Question
A study conducted by a technology company showed that the mean time spent per day browsing the video streaming service Netflix for something to watch was 20 minutes. Assume the standard deviation is o=5. Suppose a simple random sample of 109 visits taken this year has a sample mean of x=21 minutes. A social scientist is interested to know whether the mean time browsing Netflix has changed. Use the a=0.10 level of significance and the P-value method.
A) state the appropriate null and alternate hypothesis
Hv0: [?]
Hv1 : [?]
This hypothesis test is a [blank] test.
A) state the appropriate null and alternate hypothesis
Hv0: [?]
Hv1 : [?]
This hypothesis test is a [blank] test.
Answers
Answer
Compute the value of the test statistic. Round the answer to two decimal places.
z=?
z=?
Answer
One of your digits is wrong. Check your work
Answer
Correct!
Now compute the P-Value. Round the answer to 4 decimal places.
Now compute the P-Value. Round the answer to 4 decimal places.
Answer
Correct.
Determine whether to reject Hv0. use the a=0.10 level of significance.
Determine whether to reject Hv0. use the a=0.10 level of significance.
Answer
(e) State a conclusion.
There [is? isnt?] enough evidence to conclude that the mean time browsing Netflix has changed.
There [is? isnt?] enough evidence to conclude that the mean time browsing Netflix has changed.
Answered by
GPT-4o mini
To determine whether the mean time spent browsing Netflix has changed, we can set up our null and alternative hypotheses as follows:
**Null Hypothesis (H0):** The mean time spent browsing Netflix has not changed. Thus, we assume the population mean is equal to the previously established mean.
\[
H_0: \mu = 20 \text{ minutes}
\]
**Alternative Hypothesis (H1):** The mean time spent browsing Netflix has changed. This implies the population mean is not equal to the previously established mean.
\[
H_1: \mu \neq 20 \text{ minutes}
\]
This hypothesis test is a **two-tailed test** because we are investigating whether the mean has changed in either direction (either increased or decreased).
To summarize:
- Null Hypothesis (H0): \(\mu = 20\)
- Alternative Hypothesis (H1): \(\mu \neq 20\)
- This hypothesis test is a **two-tailed test**.
Now you can continue with your analysis using the significance level of \(\alpha = 0.10\) and apply the P-value method to draw conclusions.
**Null Hypothesis (H0):** The mean time spent browsing Netflix has not changed. Thus, we assume the population mean is equal to the previously established mean.
\[
H_0: \mu = 20 \text{ minutes}
\]
**Alternative Hypothesis (H1):** The mean time spent browsing Netflix has changed. This implies the population mean is not equal to the previously established mean.
\[
H_1: \mu \neq 20 \text{ minutes}
\]
This hypothesis test is a **two-tailed test** because we are investigating whether the mean has changed in either direction (either increased or decreased).
To summarize:
- Null Hypothesis (H0): \(\mu = 20\)
- Alternative Hypothesis (H1): \(\mu \neq 20\)
- This hypothesis test is a **two-tailed test**.
Now you can continue with your analysis using the significance level of \(\alpha = 0.10\) and apply the P-value method to draw conclusions.
Answered by
GPT-4o mini
To compute the test statistic \( z \) for the hypothesis test, we can use the formula for the \( z \)-score when the population standard deviation is known. The formula is:
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Where:
- \( \bar{x} \) is the sample mean,
- \( \mu \) is the population mean under the null hypothesis,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
From the information provided:
- \( \bar{x} = 21 \) minutes (sample mean),
- \( \mu = 20 \) minutes (population mean under null hypothesis),
- \( \sigma = 5 \) minutes (standard deviation),
- \( n = 109 \) (sample size).
Now, let's substitute the values into the formula:
\[
z = \frac{21 - 20}{5 / \sqrt{109}}
\]
Calculating the standard deviation of the sample mean:
\[
\sigma / \sqrt{n} = \frac{5}{\sqrt{109}} \approx \frac{5}{10.4403} \approx 0.4794
\]
Now compute the \( z \) value:
\[
z = \frac{1}{0.4794} \approx 2.084
\]
Rounding to two decimal places, we have:
\[
z \approx 2.08
\]
Thus, the value of the test statistic \( z \) is approximately **2.08**.
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Where:
- \( \bar{x} \) is the sample mean,
- \( \mu \) is the population mean under the null hypothesis,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
From the information provided:
- \( \bar{x} = 21 \) minutes (sample mean),
- \( \mu = 20 \) minutes (population mean under null hypothesis),
- \( \sigma = 5 \) minutes (standard deviation),
- \( n = 109 \) (sample size).
Now, let's substitute the values into the formula:
\[
z = \frac{21 - 20}{5 / \sqrt{109}}
\]
Calculating the standard deviation of the sample mean:
\[
\sigma / \sqrt{n} = \frac{5}{\sqrt{109}} \approx \frac{5}{10.4403} \approx 0.4794
\]
Now compute the \( z \) value:
\[
z = \frac{1}{0.4794} \approx 2.084
\]
Rounding to two decimal places, we have:
\[
z \approx 2.08
\]
Thus, the value of the test statistic \( z \) is approximately **2.08**.
Answered by
GPT-4o mini
Let's redo the calculations step by step to ensure we have the correct value for the test statistic \( z \).
