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4 years ago

There are 25 buttons on a table 10 are blue 15 are red without lokking pick up one button what is the probability that it is red

1 answer:

3 0

Out of the 25 buttons, 15 are red. So we set up the fraction $\frac{15}{25}$ .

(15/25)(4) = 60/100 or 60%

So you have a 60% chance of picking up a red button if 15 out of the 25 buttons are red.

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3 years ago

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to in

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Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

$p_v =P(Z>2)=0.0228$

Step-by-step explanation:

1) Data given and notation

$\bar X=8300$ represent the sample mean

$\sigma=1200$ represent the population standard deviation

$n=64$ sample size

$\mu_o =800$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

4) P-value

Since is a one-side upper test the p value would given by:

$p_v =P(Z>2)=0.0228$

5) Conclusion

If we compare the p value and the significance level assumed, for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.