Answer:
The calculated χ² = 0.57 does not fall in the critical region χ² ≥ 12.59 so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.
Step-by-step explanation:
1) We set up our null and alternative hypothesis as
H0: proportion of fatal bicycle accidents in 2015 was the same for all days of the week
against the claim
Ha: proportion of fatal bicycle accidents in 2015 was not the same for all days of the week
2) the significance level alpha is set at 0.05
3) the test statistic under H0 is
χ²= ∑ (ni - npi)²/ npi
which has an approximate chi square distribution with ( n-1)=7-1= 6 d.f
4) The critical region is χ² ≥ χ² (0.05)6 = 12.59
5) Calculations:
χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28
χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]
χ²= 1/14.28[8.1364]
χ²= 0.569= 0.57
6) Conclusion:
The calculated χ² = 0.57 does not fall in the critical region χ² ≥ 12.59 so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.
b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week
Answer:
b
Step-by-step explanation:
because there only allowing the first 50 to register that would be a contest
The answer is the function is not one to one
There are 14 fruit to start with and 2 bananas
2/14
then there are 13 fruit and 2 pears
2/13
So the probability of the two events in a row without replacement is:
(2/14)(2/13)=4/182
2/91 ( about 2.2%)
