Respuesta :
The information tends to refuse the claim that the mean of the population is equal to 42.1
Information from the sample:
sample size n = 25
Sample mean x = 47.5
Sample standard deviation s = 8.4
Normal Population and population mean μ = 42.1
Hypothesis Test:
A Hypothesis Test will be developed to see if 47.5 is an indication of the
a different μ, therefore.
Null Hypothesis H₀ x = μ
Alternative Hypothesis Hₐ x ≠ μ
We choose a confidence Interval of 95 % then a significance level is
α = 5 % α = 0,05 α = 0.025
z critical z(c) for α = 0.025 is from z- table z(c) = 1.96
Choosing the two tail-test we try to give more change for accepting H₀ since the rejection region is small ( or the z(c) is bigger)
To calculate z(s)
z(s) = ( x - μ ) / s/√n
z(s) = 47.5 - 42.1 )/8.4/√25
z(s) = 5.4 * 5 / 8.4
z(s) = 3.21
Comparing z(c) and z(s)
z(s) > z(c)
Then z(c) is in the rejection region we reject H₀.
The p-value of the test is of 0.0037, which is less than the standard significance level of 0.05, which means that the information tend to refuse the information that the population mean is equals to 42.1.
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From the given information, we build the hypothesis, find the test statistic and the p-value to reach a conclusion.
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Test if the mean of the population is 42.1.
At the null hypothesis, we test if the mean of the population is of 42.1, that is:
[tex]H_0: \mu = 42.1[/tex]
At the alternative hypothesis, we test if the mean not 42.1, that is, different of 42.1, so:
[tex]H_1: \mu \neq 42.1[/tex]
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The test statistic is:
We have the standard deviation for the sample, so the t-distribution is used.
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
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- 42.1 tested at the null hypothesis means that [tex]\mu = 42.1[/tex]
- Sample mean of 47.5 means that [tex]X = 47.5[/tex]
- Sample standard deviation of [tex]s = 8.4[/tex]
- Sample size of [tex]n = 25[/tex]
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Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{47.5 - 42.1}{\frac{8.4}{\sqrt{25}}}[/tex]
[tex]t = 3.21[/tex]
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p-value of the test and decision:
The p-value of the test is found using a two-tailed test(test if the mean is different), with t = 3.21 and 25 - 1 = 24 df.
Using a t-distribution calculator, the p-value is of 0.0037.
The p-value of the test is of 0.0037, which is less than the standard significance level of 0.05, which means that the information tend to refuse the information that the population mean is equals to 42.1.
A similar problem is given at https://brainly.com/question/24146681
