Respuesta :
Using the t-distribution, as we have the standard deviation for the sample, it is found that since the test statistic is less than the critical value for the right-tailed test, there is evidence to support the belief that the mean is less than 15. 25 ounces.
What are the hypothesis tested?
At the null hypothesis, it is tested if the mean is of 15.25 ounces, that is:
[tex]H_0: \mu = 15.25[/tex]
At the alternative hypothesis, it is tested if the mean is of less than 15.25 pounds, that is:
[tex]H_1: \mu < 15.25[/tex].
What is the test statistic?
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]\overline{x} = 15.18, \mu = 15.25, s = 0.12, n = 36[/tex].
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{15.18 - 15.25}{\frac{0.12}{\sqrt{36}}}[/tex]
[tex]t = -3.5[/tex]
What is the decision?
Considering a left-tailed test, as we are testing if the mean is less than a value, with 36 - 1 = 35 df and the standard significance level of 0.05, it is found that the critical value is [tex]t^{\ast} = -2.03[/tex].
Since the test statistic is less than the critical value for the right-tailed test, there is evidence to support the belief that the mean is less than 15. 25 ounces.
More can be learned about the t-distribution at https://brainly.com/question/16313918
