Question

a manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. a consumer advocacy group wants to perform a hypothesis test to determine whether the mean amount of juice is actually less than this. they test 109 bottles and found a sample mean of 16.07 ounces. assume that the population standard deviation is 0.10 ounces. test their claim at the 0.01 level of significance.

Answer 1

At the 0.01 significance level, with a sample mean of 16.07 ounces, a one-tailed z-test rejects the manufacturer's claim, indicating evidence that the mean juice amount is less than 16.1 ounces.

To test the manufacturer's claim that the mean amount of juice is less than 16.1 ounces, you can perform a one-sample z-test. The null hypothesis
`(\(H_0\))` and alternative hypothesis
`(\(H_1\))` can be stated as follows:

`\[ H_0: \mu = 16.1 \]\[ H_1: \mu < 16.1 \]`

Here,
`\(\mu\)` is the population mean amount of juice in 16-ounce bottles.

The test statistic z can be calculated using the formula:

`\[ z = \frac{\bar{x} - \mu_0}{(\sigma)/(√(n))} \]`

where:

-
`\(\bar{x}\)` is the sample mean,

-
`\(\mu_0\)` is the hypothesized population mean under the null hypothesis,

-
`\(\sigma\)` is the population standard deviation, and

- n is the sample size.

Given that:

-
`\(\bar{x} = 16.07\)` ounces,

-
`\(\mu_0 = 16.1\)` ounces,

-
`\(\sigma = 0.10\)` ounces,

- n = 109, and

- the level of significance
`(\(\alpha\))` is 0.01,

we can proceed with the calculations:

`\[ z = (16.07 - 16.1)/((0.10)/(√(109))) \]`

Calculating this gives the test statistic z.

`\[ z \approx (-0.03)/((0.10)/(√(109))) \]\[ z \approx (-0.03)/((0.10)/(√(109))) \approx (-0.03)/((0.10)/(10.44)) \approx (-0.03)/(0.00956) \approx -3.13 \]`

Now, you compare the test statistic z to the critical value for a one-tailed test at a 0.01 level of significance. You can find this critical value using a standard normal distribution table or a calculator.

If
`\(z < -z_(\alpha)\)`, then you reject the null hypothesis.

For a one-tailed test at a 0.01 level of significance,
`\(z_(\alpha) \approx -2.33\)`.

Since -3.13 < -2.33, you reject the null hypothesis.

Therefore, there is enough evidence to suggest that the mean amount of juice is less than 16.1 ounces at the 0.01 level of significance.