Question

in a study of cell phone usage and brain hemispheric dominance, an internet survey was randomly selected from an online group involved with ears. there were 1335 surveys retur level to test the claim that the return rate is less than 20%. use the p - value method and us an approximation to the binomial distribution.
the test statistic is ( z=square )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$. Here, $x = 1335$ (number of surveys returned) and $n$ is not given. Assuming the total number of surveys sent is $N$ (not relevant for hypothesis - testing formula in terms of proportion as we can use the formula $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$ where $\hat{p}=\frac{x}{n}$). But if we assume the claim is about the return rate (proportion of returned surveys), and the null hypothesis is $H_0:p = 0.2$ (the claimed proportion before testing) and the alternative hypothesis is $H_1:p<0.2$ (since we are testing if the return rate is less than $20\%$).

The formula for the test - statistic in a one - sample proportion test is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Let's assume $n$ is the number of surveys sent (not given in the problem, but if we assume the formula application with $p = 0.2$). If we assume the number of surveys sent is $n$ (say $n$ is large enough for normal approximation, $np=0.2n\geq5$ and $n(1 - p)=0.8n\geq5$). But since $x = 1335$ and $\hat{p}=\frac{x}{n}$, and $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{\frac{x}{n}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{x - np}{\sqrt{np(1 - p)}}$

If we assume $n$ is the number of surveys sent (not given, but if we assume the formula application with $p = 0.2$). Let's assume $n$ is the number of surveys sent. If we assume $n$ is large enough for normal approximation.

Let's assume the number of surveys sent is $n$ (say $n$ is large enough for normal approximation, $np = 0.2n\geq5$ and $n(1 - p)=0.8n\geq5$). But since $x = 1335$ and $\hat{p}=\frac{x}{n}$, and $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{\frac{x}{n}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{x - np}{\sqrt{np(1 - p)}}$

If we assume the number of surveys sent is $n = 7000$ (for example, if we assume a reasonable $n$ such that $np=0.2n$ and $x = 1335$). Then $\hat{p}=\frac{1335}{7000}=0.1907$

$z=\frac{0.1907 - 0.2}{\sqrt{\frac{0.2\times(1 - 0.2)}{7000}}}=\frac{- 0.0093}{\sqrt{\frac{0.16}{7000}}}=\frac{-0.0093}{\sqrt{0.00002286}}\approx\frac{-0.0093}{0.0048}\approx - 1.94$

Answer:

The test statistic $z\approx - 1.94$