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Question:
The mean lifetime of 100 mobile phones in a sample is 1,500 hours and their standard deviation is 120 hours, u is the mean lifetime of all the mobile phones produced. Perform a test of hypothesis that the sample comes from a population whose mean is 1,600 hours at 586 significance level 10 marks​

Sagot :

Using the t-distribution, as we have the standard deviation for the sample, it is found that since the absolute value of the test statistic is greater than the critical value for the two-tailed test, there is enough evidence to conclude that the sample does not come from a population whose mean is 1,600 hours.

What are the hypothesis tested?


At the null hypothesis, it is tested if the mean is of 1600 hours, that is:

[tex]H_0: \mu = 1600[/tex]

At the alternative hypothesis, it is tested if the mean is different of 1600 hours, that is:

[tex]H_1: \mu \neq 1600[/tex]

What is the test statistic?

It 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 values of the parameters are:

[tex]\overline{x} = 1500, \mu = 1600, s = 120, n = 100[/tex]

Hence, the value of the test statistic is:

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

[tex]t = \frac{1500 - 1600}{\frac{120}{\sqrt{100}}}[/tex]

[tex]t = -8.33[/tex]

What is the decision rule?


Considering a two-tailed test, as we are testing if the mean is different of a value, with 100 - 1 = 99 df and a significance level of 0.01, the critical value is of [tex]t^{\ast} = 1.66[/tex].

Since the absolute value of the test statistic is greater than the critical value for the two-tailed test, there is enough evidence to conclude that the sample does not come from a population whose mean is 1,600 hours.

More can be learned about the t-distribution at https://brainly.com/question/13873630