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:
the chance of getting heads when flipping a coin is 50%. As I cannot see the graphs, choose the one that shows 50%.
A is continuous
B is discrete
C is not a function
S is a function
Answer:
Step-by-step explanation:
The null and the alternative hypothesis is:
The t- student test statistics can be computed as:
t = -2.017
degree of freedom = (n₁ - 1) + (n₂ - 1)
= (100 - 1) + (130 - 1)
= 228
Using the data of t-value and degree of freedom;
The P-value = -0.224
Decision rule: Do not reject the null hypothesis if the p-value is greater than ∝(0.01)
Conclusion: We reject the null hypothesis since the p-value is less than ∝.
Therefore, there is enough evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United States.
