Question

A consumer group claims that more than 62% of people choose coffee rather than other beverages as their preferred drink in the morning. In the sample of 92 people, 68 reported that they prefer coffee. At α = 0.03, is there enough evidence to support the consumer group’s claim?

Answer 1

Answer on Question #79688 – Math – Statistics and Probability

Question

A consumer group claims that more than 62% of people choose coffee rather than other beverages as their preferred drink in the morning. In the sample of 92 people, 68 reported that they prefer coffee. At $\alpha = 0.03$, is there enough evidence to support the consumer group's claim?

Solution

Null hypothesis $H_0$: $p = 0.62$.

Alternative hypothesis $H_a$: $p > 0.62$.

Test statistic: $z = \frac{p - \hat{p}}{\sqrt{\frac{p(1 - p)}{n}}} = \frac{\frac{68}{92} - 0.62}{\sqrt{\frac{0.62(1 - 0.62)}{92}}} = 2.354$.

P-value (from the table): $p = 0.009$.

Since the P-value is less than 0.03 the null hypothesis should be rejected.

There is enough evidence to support the consumer group's claim.

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