Question

Suppose that you were trying to determine if the percentage of Americans that speak a second language is less than 30% . In 2016, the General Social Survey had a question that asked its participants if they spoke a second language. Out of 2867, 816 said yes.

Answer 1

Answer:

The test statistic value is -1.74.

The p-value of the test is 0.0409.

Step-by-step explanation:

A single proportion z-test can be used to test whether the proportion of American that speak a second language is less than 30%.

The hypothesis can be defined as follows:

H₀: The proportion of American that speak a second language is 0.30, i.e.      p = 0.30.

Hₐ: The proportion of American that speak a second language is less than 0.30, i.e. p < 0.30.

In a survey of n = 2867 Americans X = 816 said they speak a second language.

The sample proportion of Americans who speak a second language is:

$\hat p=\frac{X}{n}=\frac{816}{2867}=0.285$

Compute the test statistic value as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.285-0.30}{\sqrt{\frac{0.30(1-0.30)}{2867}}}=-1.74$

The test statistic value is -1.74.

Compute the p-value as follows:

$p-value=P(Z<-1.74)\\=1-P(Z<1.74)\\=1-0.9591\\=0.0409$

*Use a z-table for the probability.

The p-value of the test is 0.0409.

Answer 2

Using the z-distribution, as we are working with a proportion, it is found that this sample is enough evidence that the percentage is of less than 30%.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is of 30%, that is:

$H_0: p = 0.3$

At the alternative hypothesis, it is tested if the proportion is less than 30%, that is:

$H_1: p < 0.3$

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:

$p = 0.3, n = 2867, \overline{p} = \frac{816}{2867} = 0.2846$

Hence, the value of the test statistic is given by:

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

$z = \frac{0.2846 - 0.3}{\sqrt{\frac{0.3(0.7)}{2867}}}$

$z = -1.8$

What is the decision?

Considering a left-tailed test, as we are testing if the proportion is less than a value, and 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, it is found that this sample is enough evidence that the percentage is of less than 30%.

To learn more about the z-distribution, you can check https://brainly.com/question/26454209