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Random samples of players for two types of video games were selected, and the mean number of hours per week spent playing the games was calculated for each group. The sample means were used to construct the 90 percent confidence interval (1.5,3.8) for the difference in the mean number of hours per week spent playing the games. The maker of one of the video games claims that there is a difference in the population mean number of hours per week spent playing the two games. Is the claim supported by the interval

Respuesta :

Answer:

For this case the confidence interval for the differences of the true means is given by:

[tex] (\bar X_1 -\bar X_2 ) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}[/tex]

With the degrees of freedom given by:

[tex] df = n_1 +n_2 -2[/tex]

And for this case the confidence interval for the true difference of means is given by: [tex]1.5 \leq \mu_A -\mu_B \leq 3.8[/tex]

So then we have enough evidence to conclude that the true means are different and one of the two is significantly higher than the other one. So then the claim makes sense at 10% of significance.

Step-by-step explanation:

Assumptions

For this case we assume that we have the following info given :

[tex]\bar X_1[/tex] the sample mean for the group 1

[tex]\bar X_2[/tex] the sample mean for the group 2

[tex]s_1 [/tex] the sample deviation for the group 1

[tex]s_2 [/tex] the sample deviation for the group 2

[tex]n_1 [/tex] represent the sample size for 1

[tex]n_2 [/tex] represent the sample size for 2

Solution to the problem

For this case the confidence interval for the differences of the true means is given by:

[tex] (\bar X_1 -\bar X_2 ) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}[/tex]

With the degrees of freedom given by:

[tex] df = n_1 +n_2 -2[/tex]

And for this case the confidence interval for the true difference of means is given by: [tex]1.5 \leq \mu_A -\mu_B \leq 3.8[/tex]

So then we have enough evidence to conclude that the true means are different and one of the two is significantly higher than the other one. So then the claim makes sense at 10% of significance.

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