Question

Write the R-code for the following problem to get the solution.

A die was thrown 400 times and 6 resulted 80 times. Do the data justify the hypothesis that the die is unbiased?

Answer 1

Yes, the data justify the hypothesis that the die is unbiased.

Step-by-step explanation:

We are given that a die was thrown 400 times and 6 resulted in 80 times.

Let p = population proportion of 6 occurring on a die

So, Null Hypothesis, $H_0$ : $p=\frac{1}{6}$      {means that the die is unbiased}

Alternate Hypothesis, $H_A$ : $p\neq \frac{1}{6}$      {means that the die is biased}

The test statistics that will be used here is One-sample z-test for proportions;

                         T.S.  =  $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of 6 occurring = $\frac{80}{400}$ = 0.20

           n = sample of times die was thrown = 400

So, the test statistics =  $\frac{\frac{1}{6}-0.20 }{\sqrt{\frac{\frac{1}{6}(1-\frac{1}{6})}{400} } }$  

                                    =  1.79

The value of z-test statistics is 1.79.

Since in the question, we are not given the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the z table gives a critical value of -1.96 and 1.96  for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the die is unbiased.