frac{\bar x - \mu_0}{s/\sqrt{n}} \sim t_{n-1},$$ where $$\bar x = \frac{1}{n}\sum_{i=1}^n x_i = 0.938182$$ is the sample mean, $$s = \left(\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2 \right)^{1/2} = 0.0475012$$ is the sample standard deviation, and $t_{n-1}$ is the Student $t$ distribution with $n-1$ degrees of freedom. $n = 11$ is the sample size. The test statistic is $$T = -4.31626.$$ At a significance level of $\alpha = 0.05$, we reject $H_0$ if $T < -t_{n-1,\alpha}$ where $t_{n-1,\alpha}$ is the upper $100\alpha$ percentile of the Student $t$ distribution with $n-1$ degrees of freedom. In our case, we have $-t_{10,0.05} = -1.81246$, so our conclusion is to reject $H_0$ in favor of $H_a$: the sample contains sufficient evidence to suggest with at least $95\%$ confidence that the mean weight of caviar is less than $1$ kilogram per container.