Question

Write a confidence interval problem using the option below. Gather the appropriate data and post your problem (without solution) in the discussion board. Once your initial post is complete, respond to your own post with the solution. Think about the population mean that you may be interested in and propose a confidence interval problem for this parameter. I believe the population mean number of times that adults ate fast food each week is 2.5. My data is after speaking to with 7 people I know and found that they ate fast food 1, 4, 0, 2, 4, 0 and 3 times last week. I choose to test this hypothesis at the .08 significance level. Please help solve this!!

Answer 1

= (1 + 4 + 0 + 2 + 4 + 0 + 3)/7 = 2

s = sqrt(((1 - 2)^2 + (4 - 2)^2 + (0 - 2)^2 + (2 - 2)^2 + (4 - 2)^2 + (0 - 2)^2 + (3 - 2)^2)/6) = 1.732

H₀: = 2.5

H₁: 2.5

The test statistic t = ()/(s/)

                             = (2 - 2.5)/(1.732/)

                             = -0.764

df = 7 - 1 = 6

P-value = 2 * P(T < -0.764)

             = 2 * 0.2369 = 0.4738

Since the P-value is greater than the significance level(0.4738 > 0.08), so we should not reject the null hypothesis.

So at 0.08 significance level there is sufficient evidence to conclude that the population mean number of times that adults ate fast food each week is 2.5.