Read the sections on Hypothesis Testing in your book.

Click on these links

Think of an application where you can use hypothesis testing to test a premises. Use .05 for your alpha and give the scenario. Set up the problem, solve it and state your conclusions. See the example below. Then respond to at least 2 classmates.

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Write My Essay For MeFirst post: First summarize the 3 videos and make a note of what you learned. Then create your own hypothesis test. For instance, XYZ car company boast that it’s new car Eco Auto gets at least 59 miles to the gallon. Given a sample mean of 57 and a standard deviation of 3.5 where 35 people were tested, find the z value and p score. Hint: Use the P value calculator in the announcements to find the p score. Note: On this one you do not need to attach an excel spreadsheet.

Student Response:

Ho: u>=59

Ha: u<59

You can put this in excel or your calculator

= (59-57)/(3.5/35^.5)

=3.38

or enter this in excel

=norm.dist(59,57,3.5/35^.5,true)

and hit enter

Then take that value which we will call A

and type this in excel

=norm.s.inv(A)

and hit enter

Using the z to p value calculator with a=.05, 1 tailed test and z=3.38, you get

p=.000362

Since p < .05, you reject the null

You will want to respond to 2 classmates. Keep the conversation going by mentioning the video and tell what you learned. Also mention how the hypothesis test helped you.

ANSWER

**Summary of Videos**

There are two elements to hypothesis testing: the Null hypothesis and the alternate hypothesis. The null hypothesis represents the hypothesis to be tested. It is tested against the alternate hypothesis. The alternate hypothesis is the hypothesis that will be accepted if the null hypothesis is rejected. To reject a null hypothesis one should have a sample mean. For example, given a problem involving the height of pro basketball players. We are given their average height as 6.5 feet ten years ago. We doubt whether this is still the case today. Say we assume that the height today is greater than 6.5 feet. We will therefore set our = 6.5 and our >6.5. The sample mean should fall reasonably within the population mean. If it falls near the middle of population then we accept the mean to be 6.5, if it does not we reject it.

**Decisions**

Do not reject | Reject | |

True | Correct decision | Type I error with probability of alpha |

False | Type II error with probability of beta | Correct decision |

From the table if is true and we do not reject, then it is a correct decision. If is true and we reject, then the result is a type I error with probability of alpha. If the is false and we do not reject it then the result is a type II error with probability of beta. If is false and we reject it then it is the correct decision.

**Example**

The average IQ of a student population is 100 with a standard deviation of 15. The principle believes that his students have above average IQ. A researcher test the IQ of 30 random students with a mean score of 112. Is there sufficient evidence to support the principle’s claim? The IQ scores have a normal distribution. The level of significance is given as 0.05.

Solution:

: µ=100

: µ>100

Z= (112-100)/(15/30˄0.5)

Z=4.3818

P=0.00001

Conclusion:

Since p<0.05

Reject

Accept

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