Respuesta :
Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.
What are the hypothesis tested?
At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:
[tex]H_0: \mu = 0[/tex]
At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:
[tex]H_1: \mu > 0[/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 parameters are given as follows:
[tex]\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50[/tex].
Hence, the test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}[/tex]
t = 2.23
What is the conclusion?
Considering a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.01 and 50 - 1 = 49 df, the critical value is given by [tex]t^{\ast} = 2.4[/tex].
Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.
More can be learned about the t-distribution at https://brainly.com/question/26454209
