Question

A die was thrown 400 times and showed a six 80 times. does this data justify rhe hypothesis that the die is unbaised at 5%level of significance

Answer 1

n(S) = 400
n(6) = 400/60
So P(6) should be = 1/6
But here, P'(6) = 80/400 = 1/5
And you can then go on...

Answer 2

Answer:

yes, this data justify the rhe hypothesis and the die is unbiased at 5% of significance.

Step-by-step explanation:

You can see that the dice were rolled 400 times and 6 times 80 times.

 p = percentage of 6 populations occurring in the cube

Therefore,

the null hypothesis, p=1/6{meaning the cube is not biased}

alternative hypothesis, p≠1/6 {that the cube is biased. Means}

The following test statistic uses a one-sample z-test of ratios.

TS = P-p / √p(1-p)/n ≈ N (0,1)

where, P = sample proportion of 6 = 80/400 = 0.20

n = number of dice rolled = 400

Therefore, test statistic =1.79

value z-test The statistic is 1.79.

The question give a significance level as 5%. Here, at the significance level of 0.05, the z-table shows the critical values ​​of -1.96 and 1.96 for the two-sided test.

Since the value of the test statistic is within the critical value of z, the null hypothesis is not included in the rejection range, so there is not enough evidence to reject the null hypothesis.

therefore we ended as a die is unbiased at 5% of significance.

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