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1 year ago

A student ran a distance of 3 miles each day for 5 days. Then the student ran a distance of 4 miles each day for the next 5 days

1 answer:

azamat1 year ago

8 0

The total distance was 35 miles. Why? Well, see below:

The student ran 3 miles each day for 5 days. That can be representative of:

Day 1: 3 miles
Day 2: 3 miles
Day 3: 3 miles
Day 4: 3 miles
Day 5: 3 miles

Adding these values together, we will find that the student ran a total of 15 miles in the first five days.

As for the next set of 5 days, we can represent this by modeling like we did above:

Day 6: 4 miles
Day 7: 4 miles
Day 8: 4 miles
Day 9: 4 miles
Day 10: 4 miles

Adding the values from the second set of days, we will find that in days 6-10, the student ran a total of 20 miles.

We can add the total values together to identify the total distance the student ran during these 10 days.

15 + 20 = 35 total miles

Hence, the student ran a total of 35 miles over the course of 10 days. If you need help, let me know and I will go ally assist you.

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3 years ago

The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o

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Answer:

a) Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

b) $p_v =2*P(t_{(24)}$

c) If we compare the p value and the significance level given $\alpha=0.05$ 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 true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

The values obtained are:

$\bar X=85272$ represent the mean annual salary for the sample

$s=11039.23$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =90000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:

Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

Part b: Calculate the statistic

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

$t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(24)}$

Part c: Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ 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 true mean for the salary differs from 9000 at 5% of significance.

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3 years ago