Respuesta :
Answer:
We fail to reject the null hypothesis that the average content of containers of the lubricant is 10 liters, this at the significance level of 0.01
Step-by-step explanation:
Let X be the random variable that represents the content of a container of the lubricant. We have observed n = 10 values, [tex]\bar{x}[/tex] = 10.06 and s = 0.2459. We assume that X is normally distributed.
We have the following null and alternative hypothesis
[tex]H_{0}: \mu = 10[/tex] vs [tex]H_{1}: \mu \neq 10[/tex] (two-tailed alternative)
We will use the test statistic
[tex]T = \frac{\bar{X}-10}{S/\sqrt{10}}[/tex] because we have a small sample size. And the observed value is
[tex]t = \frac{10.06-10}{0.2459/\sqrt{10}} = 0.7716[/tex]
if [tex]H_{0}[/tex] is true, then T has a t distribution with n-1 = 9 degrees of freedom.
The rejection region for a two-tailed alternative and a significance level of 0.01 is given by RR = {t | t < -3.2498 or t > 3.2498}, where 3.2498 is the value such that there is an area of 0.005 above this number and under the density of the t distribution with 9 df.
Because the observed value 0.7716 does not fall inside RR, we fail to reject the null hypothesis.
The hypothesis that the average content of containers of a particular lubricant is 10 liters is not acceptable.
How to classify the hypotheses?
There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
When to use the z test and when to use the t test?
If the sample taken is of size less than t test, then you can use the t test.
If the sample size is larger or equal to 30, you can use the z test.
It is because as the sample size grows more and more, the z test statistic approximates more and more the normal distribution.
For the considered case, we can use:
- [tex]\mu[/tex] = average content of containers of a particular lubricant (of population)
- Null Hypothesis: [tex]H_0: \mu \neq 10[/tex]
- Alternative hypothesis: [tex]H_1: \mu = 10[/tex]
It is because we want to show that the average content of containers of a particular lubricant is 10 liters. Thus, we took alternative hypothesis such that the is = 10
Now, since the sample size is n = 10 < 30, we will use t-test here.
We evaluate 't' as:
[tex]t = \dfrac{\overline{x} - \mu}{s/\sqrt{n}}[/tex] (sample standard deviation is used if population standard deviation is not available).
where, the symbols denote:
- [tex]\overline{x}[/tex] = sample mean
- s = sample standard deviation
- n = sample size
- [tex]\mu[/tex] = hypothesized mean of population
For this case, the sample is of size n = 10
Its observed values are: 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 9.8
The mean is obtained as:
[tex]\overline{x} = \dfrac{\sum{x_i}}{n} = \dfrac{100.6}{10} = 10.06[/tex] (in litres)
The standard deviation of sample would be:
[tex]s = \dfrac{\sum(x_i - \overline{x})^2}{n} = \dfrac{\sum(x_i - 10.06)^2}{10} = 0.544[/tex]
The values for this case are:
- [tex]\overline{x} = 10.06[/tex]
- [tex]s= 0.544[/tex]
- [tex]n = 10[/tex]
- [tex]\mu = 10[/tex]
Thus, we get:
[tex]t = \dfrac{\overline{x} - \mu}{s/\sqrt{n}} = \dfrac{10 - 10.06}{0.544/\sqrt{10}} \approx 0.349[/tex]
The level of significance here is 0.01
At this degree of freedom and level of significance, the critical value of t-test statistic is [tex]t_{\alpha/2} = \pm3.2498[/tex] (two tailed)
Since [tex]t < |t_{\alpha/2}|[/tex] we may accept the null hypothesis, and thus, haven't got significant evidence to accept the alternative hypothesis.
(if the obtained value would be bigger than critical value, then we'd reject null hypothesis).
Therefore, the hypothesis that the average content of containers of a particular lubricant is 10 liters is not acceptable.
Learn more about z-test for single mean here:
https://brainly.com/question/25316952
