Answer to Question #147909 in Statistics and Probability for Zeenet

Question

2. It is known that the weights of apples from a farm are normally distributed. In order to estimate the mean weight, a random sample of 150 apples is considered and the sample mean and population standard deviation are 6 kg and 0.8 kg respectively.

(a) Construct a 95% confidence interval estimate for the population mean weight of apples.
(2 marks)
(b) The researcher suggests doing the study again so that 98% confidence interval estimate for the population mean weight of apples is (5.8835,6.1165) kg. How large should the sample size be?
(3 marks)

Answer 1

n = 150

$\bar x=6$

$\sigma =0.8$

a) the 95% confidence interval for population mean is given by:

$\bar x \pm Z_{\frac{0.05}{2}}\frac{\sigma}{\sqrt n} =\bar x \pm Z_{0.025}\frac{\sigma}{\sqrt n}=6\pm1.96\cdot\frac{0.8}{\sqrt{150}}=6\pm0.128$

So the 95% confidence interval estimate for the population mean weight of apples is (5.872, 6.128) kg

b) the 98% confidence interval estimate for the population mean weight of apples is (5.8835, 6.1165) kg. Thus:

$(5.8835, 6.1165)=(\bar x-MOE, \bar x+MOE)$

$\bar x-MOE=5.8835,\ \bar x+MOE=6.1165\implies MOE=0.1165$

We need to find such n so margin of error is 0.1165

$Z_{\frac{0.02}{2}}\frac{\sigma}{\sqrt n}=0.1165$

$Z_{0.01}\frac{\sigma}{\sqrt n}=0.1165$

$2.33\cdot\frac{0.8}{\sqrt n}=0.1165 \implies n=256$

Answer:

a) (5.872, 6.128)

b) 256