Question

1. An automobile dealership is considering lowering the price it charges drivers for a basic oil change from $45 to $40 in order to attract new customers. But, it will only do so if it can conclude, at the .01 level of significance, that the average labour costs involved with this service is less than $10.00. It randomly selected 20 automobiles coming in for this basic oil change and determined what the labour costs were for each of these automobiles. Based on these costs, the average cost was calculated to be $9.35 and the standard deviation in these costs was calculated to be $1.20. Based on this evidence, would the dealership lower its cost for a basic oil change?
2. In many bank branches across Canada, new ATM machines have been installed which no longer require users to place cheques and money in envelopes before inserting them into the machines. Did these new ATM machines affect how long customers spent depositing cheques and money? With the old ATM machines, it was determined that the true average amount of time spent depositing cheques and currency was 1.30 minutes. A random sample of 30 customers at a randomly selecting branch which had these new machines was observed depositing cheques and currency. The average time this sample of customers took was calculated to be 1.43 minutes and their standard deviation in times was calculated to be 0.40 minutes. At the .10 level of significance, can we conclude that the average time with the new ATM machines is different than the average time with the old ATM machines?

• the two hypotheses
• the value of the test statistic along with your calculations
• the p-value of the test along with the appropriate graph
• the decision rule along with the appropriate graph
• a two-part conclusion: whether you reject or don’t reject the null hypothesis, and, what does this rejection mean to the particular question

Answer 1

1.

Suppose, random variable X denotes labor cost (in dollars).

We have sample values. But we do not know population standard deviation (or variance). So, we have to perform one sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

[Using R-code 'pt(-2.422407,19)']

Level of significance

We reject our null hypothesis if

Here, we observe that

So, we cannot reject our null hypothesis.

Based on the given data we can conclude that there is no significant evidence that that average labor costs involved with this service is less than $10.00.

Hence, based on this evidence, the dealership would not lower its cost for a basic oil change.

P-value (probability of being extreme than test statistic) is as follows.

Decision rule for this hypothesis test is as follows.

Critical value is given by [Using R-code 'qt(0.01,19)']

We reject our null hypothesis if

So, critical region (rejection region) is as follows.

2.

Suppose, random variable X denotes time (in minutes) required by a customer.

We have sample values. But we do not know population standard deviation (or variance). So, we have to perform one sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

[Using R-code 'pt(-1.780098,29)+1-pt(1.780098,29)']

Level of significance

We reject our null hypothesis if

Here, we observe that

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that average time with the new ATM machines is different than the average time with the old ATM machines.

P-value (probability of being extreme than test statistic) is as follows.

P-value is equally distributed in two tails.

Decision rule for this hypothesis test is as follows.

Critical value is given by [Using R-code 'qt(1-0.10/2,29)']

We reject our null hypothesis if

So, critical region (rejection region) is as follows.