Respuesta :
Using the z-distribution, it is found that we reject the null hypothesis, which means that her claim is not validated.
What are the hypotheses tested?
At the null hypotheses, it is tested if the mean is of 110, that is:
[tex]H_0: \mu = 110[/tex].
At the alternative hypotheses, it is tested if the mean is less than 110, that is:
[tex]H_1: \mu < 110[/tex].
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\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.
- [tex]\sigma[/tex] is the standard deviation of the population.
- n is the sample size.
In this problem, we have that the parameters are given as follows:
[tex]\overline{x} = 105, \mu = 110, \sigma = 15, n = 50[/tex]
Hence the test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{105 - 110}{\frac{15}{\sqrt{50}}}[/tex]
z = -2.36.
What is the decision?
Using a z-distribution calculator, z = -2.36 has a p-value that is less than the standard significance level of 0.05, hence we reject the null hypothesis, which means that her claim is not validated.
More can be learned about the z-distribution at https://brainly.com/question/16313918
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