Consider the times taken by a ferry to complete a trip between 2 locations in a harbour.

- In good weather the times taken are known to be normally distributed with a mean of 47 minutes and a standard deviation of 7 minutes.

- What is the probability that in a random sample of 40 trips in good weather the mean time taken is greater than 45 minutes?

According to question, the time of good weather, denoted by T, has a normal distribution with a mean of 47 and a standard deviation of 7, say Hence, the mean of samples also has a normal distribution. Generally, for a normal distributed population , the mean of n-size sample also has a normal distribution Particularly, the mean of 40-trip sample, denoted by , is distributed as . Thus the wanted probability P is

- In a random sample of 40 trips in good weather how many trips would be to take less than 43 minutes?

According to question, the time of good weather, denoted by T as well, is either less than 43 minutes or not less than 43 minutes. For 40 trails, the expected success (less than 43 minutes) time is distributed as Bernoulli distribution . On the other hand, . Thus,

And for a Bernoulli distribution, the expectation . Hence, it is expected that there are 11 times less than 43 minutes good weather in 40 trips.

- In Poor weather 14% of the ferry’s departures are delayed. What is the probability that in a random sample of 100 of the ferry’s trips in poor weather at least 17 of them are delayed at departures?

Exactly Calculation:

The number of trips (are delayed in poor weather is distributed as Bernoulli, say . Therefore, the wanted probability ,

Approximation Calculation:

Since the sample is large enough to apply Large Number Law (LNL), it is supposed to consider the distribution is normally distributed as . . Hence the wanted probability

Question 2

In this question, the difference in the proportions of respondents whose reason for purchase was reliability between those who preferred a medium-sized and those who preferred a small-sized car will be tested with the methodology ofChi-square test.

Null hypothesis: there is no difference between referred groups.

Alternative hypothesis: there is difference between referred groups.

For expression, some denotations are introduced as below:

expected value of those who purchase reliability with medium-size preference

expected value of those who purchase other with medium-size preference

expected value of those who purchase reliability with small-size preference

expected value of those who purchase other with small-size preference

Hence, , ,,. , hence the p-value of the test is 0.210. It is not rather significant to reject null hypothesis and considered that there is no great difference between referred proportions.

Question 3

Null hypothesis (): Rental income has no significant effect on capital value of the property, say the coefficient of rental income .

Alternative hypothesis (): Rental income has significant effect on capital value of the property, say the coefficient of rental income .

According to the output of Minitab, the standard deviation of coefficient of Rental income is 1.519 and the related coefficient is 11.7. Under the null hypothesis, the t-statistics has the formula below:

Hence the p-value of the hypothesis test is 0.00. Null hypothesis is rejected at a low significant level. It is nearly ensured rental income has a significant level effect on the property.

Question 4

Null hypothesis (): The mean height of river red gums grown in ideal condition is equal to 50 metres, say the mean .

Alternative hypothesis (): The mean height of river red gums grown in ideal conditions is higher than 50 metres.

T-test will be used for the hypothesis.

Hence the p-value of the test is 0.00. Null hypothesis is rejected at a low significant level, and it is almost ensured that the mean height of river red gums grown in ideal condition is larger than 50 metres.

Question 5

Introduction

In hospital, many patients are prescribed antibiotics. However, the effect might have difference in female patients and male patients, resulting from various physical conditions. And now 20 questions relating to satisfaction with their doctor and 20 questions relating to satisfaction with their lives are asked for a research. It is interesting to explore whether there is difference in mean score of satisfaction between female patients and male patients.

Methods

According to the requirement, the t-test will be used for the test the difference between two groups. Furthermore, since the data is collected with much information, the related information has to be selected. Patients are prescribed for antibiotics are selected, say Medication = 1. Next, the remained data is classified into two groups: female and male. The mean, size and standard deviation are necessary in the test. Because the equality of variance in two groups influences the formulation the t-statistics, the equality of variance will also be tested with F-test. In the end, corresponding t-statistics will be worked out to test the difference.

Results

Seen from table 1, the descriptive of two groups are listed. The mean of male is 36.52 and that of female is 42.34. It indicates that the mean of male is higher than that of female. However, the difference cannot be ensured for the accident of sampling. The equality of variance of two groups seems certified through F-test and Levene’s test, both of which have a rather high p-value. In other words, the null hypothesis that there is no difference of variance is accepted. Under such condition, the equal variance t-test is applied to test the difference of means. And it indicates that the difference significantly exists in terms of statistics.

Conclusion

For patients prescribed antibiotic, there is a significant difference between male patients and female patients in their satisfaction with their doctors.

Table 1The descriptive statistics of satisfaction of doctors in two groups

Table 2The test for the equality of variance of two groups

Table 3T test for the two independent group with equal variance