A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A

A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the test statistic for testing the null hypothesis H0: p = 0.5 versus the alternative hypothesis Ha: p > 0.5 is…

The Correct Answer and Explanation is:

To find the observed value of the test statistic for testing the null hypothesis (H_0: p = 0.5) versus the alternative hypothesis (H_a: p > 0.5), we use a one-sample proportion z-test.

Step 1: Define the null hypothesis and alternative hypothesis

  • Null hypothesis: ( H_0: p = 0.5 ) (The proportion of voters in favor of Candidate A is 0.5)
  • Alternative hypothesis: ( H_a: p > 0.5 ) (The proportion of voters in favor of Candidate A is greater than 0.5)

Step 2: Find the test statistic formula

The formula for the z-test statistic for proportions is:

[
z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0(1 – p_0)}{n}}}
]

Where:

  • ( \hat{p} ) is the sample proportion (the proportion of voters in favor of Candidate A in the sample),
  • ( p_0 ) is the hypothesized population proportion (0.5 under the null hypothesis),
  • ( n ) is the sample size.

Step 3: Calculate the sample proportion and substitute the values

  • The sample size ( n = 100 ),
  • The number of voters in favor of Candidate A is 59, so the sample proportion ( \hat{p} = \frac{59}{100} = 0.59 ),
  • The hypothesized population proportion ( p_0 = 0.5 ).

Now, substitute these values into the formula:

[
z = \frac{0.59 – 0.5}{\sqrt{\frac{0.5(1 – 0.5)}{100}}}
]

[
z = \frac{0.09}{\sqrt{\frac{0.25}{100}}}
]

[
z = \frac{0.09}{\sqrt{0.0025}} = \frac{0.09}{0.05} = 1.8
]

Step 4: Interpret the result

The observed value of the test statistic is 1.8.

This z-test statistic can be used to compare against critical values from the standard normal distribution (z-table) to make a decision. For a right-tailed test with a significance level of 0.05, the critical z-value is approximately 1.645. Since 1.8 > 1.645, we would reject the null hypothesis at the 0.05 significance level, suggesting that there is sufficient evidence to support the claim that the proportion of voters in favor of Candidate A is greater than 0.5.

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