Answered: An advertising executive claims that…

Practical Management Science

6th Edition

ISBN:9781337406659

Author:WINSTON, Wayne L.

Publisher:WINSTON, Wayne L.

Chapter2: Introduction To Spreadsheet Modeling

Section: Chapter Questions

Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...

See similar textbooks

Related questions

Question

An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 15 Visa Gold cardholders resulted in a mean household income of $⁢68,420 with a standard deviation of $⁢9100. A random survey of 6 MasterCard Gold cardholders resulted in a mean household income of $⁢59,200 with a standard deviation of $⁢9200. Is there enough evidence to support the executive's claim? Let μ1 be the true mean household income for Visa Gold cardholders and μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.05 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 1 of 4:

State the null and alternative hypotheses for the test.

Step 2 of 4:

Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4:

Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4 of 4:

State the test's conclusion.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution