Question

(a)   A study of 35 gamers showed that their average score on a particular game was 90 and the population standard deviation is 5.

(i)                    Find the best point estimate of the population mean.

 

(ii)                 Find the 95% confidence interval of the mean score for all gamers.

(iii)               Find the 95% confidence interval of the mean score if a sample of 70 gamers is used instead of a sample of 35.

 

(iv)               From your answer in part (ii) and (iii), which interval is smaller?

Answer 1

(i) The best point estimate for the population mean is the sample mean: $x=90.$

(ii) 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:

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

$$=(90-1.96\times\dfrac{5}{\sqrt{35}}, 90+1.96\times\dfrac{5}{\sqrt{35}})$$

$$=(88.344, 91.656)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $88.344<\mu<91.656,$ which indicates that we are 95% confident that the true population mean $\mu$

is contained by the interval $(88.344, 91.656).$

(iii) 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:

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

$$=(90-1.96\times\dfrac{5}{\sqrt{70}}, 90+1.96\times\dfrac{5}{\sqrt{70}})$$

$$=(88.829, 91.171)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $88.829<\mu<91.171,$ which indicates that we are 95% confident that the true population mean $\mu$

is contained by the interval $(88.829, 91.171).$

(iv) The 95% confidence interval for $n=70$ is narrower than the 95% confidence interval for $n=35.$