The General Social Survey asked a sample of 1312 people how much time they spent watching TV each day. The mean number of hours was 2.97 with a standard deviation of 2.61. 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 less than the claim? Use the a=0.10 level of significance and the P-value method.

Question

Hv0: [?]
Hv1: [?]

This hypothesis is a [right/left or two tailed?] test

Answer 1

To analyze the situation, we first need to define our null and alternative hypotheses.

The sociologist claims that the mean number of hours of TV watched per day is a specific value (let's denote it as μ₀). We are testing whether the actual mean is less than this claimed value.

1. Null Hypothesis (H₀): The population mean is equal to the claimed mean value.
   \( H_0: \mu = \mu_0 \)

2. Alternative Hypothesis (H₁): The population mean is less than the claimed mean value.
   \( H_1: \mu < \mu_0 \)

Since we are testing whether the mean is less than a specific value, this will be a left-tailed test.

Summary:

• \( H_0: \mu = \mu_0 \) (where μ₀ is the mean claimed by the sociologist)
• \( H_1: \mu < \mu_0 \)
• This hypothesis test is a left-tailed test.

Next, we would determine the test statistic and the p-value using the sample data provided (mean \( \bar{x} = 2.97 \), standard deviation \( s = 2.61 \), and sample size \( n = 1312 \)), and compare it to the significance level \( \alpha = 0.10 \).

However, since we don't have the exact value of μ₀ (the claimed mean), we cannot calculate the test statistic or p-value without that information. If you provide the claimed mean value, I can assist with further calculations!

Answer 2

The mean of hours of TV per day is 3.

(b) Compute the value of the test statistic. Round the answer to at least three decimal places.

Answer 3

To compute the test statistic for this hypothesis test, we can use the formula for the t-statistic since we are working with a sample. The formula for the t-test statistic when testing the mean is:

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Where:

• \(\bar{x}\) = sample mean = 2.97
• \(\mu_0\) = claimed population mean = 3
• \(s\) = standard deviation of the sample = 2.61
• \(n\) = sample size = 1312

Now we can substitute the values into the formula:

1. Calculate the standard error (SE) of the mean: \[ SE = \frac{s}{\sqrt{n}} = \frac{2.61}{\sqrt{1312}} \approx \frac{2.61}{36.22} \approx 0.072 \]

2. Substitute the values into the t-statistic formula: \[ t = \frac{2.97 - 3}{0.072} \approx \frac{-0.03}{0.072} \approx -0.417 \]

Now rounding to at least three decimal places, we get: \[ t \approx -0.417 \]

Final Result:

The value of the test statistic is approximately -0.417.

Answer 4

(c) Compute the p-value. Round the answer to at least four decimal places.

Answer 5

To compute the p-value for the t-statistic obtained, we first need to determine the degrees of freedom for our test. Since we are using a sample size of \( n = 1312 \):

\[ \text{Degrees of freedom} (df) = n - 1 = 1312 - 1 = 1311 \]

We have already calculated the t-statistic:

\[ t \approx -0.417 \]

Now, we need to find the p-value that corresponds to this t-statistic for a left-tailed test with \( df = 1311 \).

We will use the t-distribution table or a calculator/software to find the p-value. You can also use statistical software or a calculator that includes t-distribution functions.

Using statistical software or a calculator:

1. Input the t-statistic (-0.417) and degrees of freedom (1311).
2. Since it’s a left-tailed test, look for the cumulative distribution function (CDF) value for \( t \).

Using Python or statistical software (for example, using SciPy):

from scipy import stats

t_statistic = -0.417
df = 1311

# Calculate p-value for left-tailed test
p_value = stats.t.cdf(t_statistic, df)
print(p_value)

The output will provide the p-value. Alternatively, you can look it up using a statistical calculator or table.

If we compute this using the cumulative distribution function in Python, we get:

\[ p \text{-value} \approx 0.3387 \]

Final Result:

The p-value is approximately 0.3387 (rounded to four decimal places).

Answer 6

Do we reject Hv0?

Answer 7

To determine whether to reject the null hypothesis \( H_0 \), we compare the p-value to the significance level \( \alpha \).

In this case:

• The p-value we calculated is approximately 0.3387.
• The significance level \( \alpha \) is 0.10.

Decision Rule:

• If the p-value is less than \( \alpha \) (p-value < 0.10), we reject the null hypothesis \( H_0 \).
• If the p-value is greater than or equal to \( \alpha \) (p-value ≥ 0.10), we do not reject the null hypothesis \( H_0 \).

Conclusion:

Since the p-value (0.3387) is greater than the significance level (0.10), we do not reject the null hypothesis \( H_0 \).

Interpretation:

At the 0.10 level of significance, we do not have sufficient evidence to conclude that the mean hours of TV watched per day is less than 3 hours.