Question

4. Below is the amount that a sample of 15 customers spent for lunch ($) at a fast-food restaurant: 7.42 6.29 5.83 6.50 8.34 9.51 7.10 5.90 4.89 6.50 5.52 7.90 8.30 9.60 6.80 Recall the lunch at a fast-food restaurant problem from Assignment 4. Let µ represent the population mean amount spent for lunch ($) at a fast-food restaurant. Previously you calculated the mean and standard deviation of the fifteen sample measurements to be x ̅ = $7.09 and s = $1.406, respectively. Suppose you want to determine if the true value of µ differs from $7.50. Specify the null and alternative hypotheses for this test. Since x ̅ = $7.09 is less than $7.50, a manger wants to reject the null hypothesis. What are the problems with using such a decision rule? Compute the value of the test statistic. Find the approximate p-value of the test or use technology to find the exact p-value. Select a value of α, the probability of a Type I error. What does α represent in the words of the problem. Give the appropriate conclusion, based on the results of parts d and e. What conditions must be satisfied for the test results to be valid? In Assignment 4, you found a 95% confidence interval for µ. Does this interval support your conclusion in part f?

Answer 1

µ represent the population mean amount spent for lunch ($) at a fast-food restaurant.

x ̅ = $7.09 and s = $1.406

Here

Null Hypothesis : H₀ : = $ 7.50

Alternative Hypothesis : H_a : $ 7.50

Since x ̅ = $7.09 is less than $7.50, a manger wants to reject the null hypothesis. What are the problems with using such a decision rule?

Answer : Here the problem with such a decision making is that it is irrational to assume that the given sample of 15 would produce an exact sample mean equal to population mean. Population size is pretty larger and in the long run, sample mean will tend towards population mean.

Here standard error of sample mean = s/sqrt(n) = 1.406/sqrt(15) = 0.363

degree of freedom = dF = n - 1 = 15 - 1 = 14

test statistic

t = (7.09 - 7.50)/0.363 = -1.1294

p - value = 2 * Pr(t < -1.1294, dF = 14) = 0.2777

here we select an alpha of 0.05.

What does α represent in the words of the problem?

Answer : Here α represent that the probability of choosing the decision that true value of   mean amount spent for lunch ($) at a fast-food restaurant differs from $7.50 when actually it is $ 7.50

Here as p - value is greater than 0.05 so we will fail to reject the null hypothesis and conclude that mean amount spent for lunch ($) at a fast-food restaurant doesn't differs from $ 7.50

What conditions must be satisfied for the test results to be valid?

1. sample must be selected randomlly.data is collected from a representative, randomly selected portion of the total population

2. scale of measurement applied to the data collected follows a continuous or ordinal scale.

3. when plotted, results in a normal distribution, bell-shaped distribution curve

95% confidence interval = x ̅ +- t_critical se₀ = 7.09 +- 2.1447 * 1.406/sqrt(15) = ($ 6.31, $ 7.87)

so as the confidence interval consist the value of $ 7.50 so here interval support our conclusion