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Allisa [31]
1 year ago
9

We want to conduct a hypothesis test of the claim that the population mean reading speed of second graders is different from 28.

3 words per minute. So, we choose a random sample of students' reading speeds. The sample has a mean of 28.2 words per minute and a standard deviation of 4 words per minute.
For each of the following sampling scenarios, choose an appropriate test statistic for our hypothesis test on the population mean. Then calculate that statistic. Round your answers to two decimal places.
Mathematics
1 answer:
stellarik [79]1 year ago
5 0

The test statistics based on the information given will be 0.238.

<h3>How to compute the test statistic?</h3>

Population mean = 28.3

Sample mean = 28.2

Standard deviation = 4

Sample size = 90

In this case, the z test will be used since the standard deviation is known. This will be:

= (28.2 - 28.3)/4/✓90

= -0.1/(4/9.48)

= -0.1/0.42

= -0.238

Learn more about hypothesis on:

brainly.com/question/11555274

#SPJ1

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A room has eight switches, each of which controls a different light. Initially, exactly five of the lights are on. Three people
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Answer:

The probability that 3 lights are on after the third person exited the room is 39/128

Step-by-step explanation:

In order for 6 switches to be on at the end, we need exactly 2 people turning on a switch and the other one turning one off. There are 3 possibilities:

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Note that after turning off one switch one more switch will be available to be switched in and one less will be available to be switch off. The contrary happens when someone turns in a switch.

Lets calculate the probability for the first scenario. The probability for the first person to turn the switch in is 3/8, because there are 3 lights off. For the second person, there will be only 2 lights off, thus, the probability for him or her to turn the switch in is only 2/8, leaving only 1 light off and 7 on. The third person will have, as a consecuence, a probability of 7/8 to turn off one of the 7 switches. This gives us a probability of 3/8 * 2/8 * 7/8 = 21/256 for the first scenario.

For the second scenario we will have a probability of 3/8 for the first person, a probability of 6/8 for the second one (he has to turn a switch off this time), and a probability, again, of 3/8 for the third one, giving us a probability of 3/8*6/8*3/8 = 27/256 for the second scenario.

For the third scenario, the first person has to turn off the switch, and it has a probability of 5/8 of doing so. The second person will have 4 switches to turn on, so it has a probability of 4/8 = 1/2, and the third person will have one switch less, thus, a probability of 3/8 of turning a switch on. Therefore, the probability of the third scenario is 5/8*1/2*3/8 = 15/128 = 30/256

By summing all the three disjoint scenarios, the probability that six lights are on is 21/256+27/256+30/256 = 78/256 = 39/128.

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