Question

A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decrease in the amount of food consumption. To test this hypothesis, they monitored food intake for a group of 44 patients who were randomized into two equal groups. The treatment group ate lunch while playing solitaire, and the control group ate lunch without any added distractions. Patients in the treatment group ate an average of 52.1 grams of biscuits, with a standard deviation of 45.1 grams, and patients in the control group ate an average of 27.1 grams of biscuits, with a standard deviation of 26.4 grams. Do these data provide convincing evidence that the average food intake (measured in grams of biscuits consumed) is different for the patients in the treatment group? Assume that conditions for inference are satisfied.

Answer 1

Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is tested if the consumption is not different, that is, if the subtraction of the means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, it is tested if the consumption is different, that is, if the subtraction of the means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

Two groups of 22 patients, hence, the standard errors are:

$s_1 = \frac{45.1}{\sqrt{22}} = 9.6154$

$s_2 = \frac{26.4}{\sqrt{22}} = 5.6285$

The distribution of the differences is has:

$\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14$

The test statistic is given by:

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

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{25 - 0}{11.14}$

$t = 2.2438$

The p-value of the test is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 2.2438 and 22 + 22 - 2 = 42 df.

• Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813