Question

10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Answer 1

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:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

What is the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ 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:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$.

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

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 $t^{\ast} = 2.4$.

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