Question

Based on an LG smartphone survey, assume that 51% of adults with smartphones use
them in theaters. In a separate survey of 250 adults with smartphones, it is found that
109 use them in theaters.

Answer 1

Using the z-distribution, it is found that the sample is enough evidence that the proportion is below 51%.

What are the hypotheses tested?

At the null hypotheses, it is tested if the proportion is of 51%, that is:

$H_0: p = 0.51$

At the alternative hypotheses, it is tested if the proportion is less than 51%, hence:

$H_1: p < 0.51$.

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are given as follows:

$p = 0.51, n = 250, \overline{p} = \frac{109}{250} = 0.436$

Hence, the test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.436 - 0.51}{\sqrt{\frac{0.51(0.49)}{250}}}$

z = -2.34.

What is the decision?

Considering a left-tailed test, as we are testing if the proportion is less than a value, with a standard significance level of 0.05, the critical value is of $z^{\ast} = -1.645$.

Since the test statistic is less than the critical value for the left tailed test, we reject the null hypothesis, and find that the sample is enough evidence that the proportion is below 51%.

More can be learned about the z-distribution at https://brainly.com/question/26454209

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