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A manufacturer claims that the mean lifetime of its fluorescentbulbs is 1300 hours. A homeowner selects 25 bulbs and finds themean lifetime to be 1270 hours with a ...

Question

A manufacturer claims that the mean lifetime of its fluorescentbulbs is 1300 hours. A homeowner selects 25 bulbs and finds themean lifetime to be 1270 hours with a standard deviation of 80hours. Test the manufacturerʹs claim. Use α = 0.05. Identify theclaim, state the null and alternative hypotheses, find the criticalvalue, find the standardized test statistic, make a decision on thenull hypothesis (you may use a P-Value instead of the standardizedtest statistic), write an interpretation state

A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1300 hours. A homeowner selects 25 bulbs and finds the mean lifetime to be 1270 hours with a standard deviation of 80 hours. Test the manufacturerʹs claim. Use α = 0.05. Identify the claim, state the null and alternative hypotheses, find the critical value, find the standardized test statistic, make a decision on the null hypothesis (you may use a P-Value instead of the standardized test statistic), write an interpretation statement on the decision.



Answers

(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value(s) and identify the rejection region$(s),(c)$ find the standardized test statistic $z,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. If convenient, use technology. A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 750 hours. A random sample of 25 light bulbs has a mean life of 745 hours. Assume the population is normally distributed and the population standard deviation is 60 hours. At $\alpha=0.02,$ do you have enough evidence to reject the manufacturer's claim?

Is this company truly making light bulbs that lasts on average 1000 hours? What we're going to use a confidence interval to figure that the answer out and where it's going to be a 99% confidence interval. And here's our formula for quantitative confidence interval. So let's plug in what we know the sample that they've taken. The sample averages 1015. Yeah. The standard deviation is 25. And the sample size they took was 16. It didn't. Now we need our critical T value, which I'm gonna use table on my table. That's the T distribution critical values with the degrees of freedom of 15 sample size take away one gives me with 99% confidence T critical value of 2.947 Now, when I multiply 2.947 times 25 over the square root of 16, that's going to give me a margin of error of 18.41875 Okay, And now I'm gonna subtract that from 10 15. That'll be my low end of my interval 9 96.58125 And for the high end of my interval, adding that on will be 10 33 41875 and notice 1000 is inside of this interval. So yes, this company is making light bulbs that last, on average, 1000 hours.

In this exercise, we have a manufacturer of light bulbs, claims that the average lifetime of their light bulbs is 750 hours now, based on a sample of 50. We want to test, the manufacturers claim. So are no hypothesis can be that the claim is true and the alternatives is that the true mean of the light bulb lifetimes is less than 750 hours now. From the question, we are given the sample information sample sizes 50 which means that degrees of freedom is 49. We're also told that the sample average was 738.44 hours a sample standard deviation was 38.2 hours, and we're testing our hypotheses at a significance level of 0.5 now. We don't know anything about the distribution of the population of lightbulbs. It might be reasonable to assume that's normal but were not given that. But our sample is of size 50 which means that it's reasonable to assume that based on the central limit, their that the sample averages will be approximately normally distributed and therefore the appropriate distribution for our test statistic will be the T distribution. We can calculate our test statistic with the following formula, and this comes out to minus 2.14 So now we want to calculate R P value, which is the probability of getting a test statistic at least that extreme from a sample of 50. I'm going to use our to calculate this, so we use the function. PT test statistic is minus 2.14 degrees of freedom is 49 and we get about 0.19 This is less than our significance level and therefore we would reject the null hypothesis and we can conclude that there is sufficient evidence. His suggests that the main light bulb lifetime is actually less than 750 hours. Yeah.

Let us read this question. A consumer group claims that the mean annual consumption of high fructose considered by a person in the United States is 48.8. So mean is 48 is 48.8. So this is going to be my It's not or my null hypothesis. What is my alternative hypothesis? It is going to be that this is not equal to 48 point it. Okay, a random sample of 1. 20. So let me just directly right the formula. My N is equal to 1. 20. So over here, I'm going to have root off 1 20 route off. 1. 20. Uh, a random sample of 1 20 people in the United States has mean off 49.5. So the mean is 49.5 14 9 0.5 minus. What does my hypothesized mean? 48.8 divided by what is the population standard deviation? It is 3.6. So this is 3.6. Let me use a calculator to find this. So this is 49.5 minus 48.8, divided by 3.6 multiplied by. Excuse me, sir. This is 0.194444 This is going to be multiplied by route off 1 20. So this is 2.13 myself. Statistic is coming out to be 2.13 What is my Alfa level? My Alfa level is 0.0 fight. So this is a two tailed test, my al 50.5 which means in a single tail my area should be 0.25 So if I find a critical value for 0.25 my critical values 1.96 So which means this is minus 1.96 and this is plus 1.96 And since my Z value is beyond 1.96 it falls somewhere in the rejection region. I will reject final hypothesis, So I will reject h not Okay, So I will reject. It's not so what can I say? The consumer group claims that the mean annual consumption is 48 pointed. So I will say that I have in a statistical evidence to say that the claim off the consumer group is false and this is going to be my answer

Let us read this questions. A consumer group claims that the mean annual consumption off charities by the person in the United States is at most £10.3 at most, 10.3. So it means that this is less than or equal to 10.3. This will be my null hypothesis. My alternative will be that mu is greater than 10.3 random sample off 10 people. Sorry. 100 people in the United States has a mean annual cherry cheese consumption of £9.9. So my expert is 9.9. Let me directly right the formula for the statistic It is 9.9 minus 10.3 upon What is this? Population standard deviation 2.1 divided by route hundreds. Okay, this is 2.1 divided by 200 which is nothing but which is going to become tense. So if I use a calculator 9.9 minus 10.3 divided by 2.1 multiplied by 10. So this is minus 1.90 minus 1.90 This is my Z statistic. Alright, what is my Alfa Alfa 0.5 and I can see over here that this is a single table test, right? What is going to be my? Have you made a mistake in the null and the alternative? The null is that mu is less than or equal to 10.3. And the alternative is that Mu is greater than 10.3. Okay, All right. So my Z value is turning out to be minus 1.90 My sense statistic is minus 1.90 So what is going to be my predict? Sorry. What is going to be my p value? My people, if I use a P value calculator, my people, who turns out to be 0.97 20 my p value for this turns out to be zero point. My p value turns out to be 0.9713 9713 This is my P value on. Since this is this was going to be a right tail test. I can see that my said statistic is over here, so definitely I will not be able to reject my hypothesis. I do not reject minor hypothesis, which means that I do not have enough evidence to say that my mu is greater than 10.3, right? I do not have enough evidence to say that my mu is greater than 10.3. Uh, this is going to be my answer.


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