Chapter 10 Assignment OneSample Tests Of Hypotheses

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, (See Course Schedule).
Six Steps to Conducting a Statistical Test
 The null and alternative hypotheses
 Level of significance \(\alpha\)
 Test statistics
 Compute the pvalue
 Check whether to reject the null hypothesis by comparing pvalue to \(\alpha\)
 Conclusion in words
A reminder of what is a pvalue in hypothesis testing: Pvalue is a probability of obtaining a value of the test statistic or a more extreme value of the test statistic assuming that the null hypothesis is true.
Caution: Sometimes pvalue is also referred to as the level of significance. One should be aware that \(\alpha\) (alpha) is also called level of significance. This makes for a confusion in terminology. α is the preset level of significance whereas pvalue is the observed level of significance. The pvalue, in fact, is a summary statistic which translates the observed test statistic's value to a probability which is easy to interpret.
Example: Online Purchases
An ecommerce research company claims that 60% or more graduate students have bought merchandise online. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students show that only 22 students have ever done so. Is there enough evidence to show that the true porportion is lower than 60%? Conduct the test at 10% Type I error rate, and use the pvalue and rejection region approaches.
Work out your answers to the questions below and then click on the icon to compare answers.
Set up the hypotheses for the consumer advocate, described above. Specify whether it is a lefttailed test, righttailed test, or a twotailed test.
Set up the hypotheses for the consumer advocate, described above. Specify whether it is a lefttailed test, righttailed test, or a twotailed test.
Answer:
Test on population proportion:
H_{o}: \(p =0.6\) (stands for \(p \geq 0.6\))
H_{a}: \(p <0.6\)
It is a lefttailed test.
Now at this point we want to check whether the sample size is large enough so that we can use the oneproportion ztest.
Check the conditions below now. 1) \(np_0 \geq 5\), and 2) \(n(1  p_0) \geq 5\)
np_{0} = 80 × 0.6 = 48 ≥ 5
n(1  p_{0}) = 80 × 0.4 = 32 ≥ 5
Thus, we can use the z approximation.
Next, we can calculate the test statistic.
\[\hat{p} = \frac{22}{80} = 0.275\] \[Z^{*} = \frac{\hat{p}p_{o}}{\sqrt\frac{p_{o}\times(1p_{o})}{n}} = \frac{0.2750.6}{\sqrt\frac{0.6\times0.4}{80}}=5.93\]
Next find the rejection region for the level of significance and also the pvalue for the test statistic.
The example states a 10% Type I error rate which corresponds to an alpha value of 0.10.
From the Ztable and since test is lefttailed, we want as critical value the zvalue with 0.1 to the left of it. This corresponds to  1.28 therefore the rejection region is Z* ≤ 1.28.
The pvalue would be found by \(P(Z \leq Z^{*})= P(Z \leq 5.93) \approx 0\). This Z* is off the table implying the pvalue would be close to zero.
Finally, use the test statistic and pvalue to make decision and overall conclusion.
With the test statistic Z* of 5.93 falling in the rejection region (i.e. less than 1.645) we will reject the null hypothesis.
With the pvalue close to zero and being less than \(\alpha\) of 0.10 we will reject the null hypothesis. We have statistical evidence at the 10% level of significance to conclude that fewer than 60% of graduate students have purchased merchandise online.
To determine whether the probability is small, we will compare it to the preset level of significance, which is the probability of Type I error. Recall that Type I error is the more serious error  to reject the null hypothesis when that null hypothesis is true. Think of finding guilty a person who is actually innocent.
When we specify our hypotheses, we should have some idea of what size Type I error we can tolerate. It is denoted as \(\alpha\). A conventional choice of \(\alpha\) is 0.05. Values ranging from 0.001 to 0.1 are also common and the choice of \(\alpha\) depends on the problem one is working on.
Or we can summarize the data by reporting the pvalue and let the users decide to reject \(H_0\) or not to reject \(H_0\) for their subjectively chosen \(\alpha\) values. The pvalue can also be called the observed level of significance and our book just sometimes refers to it as the level of significance. That may cause confusion and thus we recommend always calling it the pvalue and reserve the term level of significance to represent the preset \(\alpha\) value.
