Question

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various con- sumer products. Suppose five 6.0 oz tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the fol- lowing data (consistent with what the cited article reported): .53, .65, .46, .50, .37. Does it appear that the true average amount left is less than 10% of the advertised net contents?

Answer 1

Answer:

Yes. There is enough evidence to support the claim that the remaining toothpaste is significantly is less than 10% of the advertised net content.

Step-by-step explanation:

The sample of the remaining toothpaste is: [.53, .65, .46, .50, .37].

This sample has a size n=5, a mean M=0.502 and standard deviation s=0.102.

$M=\dfrac{1}{5}\sum_{i=1}^{5}(0.53+0.65+0.46+0.5+0.37)\\\\\\ M=\dfrac{2.51}{5}=0.502$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.53-(0.502))^2+...+(0.37-(0.502))^2]}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.001)+(0.022)+(0.002)+(0)+(0.017)]}\\\\\\ s=\sqrt{\dfrac{0.04188}{4}}=\sqrt{0.01047}\\\\\\s=0.102$

The 10% of the advertised content is:

$0.10\cdot 6.0\;oz=0.6\:oz$

Hypothesis test for the population mean:

The claim is that the remaining toothpaste is significantly is less than 10% of the advertised net content.

Then, the null and alternative hypothesis are:

$H_0: \mu=0.6\\\\H_a:\mu< 0.6$

The significance level is 0.05.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.102}{\sqrt{5}}=0.046$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.502-0.6}{0.046}=\dfrac{-0.098}{0.046}=-2.148$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

This test is a left-tailed test, with 4 degrees of freedom and t=-2.148, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.049) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the remaining toothpaste is significantly is less than 10% of the advertised net content.