Practise selecting random samples, calculating unbiased estimates and sampling distributions and constructing or interpreting confidence intervals for population means.
Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries.
Give a reason for using a sample rather than the whole population in carrying out this test.
A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours 2.
Batteries unusable after testing or Population too big
or too costly or too time consuming to use the whole population oe
Calculate a 94% confidence interval for the population mean life of type B batteries.
147 +/- z x 15120
Expression of correct form must be a z value
z=1.881 [or 1.882]
144 to 150 (3 s.f.)
Must be an interval
Incorrect z value can only score M1B0A0
At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15 . The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced.
It is suggested that a random sample of the students in Grade 12 should be used for the test.
Give a reason why this would not be an appropriate sample.
A suitable sample of 50 students is selected and the number not wearing correct uniform is noted. This figure is used to carry out a test at the 5\% significance level.
Not representative (of all students in the school)
OE idea of 'not being representative' e.g. different grades in the school have different characteristics/proportions....
Don't accept 'not random' or 'biased' without further explanation.
A random sample of 5 values of a variable X is given below.
Find an expression, in terms of a, for the mean of these values.
It is given that an unbiased estimate of the population variance of X, using these values, is 4. It is also given that a is positive.
13+a/5
Accept 2+3+3+5+a/5.
Do not ignore subsequent working
Find and simplify a quadratic equation in terms of a and hence find the value of a.
5/4(47+a^25-(13+a/5)^2)=4 or 1/4(47+a^2-(13+a)^25)=4
Use of correct formula using their value from (a), in terms of a, and equate to 4
2 a^2-13 a-7=0
Any correct three-term quadratic equation rearranged to a form ready to solve
a=7
Condone the other value of a(-1/2)