Respuesta :
Using the t-distribution, as we have the standard deviation for the sample, it is found that it can be concluded that the holes are being punched an average of 1. 84 cm.
What are the hypotheses tested?
At the null hypotheses, it is tested if the mean is of 1.84 cm, that is:
[tex]H_0: \mu = 1.84[/tex]
At the alternative hypothesis, it is tested if the mean is different of 1.84 cm, that is:
[tex]H_1: \mu \neq 1.84[/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.
Considering the hypothesis and the given sample, it is found that the parameters are as follows:
[tex]\overline{x} = 1.85, \mu = 1.84, s = 0.0235, n = 12[/tex].
Hence, the test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{1.85 - 1.84}{\frac{0.0235}{\sqrt{12}}}[/tex]
[tex]t = 1.47[/tex]
What is the decision?
Considering a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.1 and 12 - 1 = 11 df, the critical value is of [tex]|t^{\ast}| = 1.7959[/tex]
Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis and it can be concluded that the holes are being punched an average of 1. 84 cm.
More can be learned about the t-distribution at https://brainly.com/question/16313918
