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4 years ago

5/8 divided by 3/2 *2/3

Mathematics

1 answer:

sukhopar [10]4 years ago

4 0

Read it wrong.
5/8 divided by (2/3 * 2/3)
2/3 x 2/3 = 4/9
5/8 / 4/9 =
5/8 x 9/4 = 45/32 = 1.41

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Answer:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

Step-by-step explanation:

We have the following info given from the problem

$\bar X_1 = 52$ sample mean for this year

$s_1= 13$ sample deviation for this year

$n_1 = 75$ random sample selected for this year

$\bar X_2 = 49$ sample mean for last year

$s_2= 11$ sample deviation for last year

$n_1 = 53$ random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

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

The degrees of freedom are given by:

$df= n_1 +n_2 -2 = 75+53-2= 126$

The confidence level is 0.95 and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ so then the critical value for this case is :

$t_{\alpha/2}= 1.979$

The margin of error would be:

$ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300$

And the confidence interval would be given by:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

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4 years ago