Answer to Question #347873 in Statistics and Probability for Bum

Question

1) The researcher wants to estimate the number of hours that 5-year old children spend watching television. A sample of 50 five-year old children was observed to have a mean viewing time of 3 hours. The population is normally distributed with a population standard deviation 0.5 hour, find:

a) the best point estimate of the population mean.

b) the 90% confidence interval of the population mean.

2) The mean scores of a random sample of 17 students who took a special test is 83.5. If the sd of the scores is 4.1, the sample comes from an approximately normal population, find the point and interval estimates of the population mean adopting a confidence level of 95%.

3) In a gymnasium, physical exercise has a mean length of 30 mins with a sd of 6 mins. A PE major wants to estimate the true length of the exercise with a maximum error pegged at 0.5 adopting the 95% confidence interval. How many respondents does he need?

Answer 1

1)

a) The best point estimate of the population mean a mean viewing time.

b) The critical value for $\alpha = 0.10$ degrees of freedom is $z_c=1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(3- 1.6449\times\dfrac{0.5}{\sqrt{50}},$$ $$3+ 1.6449\times\dfrac{0.5}{\sqrt{50}})$$ $$=(2.8837, 3.1163)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $2.8837 < \mu < 3.1163,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(2.8837, 3.1163).$

2) The critical value for $\alpha = 0.05, df=n-1=16$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=2.1199.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(83.5- 2.1199\times\dfrac{4.1}{\sqrt{17}},$$ $$83.5+ 2.1199\times\dfrac{4.1}{\sqrt{17}})$$

$$=(81.392, 85.608)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $81.392 < \mu < 85.608,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(81.392, 85.608).$

$$\mu=83.5\pm2.108$$

A point estimate is a single value estimate of a parameter.

An interval estimate gives you a range of values where the parameter is expected to lie.

A confidence interval is the most common type of interval estimate.

3)

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

$$n\ge(\dfrac{z_c\sigma}{E})^2$$

The critical value for $\alpha = 0.10$ degrees of freedom is $z_c=1.6449.$

$$n\ge(\dfrac{1.6449(6)}{0.5})^2$$

$$n=390$$