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A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed wi...

Question

A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with population standard deviation of 26 hours. How many components must be sampled so that a 99% confidence interval will have margin of error of 6 hours?Write only an integer as your answer:

A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with population standard deviation of 26 hours. How many components must be sampled so that a 99% confidence interval will have margin of error of 6 hours? Write only an integer as your answer:



Answers

A simple random sample of 50 items from a population with $\sigma=6$ resulted in a sample mean of $32 .$
a. Provide a 90$\%$ confidence interval for the population mean.
b. Provide a 95$\%$ confidence interval for the population mean.
c. Provide a 99$\%$ confidence interval for the population mean.

51. What is the problem building? This sample meeting is between seven and 11. So for probability, using my normal CDF seven is mine or bound. 11 is my upper bound. I mean, is 10 on my standard deviation from problem 49 is 10/8. From that I get 0.77 99

In Exercise three, we're going to be considering a random sample that is drawn from a normally distributed population and this time the standard deviation is unknown and we've been to Use the given information to construct a 99% confidence interval for the population mean? So we have two cases in both cases, the sample uh mean and the sample standard deviation are 386 and 24 respectively. For the first case the sample size is 18 and the second cases sample size seven, So it's a smaller example for the second case, no first things. The first thing we need to do here is to determine the level of significance And also to determine the right formula that we need to use to get the 99% confidence interval. Now, since the uh population, standard division is unknown, we're going to use the following formula X. Bar plus or minus The critical value of tea for two tailed tests, multiplied by s example, standard deviation divided by the square root of n. And since it's a 99% confidence interval, it means that uh the value Of Alpha, the level of significance equals 1 -0.99 And that equals 0.01 Half of that is 0.005. So this is useful for us to determine the correct critical bottle 40. Now, if you're using the t distribution, you also need to give the degrees of freedom for each case. Now, for the first case, Because the sample size is 18 So that the number of degrees of freedom will be N -1, which equals 18 -1 and that equals 17. So you want to check for the corresponding critical value of T Where they for 0.0 Uh five as a level of significance. So that corresponds to the value two point eight 98. And now that we have all we need we can substitute um all the material and all the content that we had uh into the formula. So let's begin. So X bomb is 386 plus or minus. The critical value 40 is 2.898, Multiplied by S which is 24 Divided by the Square Root of N., which is square root of 18. And when you saw that the calculator you can to obtain 386 plus or minus 16.39. So the margin of error for this case is 16.39. Now let's proceed to the second case, We want to start by getting the degrees of freedom which will be N -1. And in this case it's going to be 7 -1 and that equal six. And so the critical value of T. There corresponds to this level of significant European 005 Is uh is obtained from the table, it is 3.70 uh 7. Then We can substitute our values into the formula again and you're going to get 386 plus or minus 3.707, multiplied by 24, divided by the Square Root of seven. and when you work that out, you're going to get 386 plus or Uh minus 33 0.63. So for the second case where we have a smaller sample, you can notice that the margin of error is greater, meaning that we're going to have a longer interval is compared to when the sample size is smaller.

So 96 wants us to define X, the variable and X bar in words. So what exes is what you collected. So you collected the length of the conferences. X bar is our sample mean? So it's the mean length of those 84 conferences we selected. So their distribution. So since everything we have is sample distribution, we have to use a tea distribution instead of the normal distribution. So the distribution would be key 83 so t for tea distribution 83 minus one. So there were 84 samples and then similar parameters So mean 3.94 and then the standard deviation one point to a over the square root of 84. So that's B see is looking for the confidence interval and the craft and in the air amount. So we would have our mean 3.94 plus or minus the critical value. There's a lot of different ways whether used the tea I calculator or the chart, um, 83 probably won't be on your chart, so I use a T I calculator that gives me 1.98 So depending home, how you have to do your critical value. This number may very times one point to a over the square root of 84. So that gives us 3.94 plus or minus point 2778 0.2778 would be your heir abounds. So we do this attraction, we get 3.662 22 actually, and the upper when we add four point 2170. So once again, if you look at the graph, we're 95% shore that the mean length is between approximately 3.66 and 4.22 hours.

Problem. 23 ex army square for degrees of freedom is 21 off. Open toe one is equal to 38.932 and the chi square left off degrees of freedom 21. And, uh 0. 00.99 is equal to eight point 897 So the boundaries are for the variance r n minus One times Sunder division squared over 38 0.932 and 20 to minus 13.6 square over 8.8 on 97 to the population. Uh oh. The population variance is also 6.99 30 point 59 and so the center division is 2.64 5.53


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