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

Given $\bar{x}=6\ kg, \sigma=0.8\ kg, n=150, \alpha=0.05$

(a)The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times \dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times \dfrac{\sigma}{\sqrt{n}})$$

$$=(6-1.96\times \dfrac{0.8}{\sqrt{150}}, 6+1.96\times \dfrac{0.8}{\sqrt{150}})$$

$$=(5.8720, 6.1280)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $5.8720<\mu<6.1280,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(5.8720, 6.1280).$

(b)The critical value for $\alpha=0.02$ is $z_c=z_{1-\alpha/2}=2.326.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times \dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times \dfrac{\sigma}{\sqrt{n}})$$

$$=(5.8835, 6.1165)$$

Then

$$2\times z_c\times \dfrac{\sigma}{\sqrt{n}}=6.1165-5.8835$$

$$z_c\times \dfrac{\sigma}{\sqrt{n}}=0.1165$$

$$n=(z_c\times \dfrac{\sigma}{0.1165})^2$$

$$n=(2.326\times \dfrac{0.8}{0.1165})^2=255$$

The sample size should be 255.