The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. A Click the icon to view the display from technology. Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? O A. Ho: = 6.00 Mbps H, p>6.00 Mbps B. Ho: u=6.00 Mbps H,:u<6.00 Mbps OC. Ho <6.00 Mbps H =6.00 Mbps: OD. Ho: = 6.00 Mbps H, p6.00 Mbps Identify the test statistic. O(Round to two decimal places as needed.)
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. A Click the icon to view the display from technology. Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? O A. Ho: = 6.00 Mbps H, p>6.00 Mbps B. Ho: u=6.00 Mbps H,:u<6.00 Mbps OC. Ho <6.00 Mbps H =6.00 Mbps: OD. Ho: = 6.00 Mbps H, p6.00 Mbps Identify the test statistic. O(Round to two decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
Determine test statistic
determine p value
reject or fail to reject?
Transcribed Image Text:### T-Test Results
The table below presents the results from an independent t-test:
- **Hypothesized Mean (\( \mu \))**: \( < 7.00 \)
- **T-Statistic (t)**: \( -2.630770 \)
- **P-Value (p)**: \( 0.005426 \)
- **Sample Mean (x)**: \( 6.35 \)
- **Standard Deviation (Sx)**: \( 1.913842 \)
- **Sample Size (n)**: \( 60 \)
#### Explanation:
- **Hypothesized Mean**: The test assumes the population mean is less than 7.
- **T-Statistic**: A negative t-value indicates that the sample mean is less than the hypothesized mean.
- **P-Value**: A p-value less than 0.05 typically means the results are statistically significant, suggesting that there is evidence to reject the null hypothesis.
- **Sample Mean**: The average of the sample data.
- **Standard Deviation**: Reflects the variability or spread of the sample data.
- **Sample Size**: Number of observations in the sample.
This information is used to assess whether there is statistical evidence to support a claim about the mean of a population.
![The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
- Option A:
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu > 6.00 \text{ Mbps} \)
- Option B: *(Selected)*
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu < 6.00 \text{ Mbps} \)
- Option C:
- \( H_0: \mu < 6.00 \text{ Mbps} \)
- \( H_1: \mu = 6.00 \text{ Mbps} \)
- Option D:
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu \neq 6.00 \text{ Mbps} \)
Identify the test statistic.
\[ \text{(Round to two decimal places as needed.)} \]
**Explanation:**
The objective of this hypothesis test is to determine whether the mean data speed of a smartphone carrier at airports is statistically significantly less than 6.00 Mbps. Based on the selection of Option B, the null hypothesis (\( H_0 \)) assumes that the mean speed is 6.00 Mbps, while the alternative hypothesis (\( H_1 \)) asserts that the mean speed is less than 6.00 Mbps. The task involves calculating the test statistic and the corresponding P-value to conclude whether to reject the null hypothesis, using a significance level of 0.05.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c1d0954-082a-48c1-b4d7-22ac9f1f697c%2F37427f4e-993e-4351-b344-a82701b03f53%2Fj5isdai_processed.png&w=3840&q=75)
Transcribed Image Text:The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
- Option A:
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu > 6.00 \text{ Mbps} \)
- Option B: *(Selected)*
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu < 6.00 \text{ Mbps} \)
- Option C:
- \( H_0: \mu < 6.00 \text{ Mbps} \)
- \( H_1: \mu = 6.00 \text{ Mbps} \)
- Option D:
- \( H_0: \mu = 6.00 \text{ Mbps} \)
- \( H_1: \mu \neq 6.00 \text{ Mbps} \)
Identify the test statistic.
\[ \text{(Round to two decimal places as needed.)} \]
**Explanation:**
The objective of this hypothesis test is to determine whether the mean data speed of a smartphone carrier at airports is statistically significantly less than 6.00 Mbps. Based on the selection of Option B, the null hypothesis (\( H_0 \)) assumes that the mean speed is 6.00 Mbps, while the alternative hypothesis (\( H_1 \)) asserts that the mean speed is less than 6.00 Mbps. The task involves calculating the test statistic and the corresponding P-value to conclude whether to reject the null hypothesis, using a significance level of 0.05.
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