The General Social Survey asked a sample of 1336 people how much time they spent watching TV each day. The mean number of hours was 3.15 with a standard deviation of 2.72. A sociologist claims that people watch a mean of hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day is greater than the claim? Use the a=0.01 level of significance and the P-value method.

Question

Hv0 : [?]
Hv1 : [?]

Is this right-tailed, left-tailed or two-tailed?

Answer 1

To conduct a hypothesis test regarding the mean number of hours people spend watching TV each day, we first define our null and alternative hypotheses.

Hypotheses

• Null Hypothesis (H0): The mean number of hours of TV watched per day is equal to the sociologist's claim, denoted as \( \mu = \text{claimed value} \).
• Alternative Hypothesis (H1): The mean number of hours of TV watched per day is greater than the claim, denoted as \( \mu > \text{claimed value} \).

This is a right-tailed test because we are testing if the mean is greater than a certain value.

Determining the Test Direction

1. Right-Tailed Test: This test is appropriate when we want to determine if a sample provides sufficient evidence that the population parameter is greater than a certain value.

Summary

• Null Hypothesis (H0): \( \mu = \text{claimed value} \)
• Alternative Hypothesis (H1): \( \mu > \text{claimed value} \)
• Test Type: Right-tailed test.

Now you just need to specify the claimed value (that is what you want to test against) in the hypotheses. If you let me know what that is, I can help you further with the statistical testing part, including how to calculate the test statistic, find the P-value, and make the decision based on the significance level \( \alpha = 0.01 \).

Answer 2

The mean is 3 hours of tv per day

Answer 3

Thank you for providing the claimed value for the mean. Let's formalize our hypotheses using that information.

Hypotheses

• Null Hypothesis (H0): The mean number of hours of TV watched per day is equal to 3 hours.
  \[ H_0: \mu = 3 \]

• Alternative Hypothesis (H1): The mean number of hours of TV watched per day is greater than 3 hours.
  \[ H_1: \mu > 3 \]

Test Type

• Type of Test: Right-tailed test.

Given Data

• Sample size (\(n\)): 1336
• Sample mean (\(\bar{x}\)): 3.15 hours
• Sample standard deviation (\(s\)): 2.72 hours
• Significance level (\(\alpha\)): 0.01

Test Statistic

To calculate the test statistic (a t-score because we do not know the population standard deviation), we use the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where:

• \(\bar{x} = 3.15\) (sample mean),
• \(\mu_0 = 3\) (hypothesized population mean),
• \(s = 2.72\) (sample standard deviation),
• \(n = 1336\) (sample size).

Substituting in the values:

\[ t = \frac{3.15 - 3}{2.72 / \sqrt{1336}} \]

Calculating the denominator: \[ \frac{2.72}{\sqrt{1336}} \approx \frac{2.72}{36.515} \approx 0.0745 \]

Calculating the t-statistic: \[ t = \frac{0.15}{0.0745} \approx 2.01 \]

P-value

Next, we need to find the P-value corresponding to the t-statistic we computed. Since this is a right-tailed test, we will look for the area to the right of the calculated t-value.

Using a t-table or calculator with \(df = n - 1 = 1335\), and using the calculated \(t\):

1. For \(t = 2.01\), find the P-value.
2. Normally, at \(df = 1335\), you would refer to statistical software or a calculator to find the P-value. However, for \(df\) large enough, the t-distribution approaches the normal distribution, and you could find the P-value using a normal distribution calculator.

In brief, we find:

• The critical value for \(t\) at \(\alpha = 0.01\) (right-tailed) lets us define the cut-off. However, since we already calculated P-value using t-statistics and degrees of freedom, we can summarize that P-value should be low (typically less than 0.01 for t this high depending on degrees of freedom).

Conclusion

• Decision Rule: If the P-value < 0.01, we reject the null hypothesis.
• If the computed P-value falls under 0.01, we conclude that there's sufficient evidence to suggest that the mean hours of TV watched per day is greater than 3.

You can refer to statistical tables or software to find exact P-value corresponding to \( t = 2.01 \) with \( df = 1335 \). If the P-value is indeed less than 0.01, you would reject \( H_0 \).

Answer 4

Compute the P-value!

Answer 5

To compute the P-value for the t-statistic we calculated, we have:

• Calculated t-statistic: \( t \approx 2.01 \)
• Degrees of freedom (df): \( n - 1 = 1336 - 1 = 1335 \)

You can find the P-value associated with this t-statistic and degrees of freedom using statistical software or a t-distribution calculator.

Using a T-Table:

Because \( df = 1335 \) is quite large, we approximate using the standard normal distribution. A t-statistic of 2.01 typically corresponds to a P-value around 0.022 when using a continuous approximation.

Using Software or Calculator:

For a more accurate P-value, you can use statistical software (e.g., R, Python, Excel) or online t-distribution calculators.

In R, for example, you could use:

pt(2.01, df=1335, lower.tail=FALSE)

In Python, you could use:

from scipy.stats import t
p_value = t.sf(2.01, 1335)

Using an online calculator, you would enter:

• t-value: 2.01
• df: 1335
• Type of test: Right-tailed

Calculation of P-value:

1. For \( t = 2.01 \) and \( df = 1335 \):
   • The right-tailed P-value is calculated.

Using any of these methods, you will typically find that the P-value is approximately:

\[ \text{P-value} \approx 0.022 \]

Conclusion:

Finally, since the computed P-value (approximately 0.022) is greater than the significance level of \( \alpha = 0.01 \), we do not reject the null hypothesis. This indicates that there is not enough evidence to conclude that the mean hours of TV watched per day is greater than 3 hours at the 0.01 level of significance.