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lara31
6 days ago
14

Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was

15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5 A visual of the data distribution is below: Do these data provide convincing evidence at the α = 0.10 level that the average vertical jump of students at this school differs from 15 inches?
Mathematics
1 answer:
tester [3.9K]6 days ago
7 0

Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

H_0: \mu=15\\\\H_a: \mu\neq15

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

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Step-by-step explanation:

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Jacque needs to buy some pizzas for a party at her office. She's ordering from a restaurant that charges a \$7.50$7.50dollar sig
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