Answer :
According to the typist claim, we build an hypothesis test, find the test statistic and the p-value relative to this test statistic, reaching a conclusion that:
The p-value of the test is 0.1333 > 0.05, which means that there is not enough evidence to reject the typist's claim.
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In an interview for a secretary position at the dealer, a typist claims a tying speed of 45 words per minute.
At the null hypothesis, we test if the mean is of at least 45, that is:
[tex]H_0: \mu \geq 45[/tex]
At the alternative hypothesis, we test if the mean is of less than 45, that is:
[tex]H_1: \mu < 45[/tex]
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The test statistic is:
We have the standard deviation for the sample, so the t-distribution is used to solve this question
[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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45 is tested at the null hypothesis:
This means that [tex]\mu = 45[/tex]
On the basis of 70 trials, she demonstrated an average speed of 43 words per minute with a standard deviation of 15 words per minute.
This means that [tex]n = 70, X = 43, s = 15[/tex]
Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{43 - 45}{\frac{15}{\sqrt{70}}}[/tex]
[tex]t = -1.12[/tex]
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P-value of the test and decision:
The p-value of the test is found using a left-tailed test(test if the mean is less than a value), with 70 - 1 = 69 degrees of freedom and t = -1.12.
Using a t-distribution calculator, the p-value is of 0.1333.
The p-value of the test is 0.1333 > 0.05, which means that there is not enough evidence to reject the typist's claim.
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