Question

The Census at Schools program collects data each year from students enrolled in primary and secondary schools in many different countries. In a random sample of 100 students selected from participating schools in Canada, the sample mean arm span of boys is 157.3 cm and the sample standard deviation is 15.43 cm. Find the 95% confidence interval for the population mean span of boys from participating schools in Canada. Use t*

Answer 1

The 95% confidence interval for the population mean span of boys from participating schools in Canada is (153.58 cm, 161.02 cm).

Step-by-step explanation:

To find the 95% confidence interval for the population mean arm span of boys from participating schools in Canada, we can use the formula:

• Lower Limit: sample mean - (t* value * (sample standard deviation / sqrt(sample size)))
• Upper Limit: sample mean + (t* value * (sample standard deviation / sqrt(sample size)))

Plugging in the given values, we have:

• Lower Limit: 157.3 - (t* value * (15.43 / sqrt(100)))
• Upper Limit: 157.3 + (t* value * (15.43 / sqrt(100)))

Since the sample size is less than 30, we will use the t-distribution. The t* value for a 95% confidence level with 99 degrees of freedom is approximately 1.984.

Calculating the lower and upper limits:

• Lower Limit: 157.3 - (1.984 * (15.43 / sqrt(100))) = 153.58 cm
• Upper Limit: 157.3 + (1.984 * (15.43 / sqrt(100))) = 161.02 cm

Therefore, the 95% confidence interval for the population mean span of boys from participating schools in Canada is (153.58 cm, 161.02 cm).

Answer 2

The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 1.9842

The margin of error is:

`M = T(s)/(√(n)) = 1.9842(15.43)/(√(100)) = 3`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 157.3 - 3 = 154.3 cm

The upper end of the interval is the sample mean added to M. So it is 157.3 + 3 = 160.3 cm.

The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.