Question

A sports physician conducts an observational study to learn the average amount of time that 3,000 swimmers in the town can hold their breath underwater. He uses 150 sampling of 60 people. The average of the means of all the samplings is 72.7, and the standard deviation is 0.92. This is a histogram of the sampling distribution of the sample mean

Answer 1

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Question:

A sports physician conducts an observational study to learn the average amount of time that 3,000 swimmers in the town can hold their breath underwater. He uses 150 sampling of 60 people. The average of the means of all the samplings is 72.7, and the standard deviation is 0.92. This is a histogram of the sampling distribution of the sample mean. Based on this data, with a 95% confidence interval the researchers can determine that the actual average amount of time the entire population can hold their breath under water is?

Given Information:

sample mean time = 72.7

sample standard deviation = 0.92

Sampling size = n = 150

Confidence level = 95%

Required Information:

95% confidence interval = ?

Answer:

`\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 72.7 \pm 0.14836\\\\\text {confidence interval} = 72.7 - 0.14836, \: 72.7 + 0.14836\\\\\text {confidence interval} = (72.552, \: 72.848)\\\\`

Explanation:

The confidence interval is given by

`\text {confidence interval} = \bar{x} \pm MoE\\\\`

Where
`\bar{x}` is the sample mean time and Margin of error is given by

`$ MoE = t_(\alpha/2)((s)/(√(n) ) ) $ \\\\`

Where n is the sampling size, s is the sample standard deviation, and
`t_(\alpha/2)` is the t-score corresponding to 95% confidence level.

The t-score corresponding to 95% confidence level is

Significance level = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 150 - 1 = 149

From the t-table at α = 0.025 and DoF = 149

t-score = 1.975

`MoE = t_(\alpha/2)((s)/(√(n) ) ) \\\\MoE = 1.975\cdot (0.92)/(√(150) ) \\\\MoE = 1.96\cdot 0.07512\\\\MoE = 0.14836\\\\`

So the required 95% confidence interval is

`\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 72.7 \pm 0.14836\\\\\text {confidence interval} = 72.7 - 0.14836, \: 72.7 + 0.14836\\\\\text {confidence interval} = (72.552, \: 72.848)\\\\`

Therefore, we are 95% confident that actual average amount of time the entire population can hold their breath under water is within the range of (72.552, 72.848)