Question

At a recent Union meeting of Foodmart staff, concern was expressed about the increasing number of hours that stores were open. Staff felt that they we being made to work longer and longer hours. One union official claimed that, on average, across all Foodmart supermarkets stores are open for more than 100 hours trading per week.

To test this claim by the union official, the 150 supermarkets were surveyd and the average trading hours per week were 104.35 with a standard deviation of 23.677 hours.

1. (a) Write down the null and alternative hypotheses in both symbols and words.

2. (b) Step 2. Write which type of test are you conducting?

3. (c) Step 3. Using a level of significance (α) of 5%, calculate the Critical Value.

4. (d) Step 4. Write your decision rule?

5. (e) Step 5. Calculate your test statistic.

6. (f) Step 6. State in plain language your conclusion.

Answer 1

(a)

on average, across all Foodmart supermarkets stores are open for 100 hours trading per week.

On average, across all Foodmart supermarkets stores are open for more than 100 hours trading per week.

(b)

So we use Right Tailed Test i.e One Tailed Test.

(c)

Given = 5% =0.05

Since it is the one tailed test; So, we have to find out the Critical value for 2 i.e 2x5% = 10% = 0.1

(d)

: then we Accepth

: then we Reject

(e)

To test the we use the Z - Statistic ( Because n = 150 which is a large number)

(f)

If you observe 2.2502 > 1.645

: So, We Reject

Therefore we Conclude that "On average, across all Foodmart supermarkets stores are open for more than 100 hours trading per week".

We can also Conclude that we Accepted  

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NOTE: To See the Critical Values of Z; we use the Standard Normal area tabulated values which i posted below.

How to see?

If alpha = 5% the confidence interval = 95%

Divide the value 95 with 100. you will get 0.95 and after that divide the value 0.95 with 2; you will get 0.475. Now Search this 0.475 in side the table ( repeating again see inside the table). You will find this at the intersection of (1.9, 0,06) which is Zcri value i.e 1.96.

Similarly If alpha = 10% the confidence interval = 90%

Divide the value 90 with 100. you will get 0.90 and after that divide the value 0.90 with 2; you will get 0.45. Now Search this 0.45 in side the table ( repeating again see inside the table). You will find this at the intersection of (1.6, 0,04) which is Zcri value i.e 1.645.

Like this you can find the Zcri for any alpha values.