hey found that 27 occurred in the Spring, 39 in the Summer, 31 in the Fall, and 53 in the Winter. Can it be concluded at the 0.05 level of significance that car accidents are not equally distributed throughout the year

This question is incomplete, the complete question is;

A Driver's Ed program is curious if the time of year has an impact on numer of car accidents in the U.S.

They assume that weather may have a significant impact on the ability of drivers to control their vehicles. They take a random sample of 150 car accidents and record the seasons each occurred in. They found that 27 occurred in the Spring, 39 in the Summer, 31 in the Fall, and 53 in the Winter. Can it be concluded at the 0.05 level of significance that car accidents are not equally distributed throughout the year

a) Yes, because the p-value = 0.0145

b) No, because the p-value = 0.0145

c) Yes, because the p-value = 0.0291

d) No, because the p-value = 0.0291

p-value = 0.0145

Since p-value ( 0.0145 ) is less than Significance level ∝ ( 0.05 ),

We reject Null hypothesis.

Hence, There is sufficient evidence to conclude that car accidents are NOT equally distributed through out the year.

[ Option a) Yes, because the p-value = 0.0145 ] is the correct answer.

Step-by-step explanation:

Given the data in the question;

number of car accident = 150

Observed Frequencies O are;

Spring = 27

Summer = 39

Fall = 31

Winter = 53

Significance level ∝ = 0.05

Hypothesis

Null hypothesis            H₀ : The car accidents are equally distributed through out the year

Alternative hypothesis H₀ : The car accidents are NOT equally distributed through out the year

Now, Expected Frequency E will be;

Spring = 150/4 = 37.5

Summer = 150/4 = 37.5

Fall = 150/4 = 37.5

Winter = 150/4 = 37.5

Test Statistics;

= ∑[ ( O-E )² / E ]

so

= [ ( 27-37.5 )² / 37.5 ] + [ ( 39-37.5 )² / 37.5 ] + [ ( 31-37.5 )² / 37.5 ] + [ ( 53-37.5 )² / 37.5 ]

= 2.94 + 0.06 + 1.1267 + 6.4067

= 2.94 + 0.06 + 1.1267 + 6.4067

= 10.5334

Degree of Freedom DF = n-1 = 4 -1 = 3

Now,

p-value = P( - ) = P( - 10.5334 ) = 0.0145

p-value = 0.0145

Since p-value ( 0.0145 ) is less than Significance level ∝ ( 0.05 ),

We reject Null hypothesis.

Hence, There is sufficient evidence to conclude that car accidents are NOT equally distributed through out the year.

[ Option a) Yes, because the p-value = 0.0145 ] is the correct answer.

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)