Certainly! To test whether the population standard deviation of TV watching times for teenagers differs from 2.97, we need to perform a chi-square test for the standard deviation.
We will use the following hypotheses:
- Null hypothesis (\( H_0 \)): The population standard deviation \(\sigma = 2.97\)
- Alternative hypothesis (\( H_a \)): The population standard deviation \(\sigma \neq 2.97\)
### Step 1: Define the chi-square statistic
The test statistic for the chi-square test of a single variance is calculated as:
\[
\chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2}
\]
Where:
- \( n \) = sample size = 30
- \( s \) = sample standard deviation = 3.2 hours
- \( \sigma_0 \) = hypothesized population standard deviation = 2.97 hours
Plugging in the values:
\[
\chi^2 = \frac{(30-1) \cdot (3.2)^2}{(2.97)^2} = \frac{29 \cdot 10.24}{8.8209} \approx 33.64
\]
### Step 2: Determine the degrees of freedom
The degrees of freedom \( \text{df} \) is:
\[
\text{df} = n - 1 = 30 - 1 = 29
\]
### Step 3: Find the critical values
Since this is a two-tailed test with \( \alpha = 0.10 \), we need to find the critical values for \( \chi^2 \) at \( \alpha/2 = 0.05 \) for the upper and lower tails.
Using a chi-square table or calculator:
- \( \chi^2_{0.05, 29} \) (Upper critical value) = 45.722
- \( \chi^2_{0.95, 29} \) (Lower critical value) = 16.047 (approx)
### Step 4: Make a decision
Compare the calculated \( \chi^2 \) statistic to the critical values:
- If \( \chi^2 \) is less than the lower critical value or greater than the upper critical value, we reject the null hypothesis.
- If \( \chi^2 \) falls within the interval defined by the critical values, we fail to reject the null hypothesis.
For our calculated \( \chi^2 = 33.64 \):
- \( 16.047 < 33.64 < 45.722 \)
Since the calculated \( \chi^2 \) falls within the critical region, we fail to reject the null hypothesis at the \( \alpha = 0.10 \) significance level.
### Conclusion
There is not enough evidence to conclude that the population standard deviation of TV watching times for teenagers differs from 2.97 hours at the 0.10 significance level.
The 2012 General Social Survey asked a large number of people how many times they spent watching TV each day. The mean number of hours was 3.09, with a standard deviation of 2.97. Assume that in a sample of 30 teenagers, the sample standard deviation of the daily TV time is 3.2 hours, and that the population of TV-watching times is normally distributed. Can we conclude that the population standard deviation of TV watching times for teenagers differs from 2.97. Use the a= 0.10 level of significance. Find the critical values
5 answers
So what are the critical values
Given that it is a two-tailed test with a significance level \( \alpha = 0.10 \), we need to find the critical values for the chi-square distribution with \( df = 29 \). These critical values can be found using a chi-square table or a statistical calculator.
To find the critical values:
- For the upper tail (\( \alpha/2 = 0.05 \)):
Look up the chi-square value corresponding to a cumulative probability of \( 1 - 0.05 = 0.95 \).
- For the lower tail (\( \alpha/2 = 0.05 \)):
Look up the chi-square value corresponding to a cumulative probability of \( 0.05 \).
Using a chi-square table or calculator, the critical values for \( df = 29 \) are:
- \( \chi^2_{0.95, 29} \approx 42.557 \) (Upper critical value)
- \( \chi^2_{0.05, 29} \approx 17.708 \) (Lower critical value)
### Summary of Critical Values:
- **Lower critical value**: 17.708
- **Upper critical value**: 42.557
These critical values define the rejection region for the test:
- Reject \( H_0 \) if \( \chi^2 < 17.708 \) or \( \chi^2 > 42.557 \)
- Fail to reject \( H_0 \) if \( 17.708 \leq \chi^2 \leq 42.557 \)
Given our calculated chi-square value of \( \chi^2 = 33.64 \), which falls between the critical values:
- \( 17.708 < 33.64 < 42.557 \)
We fail to reject the null hypothesis at the \( \alpha = 0.10 \) significance level, indicating that there is not enough evidence to conclude that the population standard deviation of TV watching times for teenagers differs from 2.97 hours.
