. The National Sleep Foundation surveyed representative samples of adults in six different countries to ask questions about sleeping habits.† Each person in a representative sample of 250 adults in each of these countries was asked how much sleep they get on a typical work night. For the United States, the sample mean was 391 minutes, and for Mexico the sample mean was 426 minutes. Suppose that the sample standard deviations were 25 minutes for the U.S. sample and 49 minutes for the Mexico sample. The report concludes that on average, adults in the United States get less sleep on work nights than adults in Mexico. Is this a reasonable conclusion? Support your answer with an appropriate hypothesis test. (Use
= 0.05.
Use 1 for Mexico and 2 for the United States.)

Answer:

Yes the conclusion in the question  is a reasonable conclusion

Step-by-step explanation:

From the question we are told that

    The sample size is  [tex]n_1 = n_2 = n = 250[/tex]

     The sample mean for USA is  [tex]\= x_1 = 391 \ minutes[/tex]

     The sample mean for Mexico is [tex]\= x_2 = 426 \ minutes[/tex]

     The sample standard deviation for USA is  [tex]s_1 = 25 \ minutes[/tex]

    The sample standard deviation for Mexico is  [tex]s_2 = 49 \ minutes[/tex]

The null hypothesis is  [tex]H_o : \mu_1 = \mu_2[/tex]

The alternative hypothesis is  [tex]H_a : \mu_1 < \mu_2[/tex]

Generally the test statistics is mathematically represented as

        [tex]z = \frac{\= x_1 - \= x_2 }{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2 } } }[/tex]

=>     [tex]z = \frac{ 391 - 426 }{ \sqrt{ \frac{25^2}{250} + \frac{49^2}{250} } }[/tex]

=>     [tex]z = -10[/tex]

From the z table  the area under the normal curve to the left corresponding to   -10  is

    [tex]p-value = P(Z < -10 ) = 0.00[/tex]

Generally from the value obtained we see that

  The p-value is  <  [tex]\alpha[/tex]  hence

The decision rule is

Reject the null hypothesis

The conclusion is  

 The is sufficient evidence to conclude that adults in the United States get less sleep on work nights than adults in Mexico

Hence the conclusion in the question is a  reasonable conclusion

The p-value is less than [tex]\alpha[/tex] hence reject the null hypothesis. This is sufficient evidence to conclude that adults in the United States get less sleep on work nights than adults in Mexico and hence it is a reasonable conclusion given in the problem.

Given :

According to the given data the sample size of the US and Mexico is [tex]\rm n_1=n_2=250[/tex].

The hypothesis are as follows:

Null Hypothesis is [tex]\rm H_0 : \mu_1=\mu_2[/tex]

Alternate Hypothesis is [tex]\rm H_1 : \mu_1<\mu_2[/tex]

Now, the mathematical representation of test statistics is:

[tex]\rm z = \dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}} }[/tex]

[tex]\rm z = \dfrac{391-496}{\sqrt{\dfrac{25^2}{250}+\dfrac{49^2}{250}} }[/tex]

z = -10

According to the z-table:

p - value = p (Z < -10) = 0

The p-value is less than [tex]\alpha[/tex] hence reject the null hypothesis.

This is sufficient evidence to conclude that adults in the United States get less sleep on work nights than adults in Mexico and hence it is a reasonable conclusion given in the problem.

For more information, refer to the link given below:

https://brainly.com/question/23044118

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