Knowledge Check Web App

Example 1: Exploring Time People Spend on Social Media

n: 50 (range 30 – 65)

x̄: 55 (range 54 – 58)

σ: 15 (range 15 – 50)

α: 0.05 (range 0.01 – 0.05)

Level 1

According to a recent Forbes article, Americans spent an average \( \mu_0 = 58 \) minutes a day on Facebook in 2020. A researcher is interested in checking if the average time has changed this year. She asks a sample of \( n = 50 \) people that resulted in an average of \( \bar{x} = 55 \) minutes daily. Assume the population standard deviation is \( \sigma = 15 \) minutes. Is there enough evidence that the average time spent on Facebook has changed at a significance level of \( \alpha = 0.05 \)?

Solution A – Finding Z-test and Comparing it to Z-critical

The first step in every statistics problem is identifying the relevant and important information. From the problem, we know we have:
- Population average: \( \mu_0 = 58 \)
- Sample size: \( n = 50 \)
- Sample average: \( \bar{x} = 55 \)
- Population standard deviation: \( \sigma = 15 \)
Next, we decide whether to use the z-score or t-score. The z-score is used when we know the population standard deviation. The t-score is used when we don’t have a population standard deviation or are calculating the sample standard deviation. Since we know the population standard deviation, we use the z-score.
Now, calculate the z-test using the formula below, and compare it with the z-critical value to decide whether to reject the null hypothesis.

\[ Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]

\[ Z = \frac{55 - 58}{\frac{15}{\sqrt{50}}} = -1.41 \]

To calculate the z-score in Excel, use the following formula:

\[ NORM.S.INV(\frac{\alpha}{2}) \rightarrow NORM.S.INV(0.05/2) \]

Finally, the z-test and the z-score can be compared to determine whether or not the null hypothesis is true. To do this, let's plot the numbers on a number line.

Number Line Showing Z-scores from -1.96 to 1.96, with a mark at -1.41

Think of the z-score as a limit. In this case, the limits are \( -1.96 \) and \( 1.96 \). Any value inside those limits confirms the null hypothesis. Using the number line, we can see that \( -1.41 \) is inside the limits, so it confirms the null hypothesis. If the null hypothesis is true (is confirmed), we do not reject it.

Change the value of Sigma to 50, What is the value of the z-test statistic:

1.414
-1.414
-0.400
-0.424

Example 1: Exploring time people spend on social media

n: 50 (range 30 – 65)

x̄: 55 (range 54 – 58)

σ: 15 (range 15 – 50)

α: 0.05 (range 0.01 – 0.05)

According to a recent Forbes Article, Americans spent an average \( \mu_{\text{Facebook}} = 58 \) minutes a day on Facebook in 2020. A researcher is interested in checking if the average time has changed this year. She takes a sample of \( n = 50 \) people that resulted in an average of \( \bar{x} = 55 \) minutes daily. Assume the population standard deviation is \( \sigma = 15 \) minutes. Is there enough evidence that the average time spent on Facebook has changed at a level of \( \alpha = 0.05 \) significance?

Solution B: Finding z-test then p-value and comparing it to the alpha

First, identify the relevant information:
- Population average \( \mu_{\text{Facebook}} = 58 \) minutes
- Sample size \( n = 50 \)
- Sample average \( \bar{x} = 55 \) minutes
- Population standard deviation \( \sigma = 15 \) minutes
The next step is to decide whether to use the z-test or t-test. In this case, the population standard deviation is given, so we use the z-test formula:
To calculate the p-value, we use the absolute value of the z-test and use the following equation in Excel:

This gives us:

Compare the p-value to the alpha \( \alpha = 0.05 \):
- Since \( 0.16 > 0.05 \), we do not reject the null hypothesis.
- Thus, there is no evidence to suggest that the average time spent on Facebook has changed.

Change the value of sample's average to 57, what is the p-value?

0.638
0.1
0.157
0.177

Example 1: Exploring time people spend on social media

n: 50 (range 30 – 65)

x̄: 55 (range 54 – 58)

σ: 15 (range 15 – 50)

α: 0.05 (range 0.01 – 0.05)

According to a recent Forbes Article, Americans spent an average \( \mu_{\text{fb}} = 58 \) minutes a day on Facebook in 2020. A researcher is interested in checking if the average time has changed this year. She takes a sample of \( n = 50 \) people that resulted in an average of \( \bar{x} = 55 \) minutes daily. Assume the population standard deviation is \( \sigma = 15 \) minutes. Is there enough evidence that the average time spent on Facebook has changed at a level of \( \alpha = 0.05 \) of significance?

Solution C: Finding the confidence interval

The first step in every statistics problem is identifying the relevant and important information. Reading the problem, we know:
- \( \mu_{\text{fb}} = 58 \) minutes
- \( n = 50 \) people
- \( \bar{x} = 55 \) minutes
- \( \sigma = 15 \) minutes
The next step is figuring out whether to use the z-score or t-score formula. Since the population standard deviation is given, we use the z-score formula:
\[ z = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} \approx 2.12 \]
We calculate the z-score critical value for a two-tailed test with \( \alpha = 0.05 \):
\[ z_{\alpha/2} = 1.96 \]
We calculate the margin of error (ME):
\[ \text{ME} = z_{\alpha/2} \times \left( \frac{\sigma}{\sqrt{n}} \right) = 1.96 \times \frac{15}{\sqrt{50}} \approx 4.16 \]
Finally, the confidence interval is:
\[ \bar{x} \pm \text{ME} = 55 \pm 4.16 = [50.84, 59.16] \]
Since the population average \( \mu_{\text{fb}} = 58 \) falls within the confidence interval, we do not reject the null hypothesis.

Using the number line, we can clearly see that the population average \( \mu_{\text{fb}} = 58 \) falls within the confidence interval. Therefore, we do not reject the null hypothesis. Hence, the null hypothesis is true.

Change the value of alpha to 0.01, what are the values of the lower and uppper limits?

49.54 and 60.46
50.64 and 59.36
50.84 and 59.16
50.07 and 59.93