Hypothesis test for a Normal mean

The masses of stones on a certain beach are normally distributed with mean 0.36 kg and standard deviation 0.12 kg.

Find the probability that two stones chosen at random from this beach both have masses

between 0.3 kg and 0.4 kg.

Determine the probability that the mean mass of a random sample of 40 stones from this

beach is less than 0.32 kg.

The mass, in kilograms, of a species of fish in the UK has population mean 4.2 and standard deviation 0.25.

An environmentalist believes that the fish in a particular river are smaller, on average, than those in other rivers in the UK.

A random sample of 100 fish of this species, taken from the river, has sample mean 4.16 kg.

Stating a necessary assumption, test at the 5% significance level whether the environmentalist is

correct.

In the past the masses of new-born babies in a certain country were normally distributed with mean 3300 g. Last year a publicity campaign was held to encourage pregnant women to improve their diet.

Following this campaign, it is required to test whether the mean mass of new-born babies has increased. A random sample of 200 new-born babies is chosen, and it is found that their mean mass is 3360 g. It is given that the standard deviation of the masses of new-born babies is 450 g.

Carry out the test at the 2.5% significance level.

Zac is planning to write a report on the music preferences of the students at his college. There is a large number of students at the college.

State one reason why Zac might wish to obtain information from a sample of students, rather

than from all the students.

Amaya suggests that Zac should use a sample that is stratified by school year.

Give one advantage of this method as compared with random sampling, in this context.

Zac decides to take a random sample of 60 students from his college. He asks each student how many hours per week, on average, they spend listening to music during term. From his results he calculates the following statistics.

Sundip tells Zac that, during term, she spends on average 30 hours per week listening to music.

Discuss briefly whether this value should be considered an outlier.

Layla claims that, during term, each student spends on average 20 hours per week listening to music. Zac believes that the true figure is higher than 20 hours. He uses his results to carry out a hypothesis test at the 5% significance level.

Assume that the time spent listening to music is normally distributed with standard deviation 4.20 hours.

Carry out the test.

In the past, the time for Jeff’s journey to work had mean 45.7 minutes and standard deviation 5.6 minutes. This year he is trying a new route. In order to test whether the new route has reduced his journey time, Jeff finds the mean time for a random sample of 30 journeys using the new route. He carries out a hypothesis test at the 2.5% significance level.

Jeff assumes that, for the new route, the journey time has a normal distribution with standard deviation 5.6 minutes.

State appropriate null and alternative hypotheses for the test.

Determine the rejection region for the test.

A certain forest contains only trees of a particular species. Dipak wished to take a random sample of 5 trees from the forest. He numbered the trees from 1 to 784. Then, using his calculator, he generated the random digits 14 781 049. Using these digits, Dipak formed 5 three-digit numbers. He took the first, second and third digits, followed by the second, third and fourth digits and so on. In this way he obtained the following list of numbers for his sample.

147 478 781 104 49

Explain why Dipak omitted the number 810 from his list.

Explain why Dipak’s sample is not random.

The mean height of all trees of this species is known to be 4.2 m. Dipak wishes to test whether the mean height of trees in the forest is less than 4.2 m. He now uses a correct method to choose a random sample of 50 trees and finds that their mean height is 4.0 m. It is given that the standard deviation of trees in the forest is 0.8 m.

Carry out the test at the 2% significance level.