A nutritionist believes that 10% of teenagers eat cereal for breakfast. To investigate this claim, she selects a random sample of 150 teenagers and finds that 25 eat cereal for breakfast. She would like to know if the data provide convincing evidence that the true proportion of teenagers who eat cereal for breakfast differs from 10%. What are the values of the test statistic and P-value for this test?
Find the z-table here.
To test the nutritionist's claim, we can conduct a hypothesis test for a population proportion.
Null hypothesis: The true proportion of teenagers who eat cereal for breakfast is 10% (p = 0.10)
Alternative hypothesis: The true proportion of teenagers who eat cereal for breakfast differs from 10% (p ≠ 0.10)
First, we calculate the test statistic:
\[ z = \frac{p_{\text{sample}} - p_{\text{null}}}{\sqrt{\frac{p_{\text{null}}(1-p_{\text{null}})}{n}}} \]
Where:
\( p_{\text{sample}} = \frac{25}{150} = 0.1667 \) (proportion in sample)
\( p_{\text{null}} = 0.10 \) (null hypothesis proportion)
\( n = 150 \) (sample size)
\[ z = \frac{0.1667 - 0.10}{\sqrt{\frac{0.10(0.90)}{150}}} \]
\[ z = \frac{0.0667}{\sqrt{\frac{0.09}{150}}} \]
\[ z = \frac{0.0667}{\sqrt{0.0006}} \]
\[ z = \frac{0.0667}{0.02449} \]
\[ z = 2.7261 \]
Next, we find the p-value associated with this z-value using the z-table. A z-value of 2.7261 corresponds to a p-value of approximately 0.0065.
Since this p-value is less than the significance level typically set at 0.05, we reject the null hypothesis. There is convincing evidence to suggest that the true proportion of teenagers who eat cereal for breakfast differs from 10%.