Problem with significance level questions

[starter90]

Hi to everyone in this forum, I'm a college student having trouble with statistics and have a final exam in 2 days, so can someone please help me out.

-Here is the question and what I came up with...

A fast-food restaunrat advertises that their delivery time is less as compared to a national fast-food chain. To determine whether this advertisement is valid, 10 orders are placed to each restaurant. Delivery time in minutes are shown below:

Local 16.8 11.7 15.6 16.7 17.5 18.1 14.1 21.8 13.9 20.8
Chain 22 15.2 18.7 15.6 20.8 19.5 17 19.5 16.5 24

At the 0.05 level of significance, is there evidence that the mean delivery time is lower for the local pizza restaurant than that of the national pizza chain?


Test Hypothesis

Step1

RV X: Delivery time.
Population 1 - N(u1, varinace(squared)
Sample 1-(X1,s1(squared)

Population 2- N(u2, variance(squared)
Sample 2-(X2,s2(squared)

Step 2

Ho=U1=U2 (mean delivery time for population is same)
H1:U1<U2 (mean delivery time for local pizza restaurant is less than that of the national pizza chain)

Step 3

n (sample size)= 10
x=find it.



Local Chain D=x1-x2 D(squared)
1 16.8 22 -5.2 27.04
2 11.7 15.2 -3.5 12.25
3 15.6 18.7 -3.1 9.61
4 16.7 15.6 1.1 1.21
5 17.5 20.8 -3.3 10.89
6 18.1 19.5 -1.4 1.96
7 14.1 17 -2.9 8.41
8 21.8 19.5 2.3 5.29
9 13.9 16.5 -2.6 6.76
10 20.8 24 -3.2 10.24
Total -21.8 93.66


Mean=21.8/10=2.18

Variance=1/9(93.66-(21.8)squared/10

Variance=5.126

Standard deviation=2.264

Step 4-Test statistic:

T=T/SD/ √n...T=2.18/2.264/3.162= 3.045- The T-value.

Step 5

0.95

1-0.95=0.05
0.05/2=0.025

Corresponding T- value i found from table is 2.2622 (t-value)


So this is what I came up with, How can i come up with a decision on this question.

Thanks....

[GretaGarbo]

It seems that you have calculated the values correctly. (I checked with R.)

Now, you know from the t-table that only 5% of the t-values will bee larger (in absolute value) than 2.26. So it would be quite rare to observe a t-value larger than 2.26 if the null hypothesis is true.

But you have observed a t-value as large as 3.04. So, how much do you believe in the null hypothesis?

The t-test is based on the normal distribution. The data does not really seems to be normally distributed (from a QQ-plot). Maybe you should ask your teacher how one really should analyse these data.

local1  <- c(16.8 ,11.7,15.6,16.7,17.5,	18.1,	14.1,	21.8,	13.9	,20.8)
chain1  <- c(22,15.2,18.7,15.6,20.8,19.5, 17,19.5,16.5,24 )

t.test(local1,chain1,paired =TRUE)
hist(local1-chain1)
qqnorm(local1-chain1)

[starter90]

Thanks for your reply,
This was an assignment our prof. gave us. I really didn't get what you meant by asking whether I believe in the null hypothesis. I probably am lacking some knowledge.
My problem is really interpreting what I find. The t- value I find on step 4 and the number i find from the t-table on step 5. I can't come up with a conclusion on whether Ho or H1 is correct.

Maybe something is wrong with my test-statistic. By the way there is a formula which I want to ask whether I can use it on this problem;

sp2=(n1-1)S1(squared)+(n2-1)s2(squared)/n1+n2-2

Thanks again....

[GretaGarbo]

I am trying to get you to think about this situation. What conclusion can you make about the null hypothesis?

Your t-values would normally be from –2.2 up to +2.2, provided that the null hypothesis is true. Now, you have observed a large t-values of 3.04. What conclusion do you make about the null hypothesis?

Suppose there is an animal in the house next to you. Your null hypothesis is that it is a dog. Your alternative hypothesis is that it is a horse. You know that a dogs weight is about 20 kg, say from 10 kg to 30 kg.

Now, someone inform you that the animals weight is 200 kg. Then, what conclusion do you make about the null hypothesis? Do you still believe it is a dog? Do you reject the null hypothesis or do you accept it?

The professor gave you an assignment. I think you should give her an assignment! How should these data be analysed? Because they don’t seem to be normally distributed. They are the teachers. They should give correct answers. I would be really interested in hearing what the professor said. We have tried to help you. You can give back information about your professor’s reply.


That formula you gave is the formula for a pooled variance. You don’t need that since you calculate a paired t-test and only need one standard deviation.

[starter90]

I finally was able to solve it. Apparently this is about comparing 2 independent populations and i am supposed to calculate each sample mean and variance separetly. Then use T=x1-x2/sp.square root of 1/n1+1/n2...

and got a value for T=1.660.

Thanks for trying to help.