To find the critical values:
- For the upper tail (\( \alpha/2 = 0.05 \)):
Look up the chi-square value corresponding to a cumulative probability of \( 1 - 0.05 = 0.95 \).
- For the lower tail (\( \alpha/2 = 0.05 \)):
Look up the chi-square value corresponding to a cumulative probability of \( 0.05 \).
Using a chi-square table or calculator, the critical values for \( df = 29 \) are:
- \( \chi^2_{0.95, 29} \approx 42.557 \) (Upper critical value)
- \( \chi^2_{0.05, 29} \approx 17.708 \) (Lower critical value)
### Summary of Critical Values:
- **Lower critical value**: 17.708
- **Upper critical value**: 42.557
These critical values define the rejection region for the test:
- Reject \( H_0 \) if \( \chi^2 < 17.708 \) or \( \chi^2 > 42.557 \)
- Fail to reject \( H_0 \) if \( 17.708 \leq \chi^2 \leq 42.557 \)
Given our calculated chi-square value of \( \chi^2 = 33.64 \), which falls between the critical values:
- \( 17.708 < 33.64 < 42.557 \)
We fail to reject the null hypothesis at the \( \alpha = 0.10 \) significance level, indicating that there is not enough evidence to conclude that the population standard deviation of TV watching times for teenagers differs from 2.97 hours.
Now compute the test statistic
Certainly! Let's compute the test statistic step by step.
### Step 1: Gather the relevant information
- Sample size (\( n \)): 30
- Sample standard deviation (\( s \)): 3.2 hours
- Hypothesized population standard deviation (\( \sigma_0 \)): 2.97 hours
### Step 2: Use the chi-square statistic formula
The test statistic for the chi-square test of a single variance is calculated as follows:
\[
\chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2}
\]
### Step 3: Plug in the values
\[
\chi^2 = \frac{(30-1) \cdot (3.2)^2}{(2.97)^2}
\]
First, compute the numerator:
\[
(n-1) \cdot s^2 = 29 \cdot (3.2)^2 = 29 \cdot 10.24 = 297.16
\]
Next, compute the denominator:
\[
\sigma_0^2 = (2.97)^2 = 8.8209
\]
Now, combine them to find the chi-square statistic:
\[
\chi^2 = \frac{297.16}{8.8209} \approx 33.69
\]
### Step 4: Verify the computation
Just to ensure our calculations are precise:
\[
29 \cdot 10.24 = 297.16
\]
\[
2.97^2 = 8.8209
\]
\[
\frac{297.16}{8.8209} = 33.69
\]
Thus, the test statistic \( \chi^2 \) is approximately 33.69.
### Step 1: Gather the relevant information
- Sample size (\( n \)): 30
- Sample standard deviation (\( s \)): 3.2 hours
- Hypothesized population standard deviation (\( \sigma_0 \)): 2.97 hours
### Step 2: Use the chi-square statistic formula
The test statistic for the chi-square test of a single variance is calculated as follows:
\[
\chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2}
\]
### Step 3: Plug in the values
\[
\chi^2 = \frac{(30-1) \cdot (3.2)^2}{(2.97)^2}
\]
First, compute the numerator:
\[
(n-1) \cdot s^2 = 29 \cdot (3.2)^2 = 29 \cdot 10.24 = 297.16
\]
Next, compute the denominator:
\[
\sigma_0^2 = (2.97)^2 = 8.8209
\]
Now, combine them to find the chi-square statistic:
\[
\chi^2 = \frac{297.16}{8.8209} \approx 33.69
\]
### Step 4: Verify the computation
Just to ensure our calculations are precise:
\[
29 \cdot 10.24 = 297.16
\]
\[
2.97^2 = 8.8209
\]
\[
\frac{297.16}{8.8209} = 33.69
\]
Thus, the test statistic \( \chi^2 \) is approximately 33.69.
3 answers