Use the following to answer questions:
A recent survey indicated that the average amount spent for breakfast by business managers was $7.58 with a standard deviation of $0.42. It was felt that breakfasts on the West Coast were higher than $7.58. A sample of 81 business managers on the West Coast had an average breakfast cost of $7.65.
-Find the P-value for the test.
A) 0.4332
B) 0.2734
C) 0.1325
D) 0.0668

Question

Quizplus · Accepted Answer

We can use a one-sample z-test to compare the sample mean to the population mean:

z = (x̄ - μ) / (σ / √n)

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x̄ = $7.65, μ = $7.58, σ = $0.42, and n = 81. Plugging in the numbers, we get:

z = (7.65 - 7.58) / (0.42 / √81) = 1.67

Using a standard normal distribution table or calculator, we can find the probability of getting a z-score of 1.67 or higher:

P(z > 1.67) = 0.0475

This is the p-value for the test, which is the probability of getting a sample mean at least as extreme as $7.65 if the true population mean is $7.58. Since this p-value is less than the usual significance level of 0.05, we reject the null hypothesis and conclude that there is evidence to suggest that breakfasts on the West Coast are indeed higher than $7.58 on average.

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