Example: Emergency Room Wait Time
The administrator at your local hospital states that on weekends the average wait time for emergency room visits is 10 minutes. Based on discussions you have had with friends who have complained on how long they waited to be seein in the ER over a weekend, you dispute the administrator's claim. You decide to test you hypothesis. Over the course of a few weekends you record the wait time for 40 randomly selected patients. The average wait time for these 40 patients is 11 minutes with a standard deviation of 3 minutes. Do you have enough evidence to support your hypothesis that the average ER wait time exceeds 10 minutes? You opt to conduct the test at a 5% level of significance.
Work out your answers to the questions below and then click on the icon to compare answers.
Set up the hypotheses for the example described above. Specify whether it is a lefttailed test, righttailed test, or a twotailed test.
Set up the hypotheses for the example described above. Specify whether it is a lefttailed test, righttailed test, or a twotailed test.
Answer:
Test on population mean:
H_{o}: \(\mu = 10\)
H_{a}: \(\mu > 10\)
It is a righttailed test.
At this point we want to check whether the data is approximately normal so we can use the onemean ttest.
Check the condition.
Since we do not have the actually wait times to check normality, we will consider the sample size. With a sample size of 40 we exceed our minimum requirement of 30 and can proceed with the test.
Next, we can calculate the test statistic.
\[t^{*}=\frac{\bar{x}\mu_{0}}{S/\sqrt{n}}=\frac{1110}{3/\sqrt{40}}=2.11\]
Next find the rejection region for the level of significance and also the pvalue for the test statistic.
The example states a 5% level of significance so \(\alpha = 0.5\).
From the ttable and since test is righttailed, we want as critical value the tvalue with 0.05 to the right of it. With degrees of freedom equal to n  1, the df are 39. Since 39 is not on the table we will use the closest without exceeding which is 35. With 35 degrees of freedom, the critical value is 1.69. Therefore the rejection region is t* \(\geq\) 1.69.
The pvalue would be found by \(P(t \geq t^{*})= P(t \geq 2.11)\). This t* is not in the table for 35 df, but does fall between table tvalues of 2.030 and 2.438. These tvalues correspond to righttail probabilities of 0.025 and 0.01 suggesting the pvalue for our t* of 2.11 is between 0.01 and 0.025
Finally, use the test statistic and pvalue to make decision and overall conclusion.
With the test statistic t* of 2.11 falling in the rejection region (i.e. greater than 1.69) we will reject the null hypothesis.
With the pvalue between 0.01 and 0.025 making the pvalue less than \(\alpha\) of 0.05, we will reject the null hypothesis. We have statistical evidence at the 5% level of significance to conclude that The average emergency wait time at the hospital is more than 10 minutes.
Statistical and Practical Significances
Our decision in this last example was to reject the null hypothesis and conclude that the average wait time exceeds 10 minutes. However, our sample mean of 11 minutes wasn't too far off from 10. So what do you think of our conclusion? Yes, statistically there was a difference at the 5% level of significance, but are that "impressed" with the results? That is, do you think 11 minutes is really that much different from 10 minutes? Since we are sampling data we have to expect some error in our results therefore even if the true wait time was 10 minutes it would be extremely unlikely for our sample data to have mean of exactly 10 minutes. This is the difference between statistical significanceandpractical significance. The former is the result produced from the sample data while the latter is the practical application of those results.
Words of Caution
Critics of hypothesistesting procedures have observed that a population mean is rarely exactly equal to the value in the null hypothesis and hence, by obtaining a large enough sample, virtually any null hypothesis can be rejected. Thus, it is important to distinguish between statistical significance and practical significance.
Statistical significance is concerned with whether an observed effect is due to chance and practical significance means that the observed effect is large enough to be useful in the real world.
To determine whether the probability is small, we will compare it to the preset level of significance, which is the probability of Type I error. Recall that Type I error is the more serious error  to reject the null hypothesis when that null hypothesis is true. Think of finding guilty a person who is actually innocent.
