Question

A restaurant chain in Ontario wants to know the level of satisfaction of their customers. From each of Toronto and Ottawa, a random sample of 14 customers was selected and asked to rate their dining experience on a scale from 1 to 10 (with 1 being not satisfied at all, 10 being extremely satisfied). A total of 28 scores was collected. Below is a summary of the data:

City	Toronto	Ottawa
Sample Mean	4.3611	5.9444
Sample std.	1.1401	2.4254
# of customers rated 8 or more	6	9
Sample Size	14	14

Which test should you run: pooled -test, or non-pooled -test (and why)? [use d.f = 21.32 if you decide to use the non-pooled test]

(c) Based on your answer in (a), test whether there are higher average satisfaction scores in Ottawa than in Toronto at the at the 5% level of significance. You should follow these instructions precisely to get full marks: state clearly the null and alternative hypotheses (in both symbols and words), compute the value of the test statistic, determine the p-value (if a precise p-value cannot be found, then you must give a range). Make your decision, and provide 2–3 lines of for your conclusions.

(d) Suppose you also want to compare the proportions of customers who rate 8 or more on their dining experience. Construct a 92% confidence interval for the difference in the proportions between Toronto and Ottawa customers. To receive full marks, you should additionally include a check on whether the statistical procedure you used is valid, and provide an interpretation of your confidence interval.

Answer 1

(c)

Let the average satisfaction score of Toronto be and the average satisfaction score of Ottawa be

Here the null and alternative hypotheses are given by

As the null hypothesis is equivalent to the difference of the two average scores being equal to 0, then we are to use pooled t test by definition.

The test statistic is given by

where s' is given by

Now a table is made as:

	Sample 1	Sample 2
Sample size	14	14
Sample mean	4.3611	5.9444
Sample SD	1.1401	2.4254

Now the test statistic is gven by

The p-value is given by 0.018018

As the p-value is less than 0.05, the null hypothesis is rejected at 5% level of significance and hence it can be concluded that there is higher average satisfaction scores in Ottawa than in Toronto at the at the 5% level of significance.

(d)

Let the proportion of customers who rated 8 or more in Toronto be denoted as p1 and that in Ottawa be denoted as p2.

From the given data,

Now the confidence interval is given by

Here n1=n2=14

Thus the 92% confidence interval is given by

The data is given to be normally distributed and heteroscedastic and hence it satisfied the assumptions.

Hopefully this will help you. In case of any query, do comment. If you are satisfied with the answer, give it a like. Thanks.