-2xSquared +4x-1 minus 6x squared+3x-9
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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:
At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:
The test statistic is given by:
The parameters are:
is the sample mean.
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:
.
Hence, the test statistic is given by:
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