Question

a political candidate believes that more than 60% of voters in a district support their campaign. to test this, the candidate surveys 150 randomly selected voters and finds that 102 of them say they support the candidate.

at the 0.05 significance level, is there enough evidence to support the claim that more than 60% of voters support the candidate?

a.) what is α? state the null and alternate hypotheses. will you use a left - tailed, right - tailed, or two - tailed test?

α =

$h_0: p = $

$h_1: p$ select an answer

the test is a select an answer test

b.) identify the sampling distribution you will use. what is the value of the test statistic?

the best sampling distribution to use is the select an answer distribution.

the test statistic (z or t value) is =

c.) find or estimate the p - value for the test.

the p - value is =

d.) conclude the test.

based on this we will select an answer the null hypothesis.

Explanation:

Part a:

Step 1: Determine α

The significance level is given as 0.05, so $\alpha = 0.05$.

Step 2: State Null Hypothesis ($H_0$)

The candidate is testing if more than 60% support, so the null hypothesis is that the proportion $p = 0.60$ (since we test against the claim, the null is the status quo or equality). So $H_0: p = 0.60$.

Step 3: State Alternate Hypothesis ($H_1$)

The claim is that more than 60% support, so $H_1: p > 0.60$.

Step 4: Determine Test Type

Since the alternate hypothesis has a "greater than" sign, it is a right - tailed test.

Part b:

Step 1: Sampling Distribution

For a proportion test, when $np_0\geq10$ and $n(1 - p_0)\geq10$ (where $n = 150$, $p_0=0.6$), we use the normal (z) distribution. Calculate $np_0=150\times0.6 = 90$ and $n(1 - p_0)=150\times(1 - 0.6)=60$, both are greater than 10, so we use the normal (z) distribution.

Step 2: Calculate Test Statistic

First, calculate the sample proportion $\hat{p}=\frac{x}{n}=\frac{102}{150}=0.68$.
The formula for the z - statistic in a proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Substitute the values: $p_0 = 0.6$, $\hat{p}=0.68$, $n = 150$.
$\sqrt{\frac{0.6\times(1 - 0.6)}{150}}=\sqrt{\frac{0.6\times0.4}{150}}=\sqrt{\frac{0.24}{150}}=\sqrt{0.0016}=0.04$.
$z=\frac{0.68 - 0.6}{0.04}=\frac{0.08}{0.04}=2$.

Part c:

Step 1: Find P - value

For a right - tailed test with $z = 2$, the P - value is $P(Z>2)$. Looking at the standard normal table, $P(Z\leq2)=0.9772$, so $P(Z > 2)=1 - 0.9772 = 0.0228$.

Part d:

Step 1: Compare P - value and α

We have $\alpha=0.05$ and P - value $= 0.0228$. Since the P - value ($0.0228$) is less than $\alpha$ ($0.05$), we reject the null hypothesis.

Answer:

Part a:

$\alpha=\boldsymbol{0.05}$
$H_0: p=\boldsymbol{0.60}$
$H_1: p\boldsymbol{> 0.60}$
The test is a $\boldsymbol{\text{right - tailed}}$ test

Part b:

The best sampling distribution to use is the $\boldsymbol{\text{normal (z)}}$ distribution.
The test statistic (z or t value) is $=\boldsymbol{2}$

Part c:

The p - value is $=\boldsymbol{0.0228}$

Part d:

Based on this we will $\boldsymbol{\text{reject}}$ the null hypothesis.