Question

A social researcher claims that the average adult listens to the radio more than 20 hours per week. He collects data on 35 individuals' radio listening habits and finds that the mean number of hours spent listening to the radio weekly was 21.4 hours. You are given that the population standard deviation is 4 hours. Test the researcher's claim at the 10% significance level. What do you find?

Answer 1

To test the researcher's claim, we will conduct a hypothesis test using the z-statistic. The main steps are:

1. **State the Hypotheses**
The null hypothesis (H₀) assumes that the average time adults listen to the radio per week is 20 hours. The alternative hypothesis (H₁) is what the researcher claims, that is, the average time is more than 20 hours.

2. **Calculate the Test Statistic**
The test statistic is calculated using the formula for the z-score:

```
z = (sample mean - claimed mean) / (population standard deviation / sqrt(sample size))
```

Plugging in the given values, we get

```
z = (21.4 - 20) / (4 / sqrt(35))
```

After performing above calculation, the obtained z-value is approximately 2.0706.

3. **Decide the significance level**
The significance level is stated in the problem as being 10%. We use this to find the critical z-value. The critical z-value is the z-score that is associated with our chosen level of significance. The corresponding critical z-value for 10% significance level is approximately 1.2816.

4. **Comparison and Decision**
Observe the calculated test statistic (z-value) and critical z-value. If the test statistic falls in the critical region (z-value > critical z-value), we reject the null hypothesis. In this case, the calculated z-value (2.0706) is greater than the critical z-value (1.2816). Therefore, we reject the null hypothesis.

5. **Conclusion**
We conclude that at a 10% level of significance, the data supports the researcher's claim that the average adult listens to the radio more than 20 hours per week.