Answer to Question #232077 in Statistics and Probability for tam

Question

(a) A machine is used to fill soft drink in bottles. 35 randomly selected bottles

are taken from the output of the machine. Those bottles have a mean

volume of 330 ml with a standard deviation of 4.8 ml. Construct a 98%

confidence interval to estimate the true mean volume of soft drink.

(5 marks)

(b) The marketing department of a smart phone store is interested in

determining the proportion of new smart phone owners who would have

spent more than or equal to RM 2000 to buy a new smart phone. Suppose

that a survey of 45 new customers, 23 indicate that they would have

purchased new smart phone which cost more than or equal to RM2000.

Construct a 99% confidence interval for the true proportion of new smart

phone owners who have spent more than or equal to RM2000 for a new

smartphone. (5 marks)

Answer 1

(a) The critical value for $\alpha = 0.02$ and $df = n-1 = 34$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.44115.$

The corresponding confidence interval is computed as shown below:

$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$

$=(330-2.44115\times\dfrac{4.8}{\sqrt{35}},$

$330+2.44115\times\dfrac{4.8}{\sqrt{35}})$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $328.0194<\mu<331.9806,$ which indicates that we are 98% confident that the true population mean $\mu$ is contained by the interval $(328.0194, 331.9806).$

(b) The sample proportion is computed as follows, based on the sample size $N = 45$ and the number of favorable cases $X=23:$

$\hat{p}=X/N=23/45$

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} =2.5758.$

The corresponding confidence interval is computed as shown below:

$CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}},$

$\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}})$

$=\bigg(\dfrac{23}{45}-2.5758\sqrt{\dfrac{\dfrac{23}{45}(1-\dfrac{23}{45})}{45}},$

$\dfrac{23}{45}+2.5758\sqrt{\dfrac{\dfrac{23}{45}(1-\dfrac{23}{45})}{45}}\bigg)$

$=(0.3192, 0.7031)$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.3192 < p < 0.7031,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.3192, 0.7031).$