Answer to Question #314141 in Statistics and Probability for Bless

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

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

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