# Solved (Free): ABC Company claims that the proportion of its employees investing in individual investment accounts is higher than national proportion

#### ByDr. Raju Chaudhari

Apr 12, 2021

ABC Company claims that the proportion of its employees investing in individual investment accounts is higher than national proportion of 50%. A survey of 100 employees in ABC Company indicated that 58 of them have invested in an individual investment account. Assume Mimi wants to use a 0.10 significance level to test the claim.

(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic.
(c) Determine the P-value for this test. Show all work; writing the correct P-value.
(d) Is there sufficient evidence to support ABC Company's claim that the proportion of its employees investing in individual investment accounts is higher than 50%?

#### Solution

Given that $n = 100$. The sample proportion is $\hat{p}=\frac{X}{n}=\frac{58}{100}=0.58$.

(a) Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : p = 0.5$ against $H_1 : p > 0.5$ ($\text{right-tailed}$)

(b) Test Statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned}
which follows $N(0,1)$ distribution.

The significance level is $\alpha = 0.1$.

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.28}$.

The rejection region (i.e. critical region) for the hypothesis testing problem is $\text{Z > 1.28}$.

The test statistic is
 \begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.58-0.5}{\sqrt{\frac{0.5* (1-0.5)}{100}}}\\ & =1.6 \end{aligned}

Decision:

The test statistic is $Z =1.6$ which falls $inside$ the critical region, we $\text{reject}$ the null hypothesis.

(c) $p$-value:

This is a $\text{right-tailed}$ test, so the p-value is the area to the right of the test statistic ($Z=1.6$). Thus the $p$-value = $P(Z>1.6) =0.0548$.

The p-value is $0.0548$ which is $\text{less than}$ the significance level of $\alpha = 0.1$, we $\text{reject}$ the null hypothesis.

(d) As the p-value is less than the level of significance, there sufficient evidence to support ABC Company's claim that the proportion of its employees investing in individual investment accounts is higher than 50% at $0.1$ level of significance.