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Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. To test DullCo's hypothesis, the test statistic is:
Select one:
a. -2.101
b. -1.980
c. -1.960
d. -1.728

Respuesta :

Answer:

lyrics d  - 1.782

Step-by-step explanation:

Assume Normal Distribution

we have  μ₀ = 40       (from Dullco claims)

And from sample   μ  = 37.8    and standard deviation of  5.4

random sample n = 18

We have to use t-student for testing the hypothesis

and we have a one tail test (left) since Dullco claims : " at least"  meaning (always bigger or at least ) not different.

Then

t (s) =( μ -  μ₀) / (σ/√n)   ⇒ t (s) = [(37.8  - 40 )* √18 ]/5.4

t (s) = - 1.7284

Using the t-distribution, as we have the standard deviation for the sample, it is found that the test statistic is given by:

d. -1.728

What are the hypothesis tested?

At the null hypothesis, it is tested if the batteries last at least 40 hours, that is:

[tex]H_0: \mu \geq 40[/tex]

At the alternative hypothesis, it is tested if they last less than 40 hours, that is:

[tex]H_1: \mu < 40[/tex]

What is the test statistic?

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given by:

[tex]\overline{x} = 37.8, \mu = 40, s = 5.4, n = 18[/tex]

Hence:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{37.8 - 40}{\frac{5.4}{\sqrt{18}}}[/tex]

[tex]t = -1.728[/tex]

Hence option d is correct.

To learn more about the t-distribution, you can check https://brainly.com/question/16313918

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