Question

A large furniture company claims that 65% of all individuals who buy chairs from its stores choose wood chairs, 20% choose plastic chairs, and 15% choose metal chairs. to investigate this claim researchers collected data from a random sample of the companys customers. the results were 305 wood, 121 plastic, and 74 metal. are the data from the sample consistent with the companys claim? conduct an appropriate statistical test at the 5% significance level to support your conclusion make sure to include parameters, check conditions, and show calculations before formulating a conclusion.

Answer 1

Answer:

In the sample, 61% chose wooden chairs (4% less than that claimed by the company), 24.2% chose plastic chairs (4.2% more than that claimed by the company), and 14.8 % chose metal chairs (0.2% less than what the company claimed). Therefore, given that these are differences of less than 5%, we can affirm that these samples are consistent with what the company said.

Step-by-step explanation:

Given that a large furniture company claims that 65% of all individuals who buy chairs from its stores choose wood chairs, 20% choose plastic chairs, and 15% choose metal chairs, and to investigate this claim researchers collected data from a random sample of the companys customers, whose results were 305 wood, 121 plastic, and 74 metal, to determine if the data from the sample consistent with the companys claim the following calculations should be performed:

305 + 121 + 74 = 500

500 = 100

305 = X

305 x 100/500 = X

30,500 / 500 = X

61 = X

500 = 100

121 = X

121 x 100/500 = X

12,100 / 500 = X

24.2 = X

100 - 61 - 24.2 = X

39 - 24.2 = X

14.8 = X

Therefore, in the sample, 61% chose wooden chairs (4% less than that claimed by the company), 24.2% chose plastic chairs (4.2% more than that claimed by the company), and 14.8 % chose metal chairs (0.2% less than what the company claimed). Therefore, given that these are differences of less than 5%, we can affirm that these are samples that are consistent with what the company said.

Answer 2

Answer:

The Chi^2 value is 5.654. The p-value is .05919. The result is not significant at p < .05.

Step-by-step explanation:

To do this we need to do a goodness of fit chi-test

You can do it by hand or use a graphing calculator, but IMO using a calculator online is the easiest. Remember one sample is frequency while the other is in proportions so convert one to the other.

In other words if you're doing a proportions test divide 305/500 to get 0.61 and then plug that in next to the expected value of 0.65.

If you/re doing a frequency test, multiply 61% * 5 =325 and then plug that in next to 305.

Both a frequency and a proportion test will yield the same result so don't worry about which you choose.

Once you've plugged the values in you'll get a response like this:

The Chi^2 value is 5.654. The p-value is .05919. The result is not significant at p < .05.