A person 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: 28, 29, 40, 41, 28, 34. Use a 0.10 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

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis,

That is the loaded die behaves as a fair die.

Alternative hypothesis, : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency,

Test statistics,

≈ 5.8

Degrees of freedom, df = n - 1

                                       = 6 - 1

                                       = 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

     = 9.236

Thus the critical value is

             = chi dist (5.80, 5)

             = 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

2,160 π

step-by-step explanation:

change of stock price/initial stock price on friday * total stock value = change of stock value

0.89/73.73 *x = 42.72

x=3539.04 = total stock value at initial stock price on friday .

total stock value at initial stock price on friday/ initial stock price on friday = number of shares

3539.04/73.73 = 48 shares

they own 48 shares of this stock