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Orlov [11]
3 years ago
13

The profit a company earns every month depends on the amount of product sold,p, for $855 each and the amount spent in rent, util

ities and other expenses, which always totals to $6,780. The ceo of the company earns 15% of this profit. How much does the ceo earn if the company sells 250 products in a given month
Mathematics
1 answer:
Rina8888 [55]3 years ago
4 0

the CEO earns about <em><u>31045.5</u></em> in a given month

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If a test of H subscript 0 colon space mu subscript D equals 0 space v s. space H subscript a colon space space mu subscript D g
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p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

The p value on this case is given  by the problem.

If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before , y = test value after

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \leq 0

Alternative hypothesis: \mu_y -\mu_x >0

The first step is calculate the difference d_i=y_i-x_i

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=t_{calculated}

The next step is calculate the degrees of freedom given by:

df=n-1

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_{(n-1)}>t_{calculated}) =0.0601

The p value on this case is given  by the problem.

If we compare the p value with a significance level assumed \alpha=0.05, we see that p_v > \alpha and we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is less or equal than 0.

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