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Pepsi [2]
3 years ago
5

Twenty students from a soccer team want to attend a match. The ticket price is $25 for each student, and the transportation and

meals for everyone will cost $500. To pay for the trip, the students design jerseys to sell for a profit of $20 per jersey. How many jerseys must the students sell so that there will be enough money to pay for the entire cost of the trip?
Mathematics
1 answer:
azamat3 years ago
5 0

Answer:

525 jerseys

Step-by-step explanation:

you multiply the 25$ and $500 ($25 x 20)+($500 x 20) by the number of students, and then add them together. after that, you multiply it by the $20 ($10,500/$20 = 525) for the jerseys to get 525 jerseys

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Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

H_0: \mu_d=0\\\\H_a: \mu_d>0

We have n=8 pairs of data. We calculate the difference as:

d_1=U_1-A_1=36.4-28.5=7.9

Then, with this procedure we get the sample for d:

d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]

The sample mean and standard deviation are:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729

The degrees of freedom for this sample size are:

df=n-1=8-1=7

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>1.729)=0.0637

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

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