Questions: In a Leichtman Research Group survey of 850 TV households, 74.6% of them had at least one Internet-connected TV device (for example, Smart TV, standalone streaming device, connected video one Internet-connected TV device is equal to 81%. Test that claim using a 0.10 significance level. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all homes with at least one Internet-connected TV device. Identify the null and alternative hypotheses. H2: p = B 1 H1: p = 81 (Type integers or decimals. Do not round.) Identify the test statistic z = -4.61 (Round to two decimal places as needed.) Identify the P-value. P -value = (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There sufficient evidence to the claim that the percentage of all homes with at least one internet-connected TV device is equal to 81%

In a Leichtman Research Group survey of 850 TV households, 74.6% of them had at least one Internet-connected TV device (for example, Smart TV, standalone streaming device, connected video one Internet-connected TV device is equal to 81%. Test that claim using a 0.10 significance level. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all homes with at least one Internet-connected TV device. Identify the null and alternative hypotheses. H2: p = B 1 H1: p = 81 (Type integers or decimals. Do not round.) Identify the test statistic z = -4.61 (Round to two decimal places as needed.) Identify the P-value. P -value = (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There sufficient evidence to the claim that the percentage of all homes with at least one internet-connected TV device is equal to 81%
Transcript text: In a Leichtman Research Group survey of 850 TV households, $74.6 \%$ of them had at least one Internet-connected TV device (for example, Smart TV, standalone streaming device, connected video one Internet-connected TV device is equal to $81 \%$. Test that claim using a 0.10 significance level. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let $p$ denote the population proportion of all homes with at least one Internet-connected TV device. Identify the null and alternative hypotheses. \[ \begin{array}{l} H_{2}: p=B 1 \\ H_{1}: p=81 \end{array} \] (Type integers or decimals. Do not round.) Identify the test statistic \[ z=-4.61 \] (Round to two decimal places as needed.) Identify the P-value. P -value $=$ $\square$ (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. $\square$ the null hypothesis. There $\square$ sufficient evidence to $\square$ the claim that the percentage of all homes with at least one internet-connected TV device is equal to $81 \%$
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Solution

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Solution Steps

Step 1: State the Hypotheses

We are testing the claim regarding the population proportion \( p \) of all homes with at least one Internet-connected TV device. The null and alternative hypotheses are defined as follows:

\[ \begin{align_} H_0: & \quad p = 0.81 \\ H_1: & \quad p \neq 0.81 \end{align_} \]

Step 2: Calculate the Test Statistic

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

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

Substituting the values:

  • Sample proportion \( \hat{p} = 0.746 \)
  • Hypothesized proportion \( p_0 = 0.81 \)
  • Sample size \( n = 850 \)

We find:

\[ Z = \frac{0.746 - 0.81}{\sqrt{\frac{0.81(1 - 0.81)}{850}}} = -4.756 \]

Step 3: Determine the P-value

The P-value associated with the calculated test statistic \( Z = -4.756 \) is:

\[ \text{P-value} = 0.0 \]

Step 4: Identify the Critical Region

For a significance level of \( \alpha = 0.10 \) in a two-tailed test, the critical values are:

\[ Z < -1.645 \quad \text{or} \quad Z > 1.645 \]

Step 5: Make a Decision

Since the calculated test statistic \( Z = -4.756 \) falls within the critical region \( Z < -1.645 \), we reject the null hypothesis.

Step 6: Conclusion

There is sufficient evidence to reject the claim that the percentage of all homes with at least one Internet-connected TV device is equal to \( 81\% \).

Final Answer

\(\boxed{\text{Reject the null hypothesis. There is sufficient evidence to reject the claim that } p = 0.81.}\)

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