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3 years ago

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Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts

/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 24 samples is 7.8 ppm with a standard deviation of 0.7. Assume the population is normally distributed. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to three decimal places.

1 answer:

Hunter-Best [27]3 years ago

7 0

Answer:

The value of the test statistic is $t = 2.1$

Step-by-step explanation:

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value, $\sigma$ is the standard deviation and n is the size of the sample.

The level of ozone normally found is 7.5 parts/million (ppm).

This means that $\mu = 7.5$

The mean of 24 samples is 7.8 ppm with a standard deviation of 0.7.

This means that $X = 7.8, \sigma = 0.7, n = 24$

Test Statistic:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$t = \frac{7.8 - 7.5}{\frac{0.7}{\sqrt{24}}}$

$t = 2.1$

The value of the test statistic is $t = 2.1$

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Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin

Artemon [7]

"<span>Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find $\bar{d}$ , $s_{d}$ , the t-test statistic, and the critical values to test the claim that $\mu_{d} = 0$ "

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find </span><span> $\bar{d}$
You are asked to find the mean difference between the two variables, which is given by the formula:
$\bar{d} = \frac{\sum (x - y)}{n}$

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example:
</span><span> $\bar{d} = \frac{6}{8} = 0.75$ </span>

B) Find <span> $s_{d}$
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span> $s_{d} = \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }$

These are the steps to follow:
1) Subtract the mean difference from each pair's difference
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
$s_{d} = \sqrt{ \frac{101.5}{8-1} }$
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
$t = \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}$

where SE = standard error is given by the formula:
$SE = \frac{ s_{d} }{ \sqrt{n} }$

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract $\mu_{d}$
6) divide by the standard error

In our example:
SE = 3.81 / √8
= 1.346

The problem gives us <span> $\mu_{d} = 0$ , therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
= 0.56

D) Find $t_{\alpha / 2}$
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find <span> $t_{\alpha / 2} = 1.895$ </span>

Since our t-value is less than <span> $t_{\alpha / 2}$ </span> we can reject our null hypothesis!!

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Volgvan

Step-by-step explanation:

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