Question

Cereal for Breakfast. A cereal company announced 80% of Americans start the day with cereal for breakfast. You believe the percentage of Americans that start the day with cereal is less. To test your claim, you randomly sample 100 Americans and ask them what they eat for breakfast. 70 of them say cereal. Test your claim that percent of Americans that start the day with cereal for breakfast is less than 80% at a 5% level of significance.

(b) Find the p-value (round to four decimal places

Answer 1

The null hypothesis is that the percentage of Americans who start the day with cereal for breakfast is 80%. The alternative hypothesis is that the percentage is less than 80%.

Let p be the true proportion of Americans who start the day with cereal. The sample proportion is 70/100 = 0.7. The standard error is √(p(1-p)/n) = √(0.8(0.2)/100) = 0.04.

The test statistic is (0.7 - 0.8) / 0.04 = -2.5. From the standard normal distribution table, the p-value for a one-tailed test with a test statistic of -2.5 is 0.0062.

Since the p-value (0.0062) is less than the significance level (0.05), we reject the null hypothesis. There is evidence to support the claim that the percentage of Americans who start the day with cereal for breakfast is less than 80%.