The American Academy of Pediatrics (AAP) recommends that kids 2-5 years old watch no more than (on average) 1.6 hours a day of quality programming. A researcher selected a sample of 101 kids in the neighborhood and asked their parents how many hours per day they allow their kids watching TV. She found out that the kids in the sample spend on average 1.75 hours watching TV with a standard deviation of 0.85. Do the kids in this neighborhood spend more hours watching TV than what recommended by the AAP? Is this a two-tailed or one-tailed test? (Test the hypothesis at 95% level).
Hypotheses:
H0: µ ≤ 1.6
H1: µ > 1.6
Test is one-tailed (H1 shows a specific direction).
Use a one-sample z-test:
z = (1.75 - 1.6)/(0.85/√101)
Finish the calculation, then determine whether or not to reject the null hypothesis (H0). If the null is rejected, you can conclude µ > 1.6 hours.
To determine whether the kids in this neighborhood spend more hours watching TV than recommended by the AAP, we can conduct a hypothesis test.
Step 1: State the hypotheses
- Null Hypothesis (H0): The kids in the neighborhood spend, on average, 1.6 hours or less watching TV per day.
- Alternative Hypothesis (H1): The kids in the neighborhood spend, on average, more than 1.6 hours watching TV per day.
Step 2: Select the significance level
The significance level, also known as alpha (α), determines the probability of rejecting the null hypothesis when it is true. In this case, we have to test the hypothesis at a 95% level, so the significance level is 0.05.
Step 3: Compute the test statistic
To compare the sample mean (1.75 hours) and the hypothesized population mean (1.6 hours), we need to compute the test statistic. We can use the formula for a one-sample t-test:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
In this case:
Sample mean (x̄) = 1.75 hours
Hypothesized mean (μ) = 1.6 hours
Sample standard deviation (s) = 0.85
Sample size (n) = 101
Plugging in these values, we get:
t = (1.75 - 1.6) / (0.85 / sqrt(101))
Step 4: Calculate the p-value
Using a t-table or a statistical calculator, we can determine the p-value associated with the test statistic calculated in Step 3. The p-value represents the probability of obtaining a test statistic as extreme as (or more extreme than) the observed sample statistic, assuming the null hypothesis is true.
Step 5: Make a decision
Compare the p-value to the significance level (α) to make a decision:
- If the p-value is less than α, reject the null hypothesis and conclude that the kids in the neighborhood spend more hours watching TV than recommended by the AAP.
- If the p-value is greater than or equal to α, fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the kids in the neighborhood spend more hours watching TV than recommended by the AAP.
Since the question asked if it is a two-tailed or one-tailed test, it is a one-tailed test because we are specifically testing for whether the kids spend more hours watching TV than recommended.