Question

PART B

1. Waiting time for checkout line at two stores of a supermarket chain were measured for a random sample of customers at each store. The chain wants to use this data to test the research (alternative) hypothesis that the mean waiting time for checkout at Store 1 is lower than that of Store 2. (12 points)

Store 1 (in Seconds)	Store 2 (in Seconds)
461	264
384	308
167	266
293	224
187	244
115	178
195	279
280	289
228	253
315	223
205
197

1. What are the null and alternative hypothesis for this research?
2. What are the sample mean waiting time for checkout line and the sample

standard deviation for the two stores?

3. Compute the test statistic t used to test the hypothesis. Note that the population standard deviations are not known and therefore you cannot use the formula in Section 10.1. Use the one in Section 10.2 instead.
4. Compute the degree of freedom for the test statistic t.
5. Can the chain conclude that the mean waiting time for checkout at Store 1 is lower than that of Store 2? Use the critical-value approach and α = 0.05 to conduct the hypothesis test.
6.    Construct a 95% confidence interval for the difference of mean waiting

time for checkout line at the two stores.

Answer 1

Step 1) Define hypothesis:

Claim: The mean waiting time for checkout at store 1 is lower than that of store 2.

a)   H₀: µ₁ = µ₂                Vs Ha : µ₁ <   µ₂    

Where      µ₁ = Population mean for store 1.

µ₂   = Population mean for store 2.

     

Step 2) Find mean and standard deviation of given data:

b) Using excel functions we get

= 252.25 and ​​ = 252.8

S₁ = 98.542 and s₂ = 37.614

n₁ = 12 and n₂ = 10

Step 3) Define and calculate test statistic:

We assume that the underlying distributions are relatively normal , as population standard deviation is unknown we use t statistic for independent samples.

t =     

Where = sample mean from population 1 and ​​ = sample mean from population 2.

S₁ = sample standard deviation from population 1 and s₂ = sample standard deviation from population 2.

n₁ = sample size from population 1 and n₂ = sample size from population 2.

t =   

t= - 0.01784

c) Test statistic | t | = 0.0178

Step 4) Finding P value:

We have n₂ = 10 smallest sample size. D.f = n₂ - 1 = 10-1= 9.

d) Degree of freedom = 9

As alternative hypothesis have < sign. This is left tail test.

For statistic t = 0.0178, with 9 d.f P value is greater than 0.250

Step 5) Conclusion using level of significance 0.05

As P value 0.250 is greater than level of significance 0.05, We fail to reject H₀ .

At 5 % level of significance, we conclude that , we do not have enough evidence to conclude that there is mean waiting time for checkout at store 1 is lower than that of store 2.

e) Confidence interval

Step 1) sample mean for store 1 is = 252.25 and sample mean for Store 2 is   = 252.8

s₁ = sample standard deviation for store 1 = 98.542 and s₂ = sample standard deviation for store 2 = 37.614

Sample size from population 1 is n₁ = 12 and sample size from population 2 is n₂ = 10.

As both population values are independent and population standard deviation σ is unknown. We assume that the samples comes from population that are approximately normally distributed and standard deviation σ is not equal, we use t statistic .

Step 2)

First we find t_0.95 , by using smallest of n₁ – 1 and n₂ - 1, we use d.f = 10- 1 = 9

Using t table t_0.95,9 = 2.262

Step 3) Margin of error E

E = tc *  

E =t_0.95,9 *   

E =2.262*  

E = 69.7449

Step 4) The confidence interval is

() – E < µ₁   - µ₂ < () + E

(252.25- 252.8 ) – 69.7449   < µ₁   - µ₂ < (252.25- 252.8 ) + 69.7449   

-70.2949 < µ₁   - µ₂ < 69.1949

Confidence interval for (µ₁   - µ₂ ) is (-70.2949, 69.1949)

Step 5) Conclusion:

Confidence interval contains one negative value and one positive value , In this case we cannot conclude that either store 1 or Store 2 have more waiting time.