The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 27, 31, 42, 40, 28, and 32. use a 0.05 significance level to test the claim that the outcomes are not equally likely. does it appear that the loaded die behaves differently than a fair die

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number:      1,    2 ,   3 ,  4 , 5    ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is

The statistic to check the hypothesis is given by:

The observed values are given:

The expected values are given by:

And now we can calculate the statistic:

Now we can calculate the degrees of freedom for the statistic given by:

And we can calculate the p value given by:

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

may i see the picture? it would a lot, xox