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AlladinOne [14]
3 years ago
13

Can kristen spend $40 on the meal before the tip? explain.

Mathematics
2 answers:
podryga [215]3 years ago
6 0

Of course! We all love a good meal!

If this is an incomplete question, please let me know how I can fully help you! :)

Sladkaya [172]3 years ago
4 0
Yes of course but it’s depends on how much money she’s coming in with and this is an incomplete question so text me if you need more help
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Answer:

a) t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.7 -0}{\frac{0.678}{\sqrt{5}}}=2.308  

p_v =P(t_{(4)}>2.308) =0.0411

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

b) The p value is illustrated on the figure attached.

If we select \alpha=0.01 we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

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  

1=test value old , 2 = test value new

1: 5.2 6.5 7.2 5.7 7.6

2: 6.4 7.8 6.8 6.5 8.2

The system of hypothesis for this case are:

Null hypothesis: \mu_2- \mu_1 \leq 0

Alternative hypothesis: \mu_2 -\mu_1 >0

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: 1.2, 1.3, -0.4, 0.8, 0.6

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}=0.7

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} =0.678

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.7 -0}{\frac{0.678}{\sqrt{5}}}=2.308

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

df=n-1=5-1=4

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

p_v =P(t_{(4)}>2.308) =0.0411

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.  

Part b

The p value is illustrated on the figure attached.

If we select \alpha=0.01 we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis.

7 0
3 years ago
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