Question

3) A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The wner of the store has determined that home delivery will be successful if the average time spent on the eliveries does not exceed 32 minutes. The owner has randomly selected 23 customers and has delivered Izzas to their homes in order to test if the mean delivery time actually exceeds 32 minutes. Suppose the -value for the test was found to be 0.0297. State the correct conclusion. (3 pts) 14) Many track hurdlers believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track hurdlers in the Hifferent starting positions. Calculate the chi-square test statistic x2 to test the claim that the probabilities of winning are the same in the different positions. Use a = 0.05. The results are based on 240 wins. (8 pts) Starting Position 1 2 3 + 5 6 Number of Wins 33 50 44 32 36 45

Answer 1

## 13 ) Answer :

Claim :

The owner of the store has determined that home delivery will be successful if the average

time spent on the deliveries does not exceed 32 minutes .

To test :

Ho : μ = 32 VS H1 : μ < 32

α = 5 % ( default )

we have given p value for the test = 0.0297

Decision :

we reject Ho or result is significant if p value is less than α value using p value approach

here p value is less than α value we reject Ho at given level of significance .

Conclusion :

There is sufficient evidence to conclude that home delivery will be successful if the average

time spent on the deliveries does not exceed 32 minutes . here claim is satisfy for α = 5 % .

## 14 ) we have to do chi square test :

Claim : To test the claim that the probabilities of winning are the same in the different position .

α = 0.05 level of significance .

N = total observation = 240

# To test :

Ho : The probabilities of winning are the same in the different position VS

H1 : The probabilities of winning are the different   in the different position

## Test statistics :

chi square =  χ2 = Σ ( Oi - Ei ) ² / Ei

where Oi : observed  Frequency and

Ei = Expected frequency .

# Observational frequency table :

Starting position	1	2	3	4	5	6
Number of wins	33	50	44	32	36	45

Expected Frequency :

Ei = N* Pi here N = 240 and Pi = 1 / 6

here we have equal probability hence  

E1 = E2 = E3 = E4 = E5 = E6 = 240 * ( 1 /6 ) = 40

χ2 = Σ ( Oi - Ei ) ² / Ei  

= [ ( 33 - 40) ² + ( 50 - 40 ) ² + ( 44 - 40) ² + ( 32 - 40) ² + ( 36 - 40 ) ² + ( 45 - 40 ) ² ] / 40

= [ 49 + 100 + 16+ 64 + 16 + 25 ] / 40

= 270 / 40

= 6.75

## df = degree of freedom = k -1 here k = 6 therefore df = 5

## p value = 0.2399 ( use statistical table )

# Decision :

We reject Ho if p value is less than α value using p value approach here p value is

greater than α value here we fail to reject Ho at given level of significance .

# Conclusion :

There is Insufficient evidence to conclude that ,

The probabilities of winning are the different   in the different position

here claim not satisfy . we accept Ho that is

the probabilities of winning are the same   in the different position