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(8 points) In 2015, the average cost of attendance at public 2-year colleges and technical schools across the United States w

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(1) Q-Q plot forms a linear line which means that the data follows approximately normal distribution.

Yes, (we can conduct the test) because the data is approximately normal.

Even thought we require large size >30 but that is the case if we have the unknown distribution of population and we assume to be normal. But here the plot is showing the distribution.

n= 15 T = 10393 S = 1590

(2) We are testing whether the average cost is less than the national avg. A left 1-tailed test.

\mathbf{H_{0}:\mu=12260\;vs\;H_{1}=\mu<12260}

(3) Test Statistic: \frac{\bar{x}-\mu_{0}}{S_{x}/\sqrt{n}}

= \frac{10393-12260}{1590/\sqrt{15}}

Test Statistic = -4.5477

We are going to use t-dist since we have unknown population standard deviation and it is 1-tailed test.

(4) p - value = P (t_{n-1} > | Test Stat| )

= P (t_{14} > 4.5477)

= \mathbf{0.00023} (using t-dist tables)

(5) We are using a 0.05 level of significance

Since p-value < 0.05

We have

Strong evidence to reject the null hypothesis

We have sufficient or strong evidence terms in hypothesis testing. Strong because p-value < 0.01

(6) (1- \alpha) confidence interval for population mean is

(\bar{x}\bar{+}t_{n-1,\alpha/2}\frac{S_{x}}{\sqrt{n}})

t_{n-1,\alpha/2}=t_{14,0.05/2} (\alpha = 1 - 0.95 = 0.05)

t_{14,0.025}=2.1448 (using t-dist tables)

Substituting the values

(10393\bar{+}2.1448\frac{1590}{\sqrt{15}})

\mathbf{(9512.487,11273.51)}

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