Respuesta :
Answer:
[tex]t=\frac{11.516-11.5}{\frac{0.0950}{\sqrt{20}}}=0.753[/tex]
[tex]p_v =2*P(t_{(19)}>0.753)=0.461[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 11.5 at 5% of signficance.
Step-by-step explanation:
Data given and notation
Data given: 11.627 11.613 11.493 11.602 11.360 11.374 11.592 11.458 11.552 11.463 11.383 11.715 11.485 11.509 11.429 11.477 11.570 11.623 11.472 11.531
We can calculate the sample mean and deviation with these formulas:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=11.516[/tex] represent the sample mean
[tex]s=0.0950[/tex] represent the sample standard deviation
[tex]n=20[/tex] sample size
[tex]\mu_o =11.5[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 11.5, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 11.5[/tex]
Alternative hypothesis:[tex]\mu \neq 11.5[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{11.516-11.5}{\frac{0.0950}{\sqrt{20}}}=0.753[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=20-1=19[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(19)}>0.753)=0.461[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 11.5 at 5% of signficance.
Using the t-distribution, it is found that since the absolute value of the test statistic is less than the absolute value of the critical value for the two-tailed test, there is not convincing evidence that the mean hardness of the tablets differs from the target value.
What are the hypothesis?
- At the null hypothesis, it is tested if the mean hardness does not differ from the target value of m = 11.5, that is:
[tex]H_0: \mu = 11.5[/tex]
At the alternative hypothesis, it is tested if it differs, that is:
[tex]H_1: \mu \neq 11.5[/tex]
Finding the test statistic
We have 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.
With the help of a calculator, it is found the values of the parameters are:
[tex]\overline{x} = 11.5164, \mu = 11.5, n = 20, s = 0.094955059[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{11.5164 - 11.5}{\frac{0.094955059}{\sqrt{20}}}[/tex]
[tex]t = 0.77[/tex]
Reaching a conclusion
- The critical value for a two-tailed test, as we are testing if the mean is different of a value, with 20 - 1 = 19 df and a significance level of 0.05, using a critical value calculator for the t-distribution, is of [tex]|t^{\ast}| = 2.093[/tex]
Since the absolute value of the test statistic is less than the absolute value of the critical value for the two-tailed test, there is not convincing evidence that the mean hardness of the tablets differs from the target value.
To learn more about the t-distribution, you can take a look at https://brainly.com/question/26062194
