Question

The following table gives the number of aircraft
accidents that occurred during the various days
of a week. Find whether the accidents are
uniformly distributed over the week.
Days
Sun Mon Tue Wed Thurs Fri
Sat
No. of
Accidents
14
16
8
12
11
9
14
Given x2 at 6 d.f=12.59​

Answer 1

Answer:

Step-by-step explanation:

i. Null hypothesis  (H0): Accidents are equally distributed over all the days of week.

Alternative Hypothesis  (Ha) : Accidents do hot occur equally.

ii. Calculation of test statics: If the accidents occur equally on all days of a week, than there will be 847=12 accident per day, i.e E=12

Day Observed frequency (O) Expected Frequency (E)  (O−E)2  X2=(O−E)2E

Sun 13 12 1 0.0833

Mon 15 12 9 0.75

Tue 11 12 1 0.0833

Wed 9 12 9 0.75

Thu 12 12 0 0

Fri 10 12 4 0.3333

Sat 14 12 4 0.3333

Total    ∑x2 = 2.33

iii. Level of significance: α = 0.05

Degree of freedom = h-1

=7-1

=6

iv Critical value ⇒ For 6 degrees of freedom at 5% level of significance table value.

x2 is 12.59

v. Decision ⇒ Since the calculated value of  x2 is less than the table value. The hypothesis is accepted.

∴ The accidents occur equally on all working days.

Answer 2

Answer:

Accept null hypothesis. That is the air accidents are uniformly distributed over the week.

Step-by-step explanation:

Given:

The table of data

To find:

whether the accidents are uniformly distributed over the week.

Solution:

Null Hypothesis $\left(H_{0}\right)$ :

The accidents are uniformly distributed over the week.

Alternative Null Hypothesis $\left(H_{\mathbf{1}}\right)$ : The accidents are not uniformly distributed over the week.

The test statistic is given by

$\begin{aligned}&\chi^{2}=\sum_{i=1}^{n} \frac{\left(O_{i}-E_{i}\right)^{2}}{E_{i}} \quad \chi^{2} \text { distribution with }(n-1) \text { d.o. } f \\&E_{i}=\frac{\text { total no. of abservations }}{n}=\frac{84}{7}=12 \quad N=84, \quad n=7 \\&E_{i}=12\end{aligned}$

From the table,

$\chi^{2}=\sum \frac{(O-E)^{2}}{E}=4.167$

(Calculated Value)

Degree of freedom $=n-1=7-1=6$

Table value of $\chi^{2} for 6 \mathrm{~d} . f . =12.592$(Table Value)

Calculated Value $<$ Table Value

We accept the null hypothesis $H_{0}$.