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.
<h3>What are the hypothesis tested?</h3>
At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:
![H_0: \mu = 0](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%200)
At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:
![H_1: \mu > 0](https://tex.z-dn.net/?f=H_1%3A%20%5Cmu%20%3E%200)
<h3>What is the test statistic?</h3>
The test statistic is given by:
![t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Coverline%7Bx%7D%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
In this problem, the parameters are given as follows:
.
Hence, the test statistic is given by:
![t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Coverline%7Bx%7D%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1.5%20-%200%7D%7B%5Cfrac%7B4.75%7D%7B%5Csqrt%7B50%7D%7D%7D)
t = 2.23
<h3>What is the conclusion?</h3>
Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by
.
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 brainly.com/question/26454209