Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The ar

Question

11 days ago

7

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The ar

ticle "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?

Mathematics

1 answer:

Inessa [12.1K] · Answer 1 · 2026-03-29T13:32:58+03:00

Answer:

Indeed. Ample evidence substantiates the assertion that the leftover toothpaste is considerably below 10% of the stated net amount.

To explain step-by-step:

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

This dataset has a size of n=5, with a mean of M=0.502 and a standard deviation of 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 calculates to:

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

Hypothesis testing regarding the population mean:

The assertion posits that the remaining toothpaste is substantially less than 10% of the advertised net amount.

Thus, the null hypothesis and alternative hypothesis can be stated as follows:

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

The significance threshold is set at 0.05.

To find the estimated standard error of the mean, we utilize the formula:

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

Subsequently, we can compute the t-statistic as follows:

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

The degrees of freedom applicable to this sample size is:

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

This is a left-tailed test with 4 degrees of freedom resulting in t=-2.148. Thus, the P-value is evaluated as (using a t-table):

$P-value=P(t$

Given that the P-value (0.049) falls below the significance threshold (0.05), we conclude that the effect is significant.

The null hypothesis gets rejected.

There is ample evidence that affirms the claim that the remaining toothpaste is considerably below 10% of the stated net content.