We use the formula for the \( z \)-score:
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Given:
- Sample mean, \( \bar{x} = 21 \) minutes
- Population mean under the null hypothesis, \( \mu = 20 \) minutes
- Population standard deviation, \( \sigma = 5 \) minutes
- Sample size, \( n = 109 \)
### Step 1: Calculate the standard error (SE)
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{109}}
\]
Calculating \( \sqrt{109} \):
\[
\sqrt{109} \approx 10.4403
\]
Now calculate the standard error:
\[
SE = \frac{5}{10.4403} \approx 0.4787
\]
### Step 2: Calculate the \( z \) statistic
Now, substitute the values back into the formula for \( z \):
\[
z = \frac{21 - 20}{0.4787} = \frac{1}{0.4787} \approx 2.091
\]
### Rounding
When rounded to two decimal places:
\[
z \approx 2.09
\]
Thus, the corrected value of the test statistic \( z \) is approximately **2.09**. Thank you for your patience, and I apologize for the earlier oversight.
We use the formula for the \( z \)-score:
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Given:
- Sample mean, \( \bar{x} = 21 \) minutes
- Population mean under the null hypothesis, \( \mu = 20 \) minutes
- Population standard deviation, \( \sigma = 5 \) minutes
- Sample size, \( n = 109 \)
### Step 1: Calculate the standard error (SE)
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{109}}
\]
Calculating \( \sqrt{109} \):
\[
\sqrt{109} \approx 10.4403
\]
Now calculate the standard error:
\[
SE = \frac{5}{10.4403} \approx 0.4787
\]
### Step 2: Calculate the \( z \) statistic
Now, substitute the values back into the formula for \( z \):
\[
z = \frac{21 - 20}{0.4787} = \frac{1}{0.4787} \approx 2.091
\]
### Rounding
When rounded to two decimal places:
\[
z \approx 2.09
\]
Thus, the corrected value of the test statistic \( z \) is approximately **2.09**. Thank you for your patience, and I apologize for the earlier oversight.
Answered by
GPT-4o mini
To compute the P-value for the test statistic \( z \approx 2.09 \) in a two-tailed test, we can follow these steps:
1. **Find the P-value for \( z = 2.09 \)**:
We will first find the area to the right of \( z = 2.09 \) under the standard normal distribution.
2. **Use the standard normal distribution table** (or a calculator) to find the P-value for \( z = 2.09 \):
The area to the left of \( z = 2.09 \) (denoted as \( P(Z < 2.09) \)) can be found using statistics software, a calculator, or a standard normal distribution table.
- For \( z = 2.09\), the cumulative probability (area to the left) is approximately \( 0.9817 \).
3. **Calculate the area to the right**:
The area to the right (or the tail) is:
\[
P(Z > 2.09) = 1 - P(Z < 2.09) = 1 - 0.9817 = 0.0183
\]
4. **Since this is a two-tailed test**, we need to multiply the one-tailed P-value by 2:
\[
P\text{-value} = 2 \times P(Z > 2.09) = 2 \times 0.0183 = 0.0366
\]
Finally, rounding to four decimal places, the P-value is:
\[
\text{P-value} \approx 0.0366
\]
Thus, the computed P-value is **0.0366**.
1. **Find the P-value for \( z = 2.09 \)**:
We will first find the area to the right of \( z = 2.09 \) under the standard normal distribution.
2. **Use the standard normal distribution table** (or a calculator) to find the P-value for \( z = 2.09 \):
The area to the left of \( z = 2.09 \) (denoted as \( P(Z < 2.09) \)) can be found using statistics software, a calculator, or a standard normal distribution table.
- For \( z = 2.09\), the cumulative probability (area to the left) is approximately \( 0.9817 \).
3. **Calculate the area to the right**:
The area to the right (or the tail) is:
\[
P(Z > 2.09) = 1 - P(Z < 2.09) = 1 - 0.9817 = 0.0183
\]
4. **Since this is a two-tailed test**, we need to multiply the one-tailed P-value by 2:
\[
P\text{-value} = 2 \times P(Z > 2.09) = 2 \times 0.0183 = 0.0366
\]
Finally, rounding to four decimal places, the P-value is:
\[
\text{P-value} \approx 0.0366
\]
Thus, the computed P-value is **0.0366**.
Answered by
GPT-4o mini
To determine whether to reject the null hypothesis \( H_0 \) using the significance level \( \alpha = 0.10 \), we follow these steps:
1. **Compare the P-value to the significance level**:
- We computed the P-value to be **0.0366**.
- The significance level is \( \alpha = 0.10 \).
2. **Decision Rule**:
- If the P-value is less than \( \alpha \), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than or equal to \( \alpha \), we fail to reject the null hypothesis \( H_0 \).
3. **Make the Decision**:
Since \( 0.0366 < 0.10 \), we reject the null hypothesis \( H_0 \).
### Conclusion:
At the \( \alpha = 0.10 \) level of significance, we reject the null hypothesis. This suggests that there is sufficient evidence to conclude that the mean time spent browsing Netflix has changed.
1. **Compare the P-value to the significance level**:
- We computed the P-value to be **0.0366**.
- The significance level is \( \alpha = 0.10 \).
2. **Decision Rule**:
- If the P-value is less than \( \alpha \), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than or equal to \( \alpha \), we fail to reject the null hypothesis \( H_0 \).
3. **Make the Decision**:
Since \( 0.0366 < 0.10 \), we reject the null hypothesis \( H_0 \).
### Conclusion:
At the \( \alpha = 0.10 \) level of significance, we reject the null hypothesis. This suggests that there is sufficient evidence to conclude that the mean time spent browsing Netflix has changed.
Answered by
GPT-4o mini
Based on the analysis and our decision to reject the null hypothesis, the conclusion would be:
There **is** enough evidence to conclude that the mean time browsing Netflix has changed.
This indicates that the sample data provides sufficient statistical evidence to suggest a significant change in the average amount of time spent browsing for something to watch on Netflix.
There **is** enough evidence to conclude that the mean time browsing Netflix has changed.
This indicates that the sample data provides sufficient statistical evidence to suggest a significant change in the average amount of time spent browsing for something to watch on Netflix.
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