For Exercises I through 25, perform each of the following...

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Sick Days in Bed A researcher wishes to see if the average number of sick days a worker takes per year is greater than 5. A random sample of 32 workers at a large department store had a mean of 5.6. The standard deviation of the population is 1.2. Is there enough evidence to support the researcher's claim at $\alpha=0.01 ?$

For Exercises I through 25, perform each of the following steps.
a. State the hypotheses and identify the claim.
b. Find the critical value(s).
c. Compute the test value.
d. Make the decision.
e. Summarize the results.
Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Sick Days in Bed A researcher wishes to see if the average number of sick days a worker takes per year is greater than 5. A random sample of 32 workers at a large department store had a mean of 5.6. The standard deviation of the population is 1.2. Is there enough evidence to support the researcher's claim at $\alpha=0.01 ?$

Key Concepts

Summary of Results

The summary of results provides a clear statement on whether the statistical analysis supports or fails to support the research claim. It involves interpreting the test statistic in the context of the critical region and significance level, effectively communicating the conclusion drawn from the hypothesis test.

Decision Rule

The decision rule establishes the criteria for rejecting or failing to reject the null hypothesis based on the calculated test statistic and the predetermined critical value. It ensures that a conclusion is reached in a consistent, objective manner, aligning the observed evidence with the significance level.

Critical Value

The critical value is the cutoff point or boundary value derived from the chosen significance level and the type of test (one-tailed or two-tailed). It divides the range of the test statistic into a region where the null hypothesis is rejected and a region where it is not.

Test Statistic

The test statistic is a standardized value computed from sample data that quantifies how far the sample statistic deviates from the null hypothesis parameter, relative to the expected variability. Its calculation is based on the sampling distribution and is used to determine whether the observed data falls within a critical region.

One-Tailed Test

A one-tailed test is used when the research claim specifies a direction (for example, greater than or less than a certain value). This type of test concentrates the rejection region in one tail of the probability distribution, increasing the test's power to detect an effect in the specified direction.

Null and Alternative Hypotheses

The null hypothesis (H?) represents the default or status quo assumption about a parameter, while the alternative hypothesis (H? or Ha) reflects the claim or change that the study aims to support. Correctly formulating these hypotheses is crucial as they frame the entire testing procedure and affect the type of statistical test conducted.

Hypothesis Testing

Hypothesis testing is an inferential statistical method used to decide whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. It involves formulating a null hypothesis (typically a statement of 'no effect' or a baseline condition) and an alternative hypothesis (which reflects the research claim or the effect being tested).

Significance Level

The significance level, denoted by ?, is the probability threshold set by the researcher for rejecting the null hypothesis. It defines the risk of committing a Type I error, which is falsely rejecting the null hypothesis when it is true. A smaller ? indicates more stringent criteria for evidence.

Transcript

00:01 Alright, so this is the question they asked in my interview, so i'll just quickly solve this question here so in this question, they want to figure out a problem involving sick days for workers and the research of which to see if the average number of sick days a worker takes per year is greater than five so let's plug in all the data they give us so first we know that n the sample size equals 32 next we have that the same deviation of population so we call that sigma equals 1 .2.

00:35 The alpha level for this test is going to be 0 .01 and the mean for our sample so x bar equals 5 .6.

00:50 So let's look at our hypotheses for this test now.

00:53 So researcher wants to see if the sick days of work is an average is greater than 5.

00:59 But the no hypothesis is basically the given.

01:01 So here the given would be, that mu which is the average number of six days equals five and the alternative hypothesis would be that mu is greater than five now to move on to the test we have to calculate a z statistic so they calculate this we just use as formula right here which is just the x bar minus mu not divided by the steady variation population divided by the square root of the sample size so plugging that in we get that z equals x bar which is 5 .6 so 5 .6 minus b not which is 5 divided by the side deviation which is 1 .2 divided by the square root of 32 so next just plug us into a calculator and if you plug us into a calculator you will get i should i've probably done this beforehand so i'm just quickly calculate this right now you will get 2 .8 so this equals to approximately 2 .83.

02:18 Now now that we have our z value that we've calculated here we have to compare it to the z value for an alpha of 0 .01 so what's actually going on here is that for the test we're going to assume that our no hypothesis is true.

02:33 That's how they work so if we plot our normal distribution, it will look like this with the center being our mu not of 5 and the alpha of 0 .001 refers to this area right here, which is an area of 0 .01, which means that all of this area is 0 .99.

02:58 So the z value we found is 2 .83 and we need to compare it to the z value of an alpha of 0 .01, which is the number we find right here.

03:07 How do we get that value? well, there's two main ways i use.

03:11 The first way, the most common way is to look at a z table and we'll look at the z table and look for area of 0 .99 because this is a one -sided test.

03:21 Now this was a two -sided test which would mean that instead of having greater than five here we would have not equal to then this should be a two -sided test but here because it's greater than directional is one -sided...

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