An article reported that 74% of companies have Twitter accounts. An economist thinks the percentage is higher at technology companies. She samples 70 technology companies and finds that 55 of them have Twitter accounts. Can she conclude that more than 77% of technology companies have Twitter accounts? Use the a=0.10 level of significance and the P-value method.

Question

Hv0 : [?]
Hv1 : [?]

Is this left tailed, right-tailed or two-tailed?

Answer 1

To conduct a hypothesis test regarding the proportion of technology companies with Twitter accounts, we first need to state the null and alternative hypotheses.

Hypotheses

• Null Hypothesis (H0): The proportion of technology companies with Twitter accounts is less than or equal to 0.77.

  This can be mathematically represented as: \[ H_0: p \leq 0.77 \]

• Alternative Hypothesis (H1): The proportion of technology companies with Twitter accounts is greater than 0.77.

  This can be mathematically represented as: \[ H_1: p > 0.77 \]

Direction of the Test

Since the alternative hypothesis is testing whether the proportion is greater than a certain value, this is a right-tailed test.

Summary of Hypotheses

• H0: \( p \leq 0.77 \)
• H1: \( p > 0.77 \)

Next Steps

1. Calculate the sample proportion: \[ \hat{p} = \frac{55}{70} \approx 0.7857 \]

2. Calculate the test statistic and P-value based on the sample size and hypothesized proportion.

If you need help with calculations or further steps, please let me know!

Answer 2

(b) Compute the value of the test statistic. Round the answer to two decimal places.

Answer 3

To compute the test statistic for the hypothesis test regarding the proportion of technology companies with Twitter accounts, we will use the formula for the test statistic for a proportion, which is given by:

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

Where:

• \(\hat{p}\) = sample proportion
• \(p_0\) = hypothesized proportion (0.77 in this case)
• \(n\) = sample size

Given:

• \(\hat{p} = \frac{55}{70} \approx 0.7857\)
• \(p_0 = 0.77\)
• \(n = 70\)

Step 1: Calculate the standard error (SE)

\[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.77(1 - 0.77)}{70}} = \sqrt{\frac{0.77 \times 0.23}{70}} \approx \sqrt{\frac{0.1771}{70}} \approx \sqrt{0.002530} \approx 0.0503 \]

Step 2: Compute the test statistic \(z\)

\[ z = \frac{\hat{p} - p_0}{SE} = \frac{0.7857 - 0.77}{0.0503} \approx \frac{0.0157}{0.0503} \approx 0.311 \]

Rounding

Rounding the result to two decimal places, we have:

\[ z \approx 0.31 \]

Conclusion

The value of the test statistic \(z\) is approximately 0.31 (rounded to two decimal places).

Answer 4

(c) Compute the P-value. Round the answer to at least four decimal places.

Answer 5

To compute the P-value for the right-tailed test statistic \( z \), we can use the standard normal distribution.

Given Test Statistic

We previously calculated the test statistic: \[ z \approx 0.31 \]

Step 1: Compute the P-value

The P-value for a right-tailed test can be found using the cumulative distribution function (CDF) of the standard normal distribution. The P-value is given by:

\[ \text{P-value} = P(Z > z) = 1 - P(Z \leq z) \]

Using a standard normal distribution table or calculator, we find \( P(Z \leq 0.31) \).

Finding \( P(Z \leq 0.31) \)

Using statistical software or a Z-table: \[ P(Z \leq 0.31) \approx 0.6217 \]

Step 2: Calculate the P-value

Now, we can calculate the P-value:

\[ \text{P-value} = 1 - P(Z \leq 0.31) = 1 - 0.6217 = 0.3783 \]

Result

The P-value is approximately 0.3783 when rounded to four decimal places.

Answer 6

Do we reject Hv0

Answer 7

To determine whether we reject the null hypothesis \(H_0\) at the significance level \(\alpha = 0.10\), we compare the P-value to the significance level.

Given:

• P-value: \(0.3783\)
• Significance level \(\alpha\): \(0.10\)

Decision Rule:

• If the P-value is less than or equal to \(\alpha\), we reject \(H_0\).
• If the P-value is greater than \(\alpha\), we fail to reject \(H_0\).

Comparison:

• P-value: \(0.3783\)
• Significance level \(\alpha\): \(0.10\)

Since \(0.3783 > 0.10\), we fail to reject \(H_0\).

Conclusion:

There is not enough evidence at the \(\alpha = 0.10\) level of significance to conclude that more than 77% of technology companies have Twitter accounts.