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This is a left-tailed test Lets use α=0.05 (5% significance level)PowerPoint Presentation

This is a left-tailed test Lets use α=0.05 (5% significance level)

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We want to test the claim that less than 15% of people have gotten a speeding ticket in the last year. We sample 400 people, and 13% of our sample has gotten a speeding ticket.

This is a left-tailed test

Lets use α=0.05 (5% significance level)

Is this z-value unusual (unlikely to occur by chance alone)? gotten a speeding ticket in the last year. We sample 400 people, and 13% of our sample has gotten a speeding ticket.

Traditional Method:

The bottom 5% of possibilities occur at z-values less than

-1.645

Since our test statistic does not fall below -1.645, our test statistic is not unusual (at the 5% significance level)

Test Statistic:

Traditional Method: gotten a speeding ticket in the last year. We sample 400 people, and 13% of our sample has gotten a speeding ticket.

The bottom 5% of possibilities occur at z-values less than

-1.645

Since our test statistic does not fall below -1.645 (in the critical region), our test statistic is not unusual (at the 5% significance level)

P-Value Method: gotten a speeding ticket in the last year. We sample 400 people, and 13% of our sample has gotten a speeding ticket.

What is the probability of getting a z-value less than -1.12?

P(z<-1.12) = 0.1314

So a z-value less than -1.12 would occur in 13% of samples, even if p = 0.15. This is not unusual (doesn’t occur less than 5% of the time, based on our significance level).

Conclusion: gotten a speeding ticket in the last year. We sample 400 people, and 13% of our sample has gotten a speeding ticket.

There is not sufficient sample evidence to support the claim that less than 15% of people have gotten a speeding ticket in the last year.

This doesn’t mean that we’re wrong, just that either the true proportion is not very much less than 15%, or that our sample size was not large enough to provide enough evidence.

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