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Midterm Exam 2 Review (Nonadaptive) Question 15 of 16 (1 point)|Question Attempt: 1 of Unlimited

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Question 15 Partially Correct Your answer is incorrect. (a): Your answer is incorrect. (b) Degrees of freedom: Your answer is incorrect. (d): Your answer is incorrect. Over the years, the restaurant Zoodle has had a mean customer satisfaction rating of 70 for its deliveries, with a variance of 53. The price for delivery was recently changed which, the owner claims, has caused the variance, σ2, to decrease. He phones a random sample of 30 customers who ordered deliveries at the new price and asks them to rate their satisfaction. The customers in the sample have a mean rating of 71 with a variance of 35. Assuming that customer ratings for the newly-priced deliveries are approximately normally distributed, is there significant evidence, at the 0.05 level of significance, to conclude the owner's claim is correct? Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)

(a) State the null hypothesis H0 and the alternative hypothesis H1. H0: H1: (b) Determine the type of test statistic to use. ▼Chi-square Degrees of freedom: (c) Find the value of the test statistic. (Round to three or more decimal places.) 19.151 (d) Find the p-value. (Round to three or more decimal places.)

(e) Can we support the claim that the variance of customer ratings for the newly-priced deliveries is less than 53? Yes No

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Chi-square Distribution
Variance

Formulas

Chi-square test statistic: χ² = (n-1) * s² / σ₀²
Degrees of freedom: df = n - 1

Theorems

Chi-square test for variance

Suitable Grade Level

Undergraduate (Statistics/Probability Course)