In Exercises 13–16, refer to the indicated data set in Appendi...

Question

In Exercises 13–16, refer to the indicated data set in Appendix B and use the sign test for the claim about the median of a population.

Cotinine in Smokers Data Set 15 “Passive and Active Smoke” includes cotinine measurements from 902 smokers. Cotinine is a biomarker of nicotine in the body. Use a 0.01 significance level to test the claim that smokers have cotinine levels with a median of 2.84 ng/mL, which is the median for nonsmokers not exposed to tobacco smoke.

Accepted Answer

All right, hello, everyone. So this question says, a scientist records the daily fiber intake in grams of 250 adults. The recommended median fiber intake is 25 g per day. In the sample, 130 adults consumed more than 25 g. 110 consumed less than 25 g. And 10 consumed exactly 25 g. At the 0.05 significance level, use the sign test to determine if the median fiber intake in the sample is different from the recommended value. So first, let's begin with what information we're actually given. First and foremost, the null hypothesis median is 25 g. And the sample size itself, in this case, excludes the people that eat exactly 25 g per day. So N, which is that sample size, is equal to the sum of 130 and 110, which gives you 240. Now, the number of positives, that is the number of people that eat more than 25 g of fiber. Which will represent here as X is equal to 130. And as mentioned before, the significance level or alpha. is equal to 0.05, and this is a two-tailed test. So first, let's begin with our hypotheses. The null hypothesis, or HO, is that the median daily fiber intake is 25 g. Or alternative hypothesis or H1. Is where the median. Is simply not 25 g. So now let's focus on the distribution under the null hypothesis. So here, under H knot, the number of positives, that is the number of people that consume more than 25 g, follows a binomial distribution. And so because the sample size is quite large, that 240, we can use the normal approximation. According to the normal approximation, mu, which is the population mean, is equal to N multiplied by P. So here, N is equal to 240. And we'll multiply that by 0.5. This equals 120. So, since we're using the normal approximation, 120 is equal to the square root of NP multiplied by 1 subtracted by P. So that's equal to. Let's see here, 240, which is the sample size. Multiplied by 0.5. And then one subtracted by 0.5 is 0.5. And after evaluating this expression, this comes out to approximately 7.75. Now to find the Z score, we can use the continuity correction. And the continuity correction. Which would give us z. Is equal to the number of positives that's 130. Subtracted by 0.5. And subtracted by 120. Divided by 7.5 or 7.75 rather. After evaluating this expression, this gives you about 1.23. So for the given Z score, we can now find the p-value. And so the P value. For a two-tailed test is equal to the following. It's going to be equal to 2 multiplied by the probability, oops. The probability of a Z score greater than 1.23. So we can rewrite this expression to say. That the p value is too multiplied by. That's 2 multiplied by 1 subtracted by. By 1.23. So here, let me scroll down to open up some more space. And here, this expression is going to simplify to 2. Multiplied by 1, subtracted by 0.89. 07. And now, your answer comes out to 0.2186. So what does this P-value tell us about the conclusion we can make with regards to this experiment? Well, notice how the P-value, 0.2186 is greater than our significance level of 0.05. Because of that, we failed to reject the null hypothesis. And so now, this brings us to our final answer, which reads as follows. There is not sufficient evidence to conclude that the median fiber intake in the sample is different from 25 g. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Sign Test

Median

Significance Level

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