Question

The manager of AllNeeds supermarket wants to estimate µ,, the average amount of money spent per week

for groceries by each student, assuming this amount of money follows approximately a normal distribution.

A random sample of 90 is taken, of which the mean and standard deviation of the amount of money spent

per week for groceries are GHC81.50 and GHC10.36 respectively.

(a) Calculate a 99% confidence interval for µ.

(b) Based on your result in (i), would you say on average a student spends more than GHC30?

(c) How many students must be surveyed if the manager wanted to estimate the true average spending

of students within a margin of .10 with probability .95?

Answer 1

$\alpha=0.99/2=0.495$

$z_{\alpha/2}=2.575$

$\mu=81.5\plusmn Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}=81.5\plusmn 2.575\frac{10.36}{\sqrt{90}}=81.5 \plusmn 2.8$

b. on average a student spends (78.7-84.3). It is greater then GHC30

c.$n=(\frac{Z_{\alpha/2}}{\sigma})^2$

$Z_{\alpha\2}(0.95)=1.96$

$n=(1.96/0.1)^20.5(1-0.5)=96$