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From a population of unknown mean \(\mu\) and a standard deviation \(\sigma=5.0,\) a sample of \(n=100\) is selected and the sample mean 41.5 is found. Compare the concepts of estimation and hypothesis testing by completing the following: a. Determine the \(95 \%\) confidence interval for \(\mu\) b. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the \(p\) -value approach and \(\alpha=0.05\) c. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the classical approach and \(\alpha=0.05\) d. On one sketch of the standard normal curve, locate the interval representing the confidence interval from part a; the \(z \star, p\) -value, and \(\alpha\) from part \(b ;\) and the \(z \\#\) and critical regions from part c. Describe the relationship between these three separate procedures.

Short Answer

Expert verified
The 95% confidence interval for \(\mu\) is [40.52, 42.48]. For the p-value approach hypothesis test, we reject the null hypothesis because the p-value (0.003) < alpha (0.05). For the classical approach, we again reject the null hypothesis where our calculated z (3) is outside the critical z-scores \(\pm\) 1.96. All tests indicate the population mean is significantly different from 40.

Step by step solution

01

Determine the 95% confidence interval for \(\mu\)

Calculate the standard error, which is \(\sigma_{\mu} = \sigma / \sqrt{n} = 5/\sqrt{100} = 0.5\). For a 95% confidence interval, the z-score is approximately 1.96. Therefore, the confidence interval is \(41.5 \pm 1.96(0.5) = [40.52, 42.48]\).
02

Hypothesis test using the p-value approach

Calculate the z-score first, which is \(Z = (X - \mu_0)/\sigma_{\mu} = (41.5 - 40)/0.5 = 3\). Next, calculate the \(p\)-value: for a two-tail test, \(p = 2*(1 - \Phi|Z|) = 2*(1 - 0.9985) = 0.003\). Since p-value is less than \(\alpha\), reject the null hypothesis.
03

Hypothesis test using the classical approach

With a z-score of 3 and \(\alpha=0.05\), the critical z-score for a two-tailed test is \(\pm\) 1.96. Since 3 is outside of this range, reject the null hypothesis.
04

Sketch and describe the relationship

All three procedures are determining whether the sample mean of 41.5 is significantly different from 40. The confidence interval is the range in which we expect the real mean to lie 95% of the time. The hypothesis tests are determining whether the mean could realistically be 40 given the sample data. Both lead to the same conclusion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval Calculation
Understanding confidence intervals (CI) is essential in statistics, as they give us a range within which we can be reasonably sure that a population parameter, like the mean, lies.

The formula for a confidence interval is based on the sample mean, the standard deviation of the population (which, in some cases, we might have to estimate from the sample), and a multiplier that comes from the standard normal distribution, often termed as the 'z-score'. For a 95% CI, this z-score is approximately 1.96, reflecting the fact that about 95% of the area under the normal curve falls within 1.96 standard deviations of the mean.

In the exercise, we used the formula to calculate the standard error, which is the standard deviation of the sampling distribution of the sample mean. The standard error tells us how much the sample mean might vary from the true population mean. We then used the z-score for the desired confidence level to construct the interval: the sample mean plus or minus the product of the z-score and the standard error. Thus, the 95% confidence interval for the population mean \( \mu \) stated that we are 95% confident that the true mean falls between 40.52 and 42.48.
P-Value Approach
The p-value is a concept that helps statisticians determine the significance of their results in a hypothesis test. It's defined as the probability of observing a test statistic as extreme as, or more extreme than, the value observed, assuming that the null hypothesis is true.

In the context of our example, the null hypothesis (\( H_0 \: \mu = 40 \) ) is being tested against the alternative hypothesis (\( H_a \: \mu eq 40 \) ). The z-score calculated from the sample data tells us how many standard errors our sample mean is from the hypothesized mean under \( H_0 \). The p-value, then, provides the probability that this difference or a more extreme one could occur if \( H_0 \) were true. If this p-value is lower than the predetermined threshold \( \alpha \), usually 0.05 or 5%, we have enough evidence to reject \( H_0 \). In our case, a p-value of 0.003 is well below 0.05, leading us to conclude that the true population mean is likely not 40.
Classical Approach in Hypothesis Testing
The classical approach, or the critical value approach, involves comparing the calculated test statistic to a value determined by the desired level of significance, known as the critical value. This approach tells us whether to accept or reject the null hypothesis.

In the hypothesis testing for our example, the z-score exceeds the critical z-score of \( \pm 1.96 \) associated with a 95% confidence level. The areas beyond the critical values are called the 'critical regions', and if our test statistic falls into this region, we reject the null hypothesis. Since our calculated z-score of 3 falls in the critical region, we reject the null hypothesis, thus supporting the alternative hypothesis that the true mean is not equal to 40.
Standard Normal Curve
The standard normal curve, or z-curve, is a graphical representation of the standard normal distribution, which is a normal distribution with a mean of 0 and a standard deviation of 1. It is crucial in hypothesis testing as it helps us understand the distribution of z-scores and the likelihood of different outcomes.

In our exercise, we use the standard normal curve to find the critical values and to understand the p-value. Both the confidence interval and the hypothesis tests relate to areas under this curve. Specifically, the confidence interval corresponds to the central area that includes 95% of the distribution, whilst the p-values and critical values are tied to the areas in the tails. The sketch from the exercise clearly shows how these concepts are visually interconnected, aiding in comprehension of their relationships and differences.

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Most popular questions from this chapter

a. Calculate the \(p\) -value, given \(H_{a}: \mu<45\) and \(z *=-2.3\) b. Calculate the \(p\) -value, given \(H_{a}: \mu>58\) and \(z *=1.8\)

a. If \(\alpha\) is assigned the value \(0.001,\) what are we saying about the type I error? b. If \(\alpha\) is assigned the value \(0.05,\) what are we saying about the type I error? c. If \(\alpha\) is assigned the value \(0.10,\) what are we saying about the type I error?

The weights of full boxes of a certain kind of cereal are normally distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. a. Find the \(95 \%\) confidence interval for the true mean weight of a box of this cereal. b. Find the \(99 \%\) confidence interval for the true mean weight of a box of this cereal. c. What effect did the increase in the level of confidence have on the width of the confidence interval?

Women own an average of 15 pairs of shoes. This is based on a survey of female adults by Kelton Research for Eneslow, the New York City-based Foot Comfort Center. Suppose a random sample of 35 newly hired female college graduates was taken and the sample mean was 18.37 pairs of shoes. If \(\sigma=6.12,\) does this sample provide sufficient evidence that young female college graduates' mean number of shoes is greater than the overall mean number for all female adults? Use a 0.10 level of significance.

The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12 minutes and a standard deviation of 2.3 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 28 customers visiting her store this week resulted in a mean of 10.9 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12 minutes? Use \(\alpha=0.02\)

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