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The output voltage of a power supply is assumed to be normallydistributed. Sixteen observations taken at random on voltage are asfollows: 10.35, 9.30, 10.00, 9.96, ...

Question

The output voltage of a power supply is assumed to be normallydistributed. Sixteen observations taken at random on voltage are asfollows: 10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58,11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, and 10.85. Test the hypothesis that the mean voltage equals 12v usinga 95% two-sided confidence interval.Test the hypothesis that the variance equals 11v using a 95%two-sided confidence interval

The output voltage of a power supply is assumed to be normally distributed. Sixteen observations taken at random on voltage are as follows: 10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, and 10.85. Test the hypothesis that the mean voltage equals 12v using a 95% two-sided confidence interval. Test the hypothesis that the variance equals 11v using a 95% two-sided confidence interval



Answers

Assume the sample is from a normally distributed population and construct the indicated confidence intervals for $(a)$ the population variance $\sigma^{2}$ and $(b)$ the population standard deviation $\sigma .$ Interpret the results. Car Batteries The reserve capacities (in hours) of 18 randomly selected automotive batteries are listed. Use a $99 \%$ level of confidence.
$$\begin{array}{llllll}1.70 & 1.60 & 1.94 & 1.58 & 1.74 & 1.60 \\1.86 & 1.72 & 1.38 & 1.46 & 1.64 & 1.49 \\1.55 & 1.70 & 1.75 & 0.88 & 1.77 & 2.07\end{array}$$

The following is a solution to # seven. And this says that a random sample of 12 people were selected. We don't really know what it is, but it's from a normal approximation or a normal distribution approximately normal. So that allows us to have this small sample size. Normally, you have to have a sample size of at least 30. But since it comes from a normal distribution, we can just say that this is okay, so in equals 12 is fine. And the uh the sample mean X bar was 45 And the Sample Standard Deviations 14. And we're asked to find the 90% confidence interval for the population mean. So, since we're estimating the population mean, we're either going to use the Z or the tea interval, and we're gonna use the tea interval, because we don't know what sigma is, we don't know that population standard deviation, because we weren't given given that we were only given the sample standard deviation. S so because of that, we have to use the T interval. Okay, so terrible. So the 90% confidence interval now you can use the tea interval formula, you can certainly go about it that way. I'm gonna use the form of the calculator just because it goes by a little bit more quickly. But if you go to stat and then tests, it's gonna be this eighth option down here, this is on a. T. I. T. For uh T. Interval, so I'm gonna click eight and make sure summary stats is Is highlighted there and I'm just gonna start filling in my stuff. So x bar is 45. The sample standard deviation is 14 And the n. was 12 And we want to be 90%. So I want to say .9 there for the sea level And we calculate and then this top band here gives us what we need. So 37 742 and 52.258. Let's go and write that down. So 37 742 All the way up to 52 258 It doesn't say to do this, but if it asks for an interpretation, you would just say we are 90% confident that the true population mean mu is between 37.742 and 52.258.

In this exercise begin to be considering a scenario where we have a random sample that is obtained from a population of unknown standard deviation. So the standard deviation for the population is unknown. And we're going to use that information To construct a 98% confidence interval for the population mean. And we have two scenarios in the first case, he were given that the sample size and Equals 225 and the sample mean X bar 92.0. The sample standard deviation is 8.4. To proceed, we have to use the following formula, you're going to have X band plus or minus The critical value for Z. In this case where the level of significance is 0.02 is equal to 2.3 uh to six. And because the sample of the population, standard division is unknown, we're going to estimate it to be as the sample standard deviation, then multiply divide that by this quality of end. So let's go ahead and do the substitution For the first case. So x bomb is 92.0 Plus or -2.3-6, Multiplied by 8.4, Divided by the Square Root of 225. So when you simplify that, You're going to obtain 92.0 plus or minus 1.30. In the 2nd case, B we have the sample size in of 64 The sample mean x bar of 92.0. And the sample standard deviation has 8.4. So we substitute the values again, it's going to be 92.0 plus or minus 2.3-6 times 8.4 Divided by the Square Root of 64. And with this one simplified becomes 92.0 Plus or -2.44. So those are the 98% confidence intervals for the population. Mean for the two cases A n B.


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