Questions: A polling agency says that over 75% of people in the United States prepare and file their income taxes before April 15th. Let p be the true proportion of people in the United States who prepare and file their income taxes before April 15th. In a random survey of 1000 people in the United States, 790 said that they prepare and file their income taxes before April 15th. (a) At 5% level of significance, to test that over 75% of people in the United States prepare and file their income taxes before April 15th, the suitable null and alternative hypotheses are [ Select ] (b) The standardized test statistic is closest to [Select] (c) Calculate the P-value and state the decision of the test at 5% level of significance. [Select]

A polling agency says that over 75% of people in the United States prepare and file their income taxes before April 15th. Let p be the true proportion of people in the United States who prepare and file their income taxes before April 15th. In a random survey of 1000 people in the United States, 790 said that they prepare and file their income taxes before April 15th. (a) At 5% level of significance, to test that over 75% of people in the United States prepare and file their income taxes before April 15th, the suitable null and alternative hypotheses are [ Select ] (b) The standardized test statistic is closest to [Select] (c) Calculate the P-value and state the decision of the test at 5% level of significance. [Select]
Transcript text: A polling agency says that over $75 \%$ of people in the United States prepare and file their income taxes before April 15th. Let $p$ be the true proportion of people in the United States who prepare and file their income taxes before April 15 th. In a random survey of 1000 people in the United States, 790 said that they prepare and file their income taxes before April 15th. (a) At 5\% level of significance, to test that over 75\% of people in the United States prepare and file their income taxes before April 15 th, the suitable null and alternative hypotheses are [ Select ] $\square$ (b) The standardized test statistic is closest to $\square$ [Select] (c) Calculate the $P$-value and state the decision of the test at $5 \%$ level of significance. [Select] $\square$
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Solution

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Solution Steps

Step 1: Hypotheses

We set up the null and alternative hypotheses as follows:

  • Null hypothesis (\(H_0\)): \(p \leq 0.75\)
  • Alternative hypothesis (\(H_1\): \(p > 0.75\)
Step 2: Calculate the Test Statistic

The standardized test statistic \(Z\) is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Where:

  • \(\hat{p} = \frac{790}{1000} = 0.79\) (sample proportion)
  • \(p_0 = 0.75\) (hypothesized population proportion)
  • \(n = 1000\) (sample size)

Substituting the values, we find:

\[ Z = \frac{0.79 - 0.75}{\sqrt{\frac{0.75(1 - 0.75)}{1000}}} = 2.9212 \]

Step 3: Calculate the P-value

The P-value associated with the test statistic \(Z = 2.9212\) is calculated to be:

\[ \text{P-value} = 0.0017 \]

Step 4: Decision of the Test

At a significance level of \(\alpha = 0.05\), we compare the P-value to \(\alpha\):

  • Since \(0.0017 < 0.05\), we reject the null hypothesis.

Final Answer

There is sufficient evidence to support that over 75% of people in the United States prepare and file their income taxes before April 15th.

\(\boxed{Z = 2.9212, \text{ P-value} = 0.0017, \text{ Reject } H_0}\)

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