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.

Question

A) state the appropriate null and alternate hypothesis

Hv0: [?]
Hv1 : [?]
This hypothesis test is a [blank] test.

Answer 1

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.

Answer 2

Compute the value of the test statistic. Round the answer to two decimal places.
z=?

Answer 3

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.

Answer 4

One of your digits is wrong. Check your work

Answer 5

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.

Answer 6

Correct!

Now compute the P-Value. Round the answer to 4 decimal places.

Answer 7

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.

Answer 8

Correct.
Determine whether to reject Hv0. use the a=0.10 level of significance.

Answer 9

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.

Answer 10

(e) State a conclusion.
There [is? isnt?] enough evidence to conclude that the mean time browsing Netflix has changed.

Answer 11

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.