Question

you receive a brochure from a large university. the brochure indicates that the mean class size for full - time faculty is fewer than 33 students. you want to test this claim. you randomly select 18 classes taught by full - time faculty and determine the class size of each. the results are shown in the table below. at \\( \alpha=0.05 \\), can you support the universitys claim? complete parts (a) through (d) below. assume the population is normally distributed.
37 30 26 33 35 42 24 22 28
29 27 39 30 32 24 27 29 28
(a) write the claim mathematically and identify \\( h_{0} \\) and \\( h_{a} \\).
which of the following correctly states \\( h_{0} \\) and \\( h_{a} \\)?
\\( \bigcirc \\) a. \\( h_{0}: \mu<33 \\)
\\( h_{a}: \mu \geq 33 \\)
\\( \bigcirc \\) b. \\( h_{0}: \mu = 33 \\)
\\( h_{a}: \mu \
eq 33 \\)
\\( \bigcirc \\) c. \\( h_{0}: \mu \leq 33 \\)
\\( h_{a}: \mu>33 \\)
\\( \bigcirc \\) d. \\( h_{0}: \mu \geq 33 \\)
\\( h_{a}: \mu<33 \\)
\\( \bigcirc \\) e. \\( h_{0}: \mu = 33 \\)
\\( h_{a}: \mu<33 \\)
\\( \bigcirc \\) f. \\( h_{0}: \mu>33 \\)
\\( h_{a}: \mu \leq 33 \\)
(b) use technology to find the p - value.
\\( p= \\) (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample mean and standard deviation

Using the data values \(37,30,26,33,35,42,24,22,28,29,27,39,30,32,24,27,29,28\)
Sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)
\(\sum_{i=1}^{18}x_{i}=37 + 30+26+33+35+42+24+22+28+29+27+39+30+32+24+27+29+28 = 549\)
\(n = 18\), so \(\bar{x}=\frac{549}{18}=30.5\)

Sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\(\sum_{i=1}^{18}(x_{i}-\bar{x})^{2}=(37 - 30.5)^{2}+(30 - 30.5)^{2}+(26 - 30.5)^{2}+(33 - 30.5)^{2}+(35 - 30.5)^{2}+(42 - 30.5)^{2}+(24 - 30.5)^{2}+(22 - 30.5)^{2}+(28 - 30.5)^{2}+(29 - 30.5)^{2}+(27 - 30.5)^{2}+(39 - 30.5)^{2}+(30 - 30.5)^{2}+(32 - 30.5)^{2}+(24 - 30.5)^{2}+(27 - 30.5)^{2}+(29 - 30.5)^{2}+(28 - 30.5)^{2}\)
\(=42.25+0.25 + 20.25+6.25+20.25+132.25+42.25+72.25+6.25+2.25+12.25+72.25+0.25+2.25+42.25+12.25+2.25+6.25=529.5\)
\(s=\sqrt{\frac{529.5}{17}}\approx5.57\)

Step2: Calculate the t - statistic

The test statistic for a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Here, \(\mu = 33\), \(\bar{x}=30.5\), \(s\approx5.57\), \(n = 18\)
\(t=\frac{30.5 - 33}{5.57/\sqrt{18}}\approx\frac{- 2.5}{1.31}\approx - 1.91\)

Step3: Calculate the P - value

Since this is a left - tailed test (\(H_{a}:\mu<33\)) with \(n-1=17\) degrees of freedom.
Using a t - distribution table or technology (e.g., TI - 84: tcdf(-100,-1.91,17)), the P - value is approximately \(0.038\)

Answer:

\(P = 0.038\)