Question

A cannery prints "weight 16 ounces" on its label. The quality control supervisor selects nine cans at random and weighs them. She finds \(\bar{x}=15.7\) and \(s=.5 .\) Do the data present sufficient evidence to indicate that the mean weight is less than that claimed on the label?

Short answer

Answer: No, there is not sufficient evidence to claim that the mean weight of the cans is less than the declared 16 ounces.

Step-by-step solution

Null and alternative hypotheses

The null hypothesis (H0) is that there is no difference between the claimed weight (16 ounces) and the true mean weight; in other words, the mean weight is equal to 16 ounces. The alternative hypothesis (H1) is that the mean weight is less than 16 ounces. Mathematically, this can be written as: H0: \(\mu = 16\) H1: \(\mu < 16\)

Choose the level of significance

The level of significance, denoted as \(\alpha\), is the probability of making a Type I error (rejecting the null hypothesis when it is true). A common level of significance is 0.05, meaning that there is a 5% chance of making a Type I error. We will use this value in this exercise.

Calculate the test statistic (t-value)

The formula for the t-value known as test statistic is: \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\) where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Plugging in the given values, we have: \(t = \frac{15.7 - 16}{\frac{0.5}{\sqrt{9}}} = \frac{-0.3}{\frac{0.5}{3}} = -1.8\)

Find the critical value

To find the critical value, we need the degrees of freedom, which is equal to the sample size (n) minus 1. In this case, we have 9 samples, so the degrees of freedom are 9 - 1 = 8. With a significance level of 0.05 and 8 degrees of freedom, we can look up the critical value for a one-sided t-test in a t-distribution table, or we can use a statistical calculator, which gives us a critical value of approximately -1.86.

Compare the test statistic to the critical value

Our calculated t-value is -1.8, and the critical value is -1.86. Since our t-value is greater than the critical value, we fail to reject the null hypothesis. In conclusion, based on the sample data, there is not sufficient evidence to claim that the mean weight of the cans is less than the declared 16 ounces.

Key concepts

Null Hypothesis

The concept of a null hypothesis is a cornerstone of statistical hypothesis testing. It's a statement posited by researchers to suggest no effect or no difference. In this exercise, the null hypothesis (\(H_0\) ) claims that the average weight of the cans is exactly 16 ounces, as printed on the label. By stating \(H_0: \mu = 16\), we propose that any apparent deviation in sample data is due to random chance rather than a real change in the population average. This hypothesis forms the basis for comparison against alternative possibilities in the testing process.

Alternative Hypothesis

Hand in hand with the null hypothesis is the alternative hypothesis (\(H_1\)), which posits what researchers try to demonstrate. For our cannery example, the alternative hypothesis is that the mean weight of the cans is actually less than 16 ounces, expressed mathematically as \(H_1: \mu < 16\). This hypothesis is one-sided, as we are specifically interested in finding if the mean weight is less, not just different. It suggests there is sufficient evidence for a claim contrary to what the null hypothesis states, based on the data collected.

Significance Level

The significance level, denoted as \(\alpha\), is like a threshold for making decisions in hypothesis testing. In this example, it’s set to 0.05, which means we accept a 5% risk of concluding there is a difference when there actually isn't. Choosing a significance level determines how strong the evidence must be to reject the null hypothesis. If the probability of observing such extreme results, assuming the null hypothesis is true, is less than \(\alpha\), we have enough evidence to go against it. This level helps to control the likelihood of making a Type I error, which is rejecting a true null hypothesis.

t-Statistic

The t-statistic is a key figure utilized to analyze data in hypothesis testing. It helps us determine if the sample mean significantly differs from the population mean as proposed in the null hypothesis. The calculation involves \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\), where:

• \(\bar{x}\) is the sample mean
• \(\mu\) is the hypothesized population mean
• \(s\) is the sample standard deviation
• \(n\) is the sample size

Substituting the values from our exercise (\(\bar{x} = 15.7\), \(\mu = 16\), \(s = 0.5\), \(n = 9\)), we get a t-value of \(-1.8\). The t-statistic lets us decide if the sample results are unusual enough to reject the null hypothesis.

Critical Value

The critical value is a benchmark in hypothesis testing used to determine the cutoff for rejecting the null hypothesis. Depending on the chosen significance level and the degrees of freedom (here, \(n-1=8\)), we refer to statistical tables or software to find the critical value. For this one-tailed test with \(\alpha = 0.05\), the critical value is approximately \(-1.86\). We compare this value with our computed t-statistic. If our t-statistic were less than the critical value, it would mean the sample data falls in the region where we'd reject the null hypothesis. In this case, because \(-1.8\)is greater than \(-1.86\), we do not have enough proof to reject the null hypothesis.