A public bus company offical claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.6 minutes with a standard deviation of 2.3 minutes. At the 0.01 signficance level, test the claim that the mean waiting time is less than 10 minutes. Use the p-value method of testing hypotheses.

Step 1:

(1)

A public bus company official claims that the mean waiting time for bus number 14 is less than 10 minutes.

Null hypothesis :

One of these statements must become the null hypothesis , and the other should be the alternate hypothesis.

The null hypothesis contains equality.

So for the above, the null hypothesis H₀ : x = 10.

Step 2:

Alternate hypothesis :

The statement that does not contain equality is the alternative hypothesis.

Mean waiting time for bus number 14 is not equal 10 minutes.

So for the above, the alternate hypothesis H_a : x ≠ 10.

Step 3:

Karen took bus number 14 during peak hours on 18 different occasions.

So the number of samples is 18.

Mean waiting time was 7.6 minutes.

Standard deviation is 2.3 minutes.

Mean = 14.02857

Standard deviation = 0.36839.

Test static : $\"\"$ .

$\"\"$

Test statics : $\"\"$ .

Step 4:

Find the critical values.

Significance value is 0.01.

$\"\"$ .

The number of samples is 18.

Degree of freedom $\"\"$ .

Calculate the critical values using calculator.

Follow these steps to evaluate critical values.

1.Select invT()

[2nd --> VARS --> 4 ]

2.Enter the values of $\"\"$ and df.

area : 0.01

df : 17

3.Now press Enter in calculator to view answer.

invT(0.01, 17)

$\"\"$

Step 5:

Find the p-value.

Follow these steps to evaluate p-value.

1.Select tcdf()

[2nd --> VARS --> 6 ]

2.Enter the values of t and df.

Lower : 4.427

Upper : 1000

df : 17 \ \

For two-tailed (non-directional), the upper value is considered as 1000.

3.Now press Enter in calculator to view answer.

tcdf(4.427, 1000, 17)

=0.000184.

$\"\"$ .

p-Value of the hypothesis is 0.000184.

Step 6:

Conclusion:

Since p -value = $\"\"$ , we should reject H₀ .

We conclude that the mean is less than 10 minutes.

$\"\"$