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Stacy recorded the number of minutes she used her
smartphone each day for one week before using an
app that awards points for decreased phone usage,
and for one week after using the app. Stacy's
alternative hypothesis is Ha:X< 114.6.
Before 103 127 87 12495 136 130
After 83 94 1201057598 96
Stacy used a simulation to randomize her data 200
times. She found that the difference between sample
means over 200 trials is approximately normally-
distributed, with a mean of -0.2 and a standard
deviation of 9.1. Which of the following conclusions
is most appropriate?
Phone use did not change.
Phone use increased and then decreased.
Phone use increased.
Phone use decreased.

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Using the t-distribution, it is found that the most appropriate conclusion for the hypotesis test is given by:

Phone use did not change.

What are the hypothesis tested?

At the null hypothesis, we test if the mean has not changed, that is:

$$H_0: \mu = 0$$

At the alternative hypothesis, we test if it has changed, that is:

$$H_1: \mu \neq 0$$.

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 values of those parameters are as follows:

$$\overline{x} = -0.2, \mu = 0, s = 9.1, s = 200$$

Hence, the test statistic is given by:

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

$$t = \frac{-0.2 - 0}{\frac{9.1}{\sqrt{200}}}$$

t = 0.31.

What is the conclusion?

Considering a two-tailed test, as we are testing if the mean is different of a value, with a standard significance level of 0.05 and 200 - 1 = 199 df, the critical value is of $$|t^{\ast}| = 1.972$$.

Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis and the conclusion is:

Phone use did not change.

More can be learned about the t-distribution at

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