Hypothesis Testing Using a P-Value In Exercises 33–38,
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Question

Hypothesis Testing Using a P-Value In Exercises 33–38,

a. identify the claim and state and .

b. find the standardized test statistic z.

c. find the corresponding P-value.

d. decide whether to reject or fail to reject the null hypothesis.

e. interpret the decision in the context of the original claim.

Sprinkler Systems A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135°F. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133°F. Assume the population standard deviation is 3.3°F. At alpha=0.10, do you have enough evidence to reject the manufacturer’s claim?

Accepted Answer

Hello everyone. Let's take a look at this question together. A company claims that the mean lifetime of its LED bulbs is at least 25,000 hours. A random sample of 35 bulbs has a mean lifetime of 24,400 hours. The population standard deviation is known to be 1200 hours. At alpha equals 0.05, do you have enough evidence to reject the company's claim? Use a P value. So, in order to solve this question, we have to recall how we can determine whether there is enough evidence to reject the company's claim that the mean lifetime of its LED bulbs is at least 25,000 hours if we have a random sample of 35 bulbs with a mean lifetime of 24,400 hours. and a population standard deviation of 1200 hours. And so looking at the information provided in the question, we should note that the sample size is and equals 35. And so to determine if there is enough evidence to reject the company's claim, we have to conduct a requirement check. We know since the population standard deviation is known, the sample is random, and our sample size N is greater than 30, we can use a P value for a Z test. And the first step is to state the null and alternative hypotheses, which we know that the company claims that the mean lifetime of its LED bulbs is at least 25,000 hours, and this can be expressed as mu is greater than or equal to 25,000. And since it is a statement of equality, it is the null hypothesis. So our null hypothesis is mu is greater than or equal to 25,000. And the complement of the null hypothesis is the alternative hypothesis. So our alternative hypothesis is is less than 25,000. So then our null and alternative hypotheses are the null hypothesis of mu is greater than or equal to 25,000, which is our claim, and the alternative hypothesis is that mu is less than 25,000. And based on the information provided in the question, we know the given level of significance, which is alpha, is alpha equals 0.05. And now we have to find the standardized test statistic Z, which is equal to the sample mean subtracted by the population mean divided by the population standard deviation, which is divided by the square root of the sample size, which substituting the values into our formula, we have 24,400 subtracted. By 25,000, which is divided by 1200 divided by the square root of 35. And when simplified, our equation gives us -2.958039892 as the value for the standardized test statistic Z. And now we have to find the area that course. response to Z, where we will use the table for standard normal distribution, and note that the table values represent the cumulative area to the left of the Z score. And since -2.958039892 is between -2.96 and 2.95, we can And interpolate. And looking at the table for standard normal distribution, we will let the area corresponding to our Z of -2.958039892 be A. And then by interpolation, we have the following equation, which we can then rewrite this equation to solve for A. So that A is equal to -2.958039892, subtracted by -2.96, divided by -2.95, subtracted by 2.96, all multiplied by 0.0016, subtracted by 0.0015, plus 0.0015. And when simplified, A is equal to 0.00152. And now we can find the p value. And since our alternative hypothesis contains the less than inequality symbol, the hypothesis test is a left-tailed test, wherefore a left-tailed test, P is equal to the area in the left tail. And since the area obtained from the previous step is a cumulative area. To the left of the Z score, we know that P is equal to A, so P is equal to 0.00152. And using this information, we can now make a decision. And since P is less than alpha, as 0.00152 is less than 0.05, we reject the null hypothesis. And since the claim is our null hypothesis, and we reject the null hypothesis. The interpretation for this decision is that there is sufficient evidence to reject the claim that the mean lifetime of the LED bulbs is at least 25,000 hours. And looking at our answer choices, we can identify that answer choice A is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

P-Value

Standardized Test Statistic (z)

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