A researcher polled a sample of 1066 adults in the year 2010, asking them how many hours per week they spent on the Internet. The sample mean was 9.42 with a standard deviation of 13.65. A second sample of 1059 adults was taken in the year 2012. For this sample, the mean was 10.49 with a standard deviation of 14.45. Assume these are simple random samples from populations of adults. Can you conclude that the mean number of hours per week spent on the Internet differs between 2010 and 2012? Let uv1 denote the mean number of hours spent on the Internet in 2010. Use the a=0.10 level and the P-value method.

Question

Hv0 : [?]
Hv1 : [?]

This is a right/left or two tailed test?

Answer 1

To determine whether the mean number of hours spent on the Internet differs between 2010 and 2012, we will set up null and alternative hypotheses as follows:

• Null Hypothesis (H₀): The mean number of hours spent on the Internet in 2010 is equal to the mean number of hours in 2012. Mathematically, this can be expressed as: \[ H_0: \mu_{2010} - \mu_{2012} = 0 \]

• Alternative Hypothesis (H₁): The mean number of hours spent on the Internet in 2010 is not equal to the mean number of hours in 2012. Mathematically, this can be expressed as: \[ H_1: \mu_{2010} - \mu_{2012} \neq 0 \]

This is a two-tailed test because we are checking for any difference (either an increase or a decrease) in the mean number of hours spent on the Internet between the two years.

Summary of hypotheses:

• \( H_0: \mu_{2010} = \mu_{2012} \)
• \( H_1: \mu_{2010} \neq \mu_{2012} \)

We will proceed with calculating a test statistic and finding the corresponding p-value to draw a conclusion at the \(\alpha = 0.10\) level.

Answer 2

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

Answer 3

To compute the P-value for the given data, we first need to conduct a two-sample t-test since we have two different sample means, standard deviations, and sample sizes.

Given Data:

1. 2010 Sample:

   • Sample size (\(n_1\)) = 1066
   • Sample mean (\(\bar{x}_1\)) = 9.42
   • Sample standard deviation (\(s_1\)) = 13.65

2. 2012 Sample:

   • Sample size (\(n_2\)) = 1059
   • Sample mean (\(\bar{x}_2\)) = 10.49
   • Sample standard deviation (\(s_2\)) = 14.45

Step 1: Calculate the test statistic (t)

We use the formula for the t-test for two independent samples: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Step 2: Plug in the values

1. Calculate the individual variances:

   • \(s_1^2 = 13.65^2 = 186.4225\)
   • \(s_2^2 = 14.45^2 = 208.6025\)

2. Calculate the standard error (SE): \[ SE = \sqrt{\frac{186.4225}{1066} + \frac{208.6025}{1059}} \] \[ SE \approx \sqrt{\frac{0.1748 + 0.1965}{1}} \approx \sqrt{0.3713} \approx 0.6093 \]

3. Now calculate the t-statistic: \[ t = \frac{9.42 - 10.49}{0.6093} = \frac{-1.07}{0.6093} \approx -1.7592 \]

Step 3: Determine degrees of freedom

The degrees of freedom for two independent samples can be calculated using the following formula (Welch's approximation): \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \] Calculating the numerator: \[ \left(\frac{186.4225}{1066} + \frac{208.6025}{1059}\right)^2 \approx (0.1748 + 0.1965)^2 \approx (0.3713)^2 \approx 0.1376 \]

Calculating the denominator: \[ = \frac{\left(\frac{186.4225}{1066}\right)^2}{1065} + \frac{\left(\frac{208.6025}{1059}\right)^2}{1058} \approx \frac{(0.1748)^2}{1065} + \frac{(0.1965)^2}{1058} \] \[ \approx \frac{0.0306}{1065} + \frac{0.0386}{1058} \approx 0.0000288 + 0.0000364 \approx 0.0000652 \]

\[ df \approx \frac{0.1376}{0.0000652} \approx 2105.57 \] Rounding gives us approximately 2105 degrees of freedom.

Step 4: Find the P-value

Now we can use a t-distribution table or software to find the P-value corresponding to the calculated t-statistic and degrees of freedom. Since we have a two-tailed test, we will find the P-value for \(t \approx -1.7592\).

Using a t-distribution calculator or software (or a t-distribution table):

1. Input t = -1.7592 and df = 2105.
2. The calculated P-value for a two-tailed test will be doubled since we are considering both tails.

After looking it up or calculating:

• The P-value associated with \(t \approx -1.76\) and \(df \approx 2105\) is approximately 0.0784.

Final P-value

Thus, the P-value is: \[ \text{P-value} \approx 0.0784 \]

Conclusion

If you round it to four decimal places: \[ \text{P-value} = 0.0784 \]

Answer 4

Do we reject Hv0?

Answer 5

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

Given:

• P-value ≈ 0.0784
• \(\alpha = 0.10\)

Decision Rule:

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

Comparison:

• Since \(0.0784 < 0.10\), we reject \(H_0\).

Conclusion:

There is sufficient evidence at the \(\alpha = 0.10\) significance level to conclude that there is a significant difference in the mean number of hours spent on the Internet between 2010 and 2012.