$$ 1 \leftarrow $$ Claim. Fewer than $$ 9.5 \% $$ of homes have only a landline telephone and no wireless phone Sample data. A survey by the National Center for Health Statistics showed that among 14,473 homes $$ 5.83 \% $$ had landline

Question

UpStudy AI · Accepted Answer

To analyze the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone, we can set up a hypothesis test.

### Step 1: Define the Hypotheses
- Null Hypothesis ($$ H_0 $$): $$ p \geq 0.095 $$ (The proportion of homes with only a landline is at least $$ 9.5\% $$)
- Alternative Hypothesis ($$ H_a $$): $$ p < 0.095 $$ (The proportion of homes with only a landline is less than $$ 9.5\% $$)

### Step 2: Gather Sample Data
From the survey:
- Sample size ($$ n $$) = 14,473
- Sample proportion ($$ \hat{p} $$) = $$ 5.83\% = 0.0583 $$

### Step 3: Calculate the Test Statistic
We can use the formula for the test statistic for a proportion:

$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
$$

Where:
- $$ p_0 = 0.095 $$ (the hypothesized population proportion)
- $$ \hat{p} = 0.0583 $$ (the sample proportion)
- $$ n = 14,473 $$ (the sample size)

### Step 4: Calculate the Test Statistic
Let's calculate the test statistic $$ z $$.

First, we need to calculate the standard error:

$$
SE = \sqrt{\frac{p_0(1 - p_0)}{n}}
$$

Now, we can compute both the standard error and the test statistic.

1. Calculate the standard error.
2. Calculate the test statistic $$ z $$.

Let's perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\frac{\left(0.095\left(1-0.095\right)\right)}{14473}}$$
- step1: Remove the parentheses:
  $$\sqrt{\frac{0.095\left(1-0.095\right)}{14473}}$$
- step2: Subtract the numbers:
  $$\sqrt{\frac{0.095\times 0.905}{14473}}$$
- step3: Multiply the numbers:
  $$\sqrt{\frac{0.085975}{14473}}$$
- step4: Divide the terms:
  $$\sqrt{\frac{3439}{578920000}}$$
- step5: Use the properties of radicals:
  $$\frac{\sqrt{3439}}{\sqrt{578920000}}$$
- step6: Simplify the expression:
  $$\frac{\sqrt{3439}}{200\sqrt{14473}}$$
- step7: Multiply by the Conjugate:
  $$\frac{\sqrt{3439}\times \sqrt{14473}}{200\sqrt{14473}\times \sqrt{14473}}$$
- step8: Multiply the numbers:
  $$\frac{\sqrt{49772647}}{200\sqrt{14473}\times \sqrt{14473}}$$
- step9: Multiply the numbers:
  $$\frac{\sqrt{49772647}}{2894600}$$
Calculate or simplify the expression $$ (0.0583 - 0.095) / \sqrt((0.095 * (1 - 0.095)) / 14473) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(0.0583-0.095\right)}{\sqrt{\frac{\left(0.095\left(1-0.095\right)\right)}{14473}}}$$
- step1: Remove the parentheses:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.095\left(1-0.095\right)}{14473}}}$$
- step2: Subtract the numbers:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.095\times 0.905}{14473}}}$$
- step3: Multiply the numbers:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.085975}{14473}}}$$
- step4: Divide the terms:
  $$\frac{0.0583-0.095}{\sqrt{\frac{3439}{578920000}}}$$
- step5: Simplify the root:
  $$\frac{0.0583-0.095}{\frac{\sqrt{49772647}}{2894600}}$$
- step6: Subtract the numbers:
  $$\frac{-0.0367}{\frac{\sqrt{49772647}}{2894600}}$$
- step7: Convert the expressions:
  $$\frac{-\frac{367}{10000}}{\frac{\sqrt{49772647}}{2894600}}$$
- step8: Multiply by the reciprocal:
  $$-\frac{367}{10000}\times \frac{2894600}{\sqrt{49772647}}$$
- step9: Reduce the numbers:
  $$-\frac{367}{50}\times \frac{14473}{\sqrt{49772647}}$$
- step10: Multiply the fractions:
  $$-\frac{367\times 14473}{50\sqrt{49772647}}$$
- step11: Multiply:
  $$-\frac{5311591}{50\sqrt{49772647}}$$
- step12: Multiply by the Conjugate:
  $$\frac{-5311591\sqrt{49772647}}{50\sqrt{49772647}\times \sqrt{49772647}}$$
- step13: Multiply the numbers:
  $$\frac{-5311591\sqrt{49772647}}{2488632350}$$
- step14: Reduce the fraction:
  $$\frac{-367\sqrt{49772647}}{171950}$$
- step15: Calculate:
  $$-\frac{367\sqrt{49772647}}{171950}$$
### Step 4: Calculate the Test Statistic

1. **Calculate the Standard Error (SE)**:
   $$
   SE \approx 0.002437
   $$

2. **Calculate the Test Statistic $$ z $$**:
   $$
   z \approx -15.057721
   $$

### Step 5: Determine the p-value
Since this is a one-tailed test, we will find the p-value corresponding to the calculated $$ z $$-value. A $$ z $$-value of $$ -15.057721 $$ is extremely low, indicating that the sample proportion is significantly less than the hypothesized proportion.

### Step 6: Conclusion
Given that the p-value will be very close to 0 (much less than any common significance level such as $$ 0.05 $$ or $$ 0.01 $$), we reject the null hypothesis.

### Final Conclusion
There is sufficient evidence to support the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone.
There is sufficient evidence to support the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone.

Claim. Fewer than of homes have only a landline telephone and no wireless phone Sample data. A survey by the National Center for Health Statistics showed that among 14,473 homes had landline

Upstudy AI Solution

Answer

Solution

Step 1: Define the Hypotheses

Step 2: Gather Sample Data

Step 3: Calculate the Test Statistic

Step 4: Calculate the Test Statistic

Step 4: Calculate the Test Statistic

Step 5: Determine the p-value

Step 6: Conclusion

Final Conclusion

The Deep Dive

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