In a certain hypothesis test, H0:p=0.4H_0:p=0.4, Ha:pH_{a}:p < 0.40.4. You collect a sample and calculate a test statistic z=−1.32z=-1.32. Find the PP-value.

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.0934 -1.32)=0.0934

In a certain hypothesis test, H 0 : p = 0.4 H_0:p=0.4 , H a : p H_{a}:p < 0.4 0.4 . You collect a sample and calculate a test statistic z = − 1.32 z=-1.32 . Find the P P -value .

Back, everyone. Let's take a look at this practice problem here. So in a certain hypothesis test, we have no idea what the context of the problem is. We have a null and alternative hypothesis, which is that the proportion is equal to 0.4 of something. And we're trying to find evidence for, that's our alternative hypothesis, that p is actually less than 0.4. Well, these are proportions, not p values. Alright? So in order to find evidence for this claim, you're gonna collect a sample and calculate a test statistic. And we find that z is equal to negative 1.32. So we're going to use the z score over here that's given to us to find the p value. Alright. So even though we actually have no idea what is going on in this problem, we have no context. We actually can still find an answer for this because remember that a p value is just an area that's under a graph, and we know exactly how to find areas from given z scores. Essentially, what this problem is asking you to do is assuming that this is a normal distribution, and you're using a z score, which that's that's the case, then what's the probability that z is less than this number, that p is less than negative 1.32? We've seen how to do this before with a z table or graphing calculator. So, basically, what I'm gonna assume in this problem is that you have access to a z table, which we have in our worksheets, or you might find 1 in your book, or you know how to do this on a graphing calculator. We have some videos on how to find probabilities from given z scores. Alright? So I'm just going to assume that you have both of these things, but basically what's going on here if I can just draw a little graph. I always like to draw a graph if everything makes a lot more sense. So what we're assuming here is that this normal distribution, we're going to assume that the peak is actually equal to p equals 0.4. That's your null hypothesis. Alright? So basically what we have what we saw from sampling distributions is that if this is actually the true number, then if you collect a bunch of samples, it should form a normal distribution around this number. Some proportions will be less, some proportions will be more, but it basically should form a normal around this number. So what happens is we're going to go out and grab a sample, which we did, and we found that from our sample, we found a z score that's negative 1.32, which is gonna be somewhere over here on this diagram. Alright? So in this, what we calculated was a sample proportion, which is p hats. We don't know what that number is, but we do know what the z score is. Right? This z score over here was given to us. It's negative 1.32. Essentially, this p value, the probability that z is less than this number is really just the area that lies beyond or to the outside of this z score. So it's basically all of this area that's over here. Now, this is an area to the left, which means you can just go directly use your z table, or again, you could use a graphing calculator. But Basically, I'm just going to assume that you know how to do that. And the number for this is going to be 0.0934. Alright. So basically, if you use a z table or a graphing calculator, that's what this number ends up being, which is the number that's inside that represents this area over here. Alright. So here's what's kind of confusing about this whole entire thing and hypothesis test, especially when you get to proportions. You have 3 different p's that are inside the problem. You have a regular p, which is your population proportion, that's your parameter, the thing that you have a null hypothesis about. There's your sample proportion, which comes from your one sample, and then you also have a p value. So there's p, p hat, and p value. All of them mean different things, and this sometimes can be very, very confusing here. So I just wanted to go ahead and, give you that. Alright. So basically what we found over here was that our p value, was equal to 0.0934. Again, this always use it's always good to use a graph because we can see here that this area is kind of small, and that's basically what we see here in our graph. It's it's kind of small sliver of area that's to the left of the graph because you have a pretty high z score of negative 1.32. Alright, Alright, folks. Thanks for watching. Let me know if you have any questions. Let's move on.

In a certain hypothesis test, H 0 : p = 0.4 H_0:p=0.4 , H a : p H_{a}:p < 0.4 0.4 . You collect a sample and calculate a test statistic z = − 1.32 z=-1.32 . Find the P P -value .

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.0934 -1.32)=0.0934

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.8132 -1.32)=0.8132

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.9066 -1.32)=0.9066

P ( z P\left(z\right. < − 1.32 ) = 0.1868 -1.32)=0.1868

9. Hypothesis Testing for One Sample

Steps in Hypothesis Testing

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9. Hypothesis Testing for One Sample

Steps in Hypothesis Testing

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Multiple Choice

In a certain hypothesis test, $H sub 0 colon p equals 0.4$ , $H_{a} : p$ < $0.4$ . You collect a sample and calculate a test statistic $z = - 1.32$ . Find the $P$ -value.

$P\left(z\right.$ < $- 1.32) = 0.8132$

$P\left(z\right.$ < $- 1.32) = 0.9066$

$P\left(z\right.$ < $- 1.32) = 0.0934$

$P (z$ < $- 1.32) = 0.1868$

Verified step by step guidance

Identify the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, H0: p = 0.4 and Ha: p < 0.4.

Recognize that the test statistic z = -1.32 is given, which is used to determine the p-value for the hypothesis test.

Understand that the p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis.

Since the alternative hypothesis is Ha: p < 0.4, this is a left-tailed test. Therefore, the p-value is the probability that z is less than -1.32.

Use the standard normal distribution table or a calculator to find P(z < -1.32). This value represents the p-value for the test.

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Multiple Choice

A popular theme park claims that their weekly attendance is around $100 comma 000$ . You believe that the weekly attendance is different than this claimed value, so you gather sample data. Write the null and alternative hypotheses.

Multiple Choice

A candy manufacturer seeking to minimize the variation in weights of their candies claims to produce candies with a standard deviation less than $0.300$ g. Write the null and alternative hypotheses.

Multiple Choice

A survey claimed that $30 percent sign$ of adults prefer electric cars over traditional cars. A car manufacturer believes the true proportion is higher than $30 percent sign$ . To test this, they survey a random sample of $50$ adults and find that $19$ say they prefer electric cars. Determine which test statistic to use & calculate it.

Multiple Choice

Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.

$H sub 0 colon$ $p equals 0.4$

$H_{a} :$ $p is not equal to 0.4$

Textbook Question

Stem Cell Survey In a Newsweek poll of 882 adults, 481 (or 55%) said that they were in favor of using federal tax money to fund medical research using stem cells obtained from human embryos. A politician claims that people don’t really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Use the following probabilities related to determining whether the result of 481 is significantly high (assuming the true rate is 50%). Is 481 significantly high? What should be concluded about the politician’s claim? Explain.

P(respondent says to use the federal tax money) = 0.5

P(among 882, exactly 481 says to use federal tax money) = 0.000713

P(among 882,481 or more say to use federal tax money) = 0.00389

Textbook Question

Final Conclusions

In Exercises 21–24, use a significance level of α = 0.05 and use the given information for the following:

State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.)

Without using technical terms or symbols, state a final conclusion that addresses the original claim

Original claim: More than 35% of air travelers would choose another airline to have access to inflight Wi-Fi. The hypothesis test results in a P-value of 0.00001.

Textbook Question

Final Conclusions

In Exercises 21–24, use a significance level of α = 0.05 and use the given information for the following:

State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.)

Without using technical terms or symbols, state a final conclusion that addresses the original claim.

Original claim: The mean pulse rate (in beats per minute) of adult males is 72 bpm. The hypothesis test results in a P-value of 0.0095.

Textbook Question

Type I and Type II Errors

In Exercises 25–28, provide statements that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1.)

The proportion of people who write with their left hand is equal to 0.1.

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