When we specify our hypotheses, we should have some idea of what size Type I error we can tolerate. It is denoted as \(\alpha\) . A conventional choice of \(\alpha\) is 0.05. Values ranging from 0.001 to 0.1 are also common and the choice of \(\alpha\) depends on the problem one is working on. Or we can summarize the data by reporting the pvalue and let the users decide to reject \(H_0\) or not to reject \(H_0\) for their subjectively chosen α values.
Another oneproportion example  this example uses \(\pi\) in place of p to represent the proportion. Note this changes nothing in the overall testing process.
A pharmaceutical company claims that a new treatment is successful in reducing fever in more than 60% of the cases. The treatment was tried on 40 randomly selected cases and 11 were successful. Do you think the company's claim is valid? (Can you reject the company's claim)
Work this our yourself and then review the video (no sound) below:
Using a Confidence Interval to Draw a Conclusion About a Twotailed Test
The primary purpose of a confidence interval is to estimate some unknown parameter. A secondary use of confidence intervals is to support decisions in hypothesis testing, especially when the test is twotailed. The essence of this method is to compare the hypothesized value to the confidence interval. If the hypothesized value falls within the interval we fail to reject the null hypothesis. If the hypothesized value falls outside the interval we reject the null hypothesis. Let's look at a couple of examples.
For the twotailed test:
\(H_0: \mu = \mu_0\)
\(H_a: \mu \ne \mu_0\)
The null hypothesis will be rejected at level α if and only if the value \(\mu_0\) does not fall within the (1  \(\alpha\)) confidence interval for \(\mu\) .
Recall our lumber example from the lesson on confidence intervals to show how to use a confidence interval to draw a conclusion about a twotailed test. A 95% confidence interval for the mean lumber length was 8.03 feet to 8.57 feet.
For our twotailed test the hypotheses were:
\(H_0: \mu = 8.5\)
\(H_a: \mu \ne 8.5\)Since 8.5 falls within the 95% confidence interval, we cannot reject the null hypothesis at level 0.05. In general, if the null value falls within the confidence interval we fail to reject the null hypothesis. If the null value falls outside the confidence interval then we would reject the null hypothesis.
It is possible to use a onesided confidence bound to draw a conclusion about a onesided test, but you have to be very careful about obtaining the onesided confidence bound.
Chapter 10 – OneSample Tests of Hypothesis Chapter 10 OneSample Tests of Hypothesis 1. a. Twotailed, because the alternate hypothesis does not indicate a direction. b Reject H o when z does not fall in the region from – 1.96 and 1.96 c. 1.2, found by 49 50 (5/ 36) z d. Fail to reject H o e. p = 0.2302, found by 2(0.5000 – 0.3849). There is a 23.02% chance of finding a z value this large by “sampling error” when H o is true. 2. a. Onetailed, because the alternate hypothesis indicates a greater than direction. b. Reject H o when z > 2.05 c. 4, found by 12 10 (3/ 36) z d. Reject H o and conclude that >10 e. The pvalue is close to 0. So there is very little chance H o is true. 3. a. Onetailed, because the alternate hypothesis indicates a greater than direction. b. Reject H o when z > 1.65 c. 1.2, found by 21 20 (5/ 36) z d. Fail to reject H o at the 0.05 significance level. e. p = 0.1151, found by 0.5000 – 0.3849. There is an 11.51% chance of finding a z value this large or larger by “sampling error” when H o is true. 4. a. Onetailed, because the alternate hypothesis indicates a less than direction. b. Reject H o when z < 1.88 c. 2.67, found by 215 220 (15/ 64) z d. Reject H o and conclude that the population mean is less than 220 at the 0.03 significance level. e. p = 0.0038, found by 0.5000 – 0.4962. There is less than 0.5% chance H o is true. 5. a. H o : = 60,000 H 1 : 60,000 b. Reject H o if z < 1.96 or z > 1.96 c. 0.69, found by 59,500 60,000 (5000/ 48) z d. Do not reject H o e. p = 0.4902, found by 2(0.5000 – 0.2549). Crosset’s experience is not different from that claimed by the manufacturer. If the H o is true, the probability of finding a value more extreme than this is 0.4902. 101
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