The 2012 general social survey asked a large number of people how much time 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 38 teenagers, the sample standard deviation of daily TV time is 3.3 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.97? Use the α = 0.10 level of significance. State the appropriate null and alternate hypothesis: H0: σ = 2.97 H1: σ ≠ 2.97 This hypothesis test is a chi-square test. Find the critical values. Round the answer to three decimal places. For α = 0.10, the critical values are 23.542 and 52.192. Compute the test statistic. Round the answer to three decimal places. χ² = 45.684 Determine whether to reject H0. At the α = 0.10 level, the null hypothesis H0 should not be rejected because the test statistic falls between the critical values.
Answer (1)
To determine if the population standard deviation of TV watching times for teenagers differs from 2.97 hours, we are conducting a hypothesis test for a population standard deviation using the chi-square distribution.
Here is a step-by-step explanation of how to approach this problem:
Define the Hypotheses :
Null hypothesis ( H 0 ): σ = 2.97 hours. This means that the population standard deviation is 2.97 hours.
Alternative hypothesis ( H 1 ): σ = 2.97 hours. This suggests that the population standard deviation is different from 2.97 hours.
Identify the Sample Statistics :
Sample standard deviation ( s ) = 3.3 hours.
Sample size ( n ) = 38.
Calculate the Test Statistic : The test statistic for a chi-square test of a standard deviation is calculated using the formula: χ 2 = σ 0 2 ( n − 1 ) s 2 where σ 0 is the standard deviation under the null hypothesis. χ 2 = ( 2.97 ) 2 ( 38 − 1 ) × ( 3.3 ) 2 χ 2 = 8.8209 37 × 10.89 χ 2 ≈ 45.684
Determine the Critical Values :
At α = 0.10 , the chi-square distribution with 37 degrees of freedom gives critical values (using a chi-square table or calculator) of approximately 23.542 and 52.192.
Decision Rule :
If the test statistic falls below the lower critical value or above the upper critical value, reject H 0 .
If the test statistic falls between the critical values, do not reject H 0 .
Conclusion :
Since the test statistic (45.684) falls between the critical values (23.542 and 52.192), we do not reject the null hypothesis at the α = 0.10 level.
Therefore, based on this test, we do not have enough evidence to conclude that the population standard deviation of TV watching times for teenagers significantly differs from 2.97 hours.
