Respuesta :
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.
[tex]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[/tex]
[tex]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[/tex]
The 10% of the advertised content is:
[tex]0.10\cdot 6.0\;oz=0.6\:oz[/tex]
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:
[tex]H_0: \mu=0.6\\\\H_a:\mu< 0.6[/tex]
The significance level is 0.05.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.102}{\sqrt{5}}=0.046[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.502-0.6}{0.046}=\dfrac{-0.098}{0.046}=-2.148[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=5-1=4[/tex]
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):
[tex]P-value=P(t<-2.148)=0.049[/tex]
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.
Using the t-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, it does appear that the true average amount left is less than 10% of the advertised net contents.
At the null hypothesis, it is tested if the true average amount left is not less than 10% of the advertised net contents, that is, not less than 0.6, hence:
[tex]H_0: \mu \geq 0.6[/tex]
At the alternative hypothesis, it is tested if it is less, hence:
[tex]H_1: \mu < 0.6[/tex]
We will find the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
In this problem, the values of the parameters are: [tex]\mu = 0.6, n = 5[/tex].
- Using a calculator, it is found that [tex]\overline{x} = 0.502, s = 0.1[/tex]
Then, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{0.502 - 0.6}{\frac{0.1}{\sqrt{5}}}[/tex]
[tex]t = -2.19[/tex]
The critical value, for a left-tailed test, as we are testing if the mean is less than value, with a 0.05 significance level and 5 - 1 = 4 df is [tex]t^{\ast} = -2.13[/tex]
Since the test statistic is less than the critical value for the left-tailed test, it does appear that the true average amount left is less than 10% of the advertised net contents.
A similar problem is given at https://brainly.com/question/13873630
