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A manufacturer of electronic kits has found that the mean time required for novices to assemble its new circuit tester is 3 hours, with a standard deviation of0.20 hours. A consultant has developed a new instruc- tional booklet intended to reduce the time an inexpe- rienced kit builder will need to assemble the device. In a test of the effectiveness of the new booklet, 15 nov- ices require a mean of 2.90 hours to complete the job. Assuming the population of times is normally distrib- uted, and using the 0.05 level of significance, should we conclude that the new booklet is effective? Determine and interpret the p-value for the test.( DATA SET ) Note: Exercises 10.33 and 10.34 require a computer and statistical software.

Sagot :

Testing the hypothesis using the z-distribution, it is found that since the p-value of the test is of 0.0262 < 0.05, it can be concluded that the new booklet is effective.

At the null hypothesis, we test if there has been no improvement, that is, the time is still of 3 hours. Hence:

[tex]H_0: \mu = 3[/tex]

At the alternative hypothesis, we test if there has been improvement, that is, the time is of less than 3 hours. Thus:

[tex]H_1: \mu < 3[/tex]

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\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.
  • [tex]\sigma[/tex] is the standard deviation of the population.
  • n is the sample size.

In this problem, the values of the parameters are: [tex]\mu = 3, \sigma = 0.2, \overline{x} = 2.9, n = 15[/tex].

Then, the value of the test statistic is:

[tex]z = \frac{2.9 - 3}{\frac{0.2}{\sqrt{15}}}[/tex]

[tex]z = -1.94[/tex]

The p-value of the test is the probability of finding a sample mean of 2.9 hours or less, which is the p-value of z = -1.94.

Looking at the z-table, z = -1.94 has a p-value of 0.0262.

Since the p-value of the test is of 0.0262 < 0.05, it can be concluded that the new booklet is effective.

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