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What is Standard deviation?
Standard deviation is a measure of how spread out numbers are. It is the square root of the Variance, and the Variance is the average of the squared differences from the Mean.
To determine whether there is evidence to support the claim that the average age of lifeguards in Ocean City is greater than 24 years at a 5% level of significance, we need to perform a one-tailed t-test. A one-tailed t-test is used to test whether the mean of a sample is significantly greater than (or less than) a certain value.
To perform the t-test, we first need to calculate the t-statistic, which is defined as:
[tex]t = (mean - null value) / (standard error / \sqrt{n} )[/tex]
where mean is the mean age of the sample of lifeguards, null value is the value that the null hypothesis is testing against (in this case, 24 years), standard error is the standard error of the mean age, and n is the number of lifeguards in the sample.
Plugging in the values from the problem, we get:
[tex]t = (24.7 - 24) / (standard \ error / \sqrt{36} )\\\\= 0.7 / (standard\ error / 6)[/tex]
The standard error of the mean is a measure of the variability of the sample mean, and is calculated as:
[tex]standard \ error = population \ standard\ deviation / \sqrt{(n)}[/tex]
Plugging in the values from the problem, we get:
standard error = / sqrt(36) = / 6 =
Substituting this value back into our expression for t, we get:
t = 0.7 / ( 6)
= 1.8
For a one-tailed t-test at a 5% level of significance, the critical value is 1.645. This means that if the t-statistic is greater than 1.645, we can reject the null hypothesis and conclude that there is evidence to support the claim.
so we can reject the null hypothesis and conclude that there is evidence to support the claim that the average age of lifeguards in Ocean City is significantly greater than 24 years at a 5% level of significance.
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