Make sure to pay particular attention to the format of the question to ensure co

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Question

Make sure to pay particular attention to the format of the question to ensure correctness.

What is the main difference in using confidence intervals and hypothesis testing (not looking for the difference in formulas you use or buttons you press on your calculator)?

Using the same scenario as on the previous page, write a question you could answer using a one sample t-test. State your null and alternate hypothesis.

Using the same scenario as on the previous page, write a question you could answer using a one sample proportion test. State your null and alternate hypothesis.

Using the same scenario as on the previous page, write a question you could answer using a two sample t-test. State your null and alternate hypothesis.

Using the same scenario as on the previous page, write a question you could answer using a two sample proportion test. State your null and alternate hypothesis.

Using the same scenario as on the previous page, write a question you could answer using a paired difference t-test. State your null and alternate hypothesis.

Explanation / Answer

Confidence interval for the mean from a small sample

Suppose we observe a value ˆp from our data, and want to express how certain we are that ˆp is close to the true parameter p. We can think about how often the random quantity ˆp will end up within some distance of the fixed but unknown p. In particular, we can ask for an interval around ˆp for any sample so that in 95% of samples, the true mean p will lie inside this interval. Such an interval is called a confidence interval.

for example

A rare congenital disease, Everley's syndrome, generally causes a reduction in concentration of blood sodium. This is thought to provide a useful diagnostic sign as well as a clue to the efficacy of treatment. Little is known about the subject, but the director of a dermatological department in a London teaching hospital is known to be interested in the disease and has seen more cases than anyone else. Even so, he has seen only 18. The patients were all aged between 20 and 44.

The mean blood sodium concentration of these 18 cases was 115 mmol/l, with standard deviation of 12 mmol/l.

Hypothesis testing

Suppose we have a baseline value p and obtain from our data a value ˆp that’s smaller than p. If we’re interested in reasoning about whether ˆp is “significantly” smaller than p, one way to quantify this would be to assume the true value were p and then compute the probability of getting a value smaller than or as small as the one we observed (we can do the same thing for the case where ˆp is larger). If this probability is “very low”, we might think the hypothesized value p is incorrect. This is the hypothesis testing framework.

We begin with a null hypothesis, which we call H0 (in this example, this is the hypothesis that the true proportion is in fact p) and an alternative hypothesis, which we call H1 or Ha (in this example, the hypothesis that the true mean is significantly smaller than p).

For example,

in our polio example, a ratio of 1 corresponded to the hypothesis that the vaccine provides no benefit or harm compared to placebo.This specific value of interest is called the null hypothesis (“null” referring to the notion that nothing is different between the two groups – the observed differences are entirely due to random chance)

The goal of hypothesis testing is to weigh the evidence and deliver a number that quantifies whether or not the null hypothesis is plausible in light of the data

All hypothesis tests are based on calculating the probability of obtaining results as extreme or more extreme than the one observed in the sample, given that the null hypothesis is true.

This probability is denoted p and called the p-value of the test

The smaller the p-value is, the stronger the evidence against the null.

The t tests

Previously we have considered how to test the null hypothesis that there is no difference between the mean of a sample and the population mean, and no difference between the means of two samples. We obtained the difference between the means by subtraction, and then divided this difference by the standard error of the difference. If the difference is 196 times its standard error, or more, it is likely to occur by chance with a frequency of only 1 in 20, or less.

With small samples, where more chance variation must be allowed for, these ratios are not entirely accurate because the uncertainty in estimating the standard error has been ignored. Some modification of the procedure of dividing the difference by its standard error is needed, and the technique to use is the t test. Its foundations were laid by WS Gosset, writing under the pseudonym "Student" so that it is sometimes known as Student's t test. The procedure does not differ greatly from the one used for large samples, but is preferable when the number of observations is less than 60, and certainly when they amount to 30 or less.

The application of the t distribution to the following four types of problem will now be considered.

In each case the problem is essentially the same - namely, to establish multiples of standard errors to which probabilities can be attached. These multiples are the number of times a difference can be divided by its standard error. We have seen that with large samples 1.96 times the standard error has a probability of 5% or less, and 2.576 times the standard error a probability of 1% or less (Appendix table A ). With small samples these multiples are larger, and the smaller the sample the larger they become.

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of null and alternative hypotheses. Each makes a statement about the difference d between the mean of one population 1 and the mean of another population 2. (In the table, the symbol means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test, since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests, since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

When the null hypothesis states that there is no difference between the two population means (i.e., d = 0), the null and alternative hypothesis are often stated in the following form.

H0: 1 = 2
Ha: 1 2

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Analyze Sample Data

Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = sqrt[ (s12/n1) + (s22/n2) ]

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }

t = [ (x1 - x2) - d ] / SE

Problem 1: Two-Tailed Test

Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.

At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15.

Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(102/30) + (152/25] = sqrt(3.33 + 9) = sqrt(12.33) = 3.51

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (102/30 + 152/25)2 / { [ (102 / 30)2 / (29) ] + [ (152 / 25)2 / (24) ] }
DF = (3.33 + 9)2 / { [ (3.33)2 / (29) ] + [ (9)2 / (24) ] } = 152.03 / (0.382 + 3.375) = 152.03/3.757 = 40.47

t = [ (x1 - x2) - d ] / SE = [ (78 - 85) - 0 ] / 3.51 = -7/3.51 = -1.99

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) = 0.027, and P(t > 1.99) = 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.

Interpret results. Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.

Problem 2: One-Tailed Test

The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at least 7 minutes longer than the old battery.

To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes.

Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (Assume that there are no outliers in either sample.)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 >= 7
Alternative hypothesis: 1 - 2 < 7

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(402/100) + (202/100] = sqrt(16 + 4) = 4.472

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (402/100 + 202/100)2 / { [ (402 / 100)2 / (99) ] + [ (202 / 100)2 / (99) ] }
DF = (20)2 / { [ (16)2 / (99) ] + [ (2)2 / (99) ] } = 400 / (2.586 + 0.162) = 145.56

t = [ (x1 - x2) - d ] / SE = [(200 - 190) - 7] / 4.472 = 3/4.472 = 0.67

Here is the logic of the analysis: Given the alternative hypothesis (1 - 2 < 7), we want to know whether the observed difference in sample means is small enough (i.e., sufficiently less than 7) to cause us to reject the null hypothesis.

The observed difference in sample means (10) produced a t statistic of 0.67. We use the t Distribution Calculator to find P(t < 0.67) = 0.75.

This means we would expect to find an observed difference in sample means of 10 or less in 75% of our samples, if the true difference were actually 7. Therefore, the P-value in this analysis is 0.75.

Interpret results. Since the P-value (0.75) is greater than the significance level (0.05), we cannot reject the null hypothesis.

proportion tests

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Analyze Sample Data

Using sample data, find the test statistic and its associated P-Value.

= sqrt[ P * ( 1 - P ) / n ]

z = (p - P) /

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Problem 1: Two-Tailed Test

The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisified. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.80
Alternative hypothesis: P 0.80

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100] = sqrt(0.0016) = 0.04
z = (p - P) / = (.73 - .80)/0.04 = -1.75

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.75 or greater than 1.75.

We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04, and P(z > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.

Interpret results. Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the sample included at least 10 successes and 10 failures, and the population size was at least 10 times the sample size.

Problem 2: One-Tailed Test

Suppose the previous example is stated a little bit differently. Suppose the CEO claims that at least 80 percent of the company's 1,000,000 customers are very satisfied. Again, 100 customers are surveyed using simple random sampling. The result: 73 percent are very satisfied. Based on these results, should we accept or reject the CEO's hypothesis? Assume a significance level of 0.05.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P >= 0.80
Alternative hypothesis: P < 0.80

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100] = sqrt(0.0016) = 0.04
z = (p - P) / = (.73 - .80)/0.04 = -1.75

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -1.75. We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04. Thus, the P-value = 0.04.

Interpret results. Since the P-value (0.04) is less than the significance level (0.05), we cannot accept the null hypothesis.

paired difference t-tests

The hypotheses concern a new variable d, which is based on the difference between paired values from two data sets.

d = x1 - x2

where x1 is the value of variable x in the first data set, and x2 is the value of the variable from the second data set that is paired with x1.

The table below shows three sets of null and alternative hypotheses. Each makes a statement about how the true difference in population values d is related to some hypothesized value D. (In the table, the symbol means " not equal to ".)

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Analyze Sample Data

Using sample data, find the standard deviation, standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

sd = sqrt [ ((di - d)2 / (n - 1) ]

SE = sd * sqrt{ ( 1/n ) * [ (N - n) / ( N - 1 ) ] }

SE = sd / sqrt( n )

t = [ (x1 - x2) - D ] / SE = (d - D) / SE

Problem

Forty-four sixth graders were randomly selected from a school district. Then, they were divided into 22 matched pairs, each pair having equal IQ's. One member of each pair was randomly selected to receive special training. Then, all of the students were given an IQ test. Test results are summarized below.

(d - d)2 = 270
d = 1

Do these results provide evidence that the special training helped or hurt student performance? Use an 0.05 level of significance. Assume that the mean differences are approximately normally distributed.

Solution

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: d = 0
Alternative hypothesis: d 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ ((di - d)2 / (n - 1) ] = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586

SE = s / sqrt(n) = 3.586 / [ sqrt(22) ] = 3.586/4.69 = 0.765

DF = n - 1 = 22 -1 = 21

t = [ (x1 - x2) - D ] / SE = (d - D)/ SE = (1 - 0)/0.765 = 1.307

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 21 degrees of freedom is more extreme than 1.307; that is, less than -1.307 or greater than 1.307.

We use the t Distribution Calculator to find P(t < -1.307) = 0.103, and P(t > 1.307) = 0.103. Thus, the P-value = 0.103 + 0.103 = 0.206.

Interpret results. Since the P-value (0.206) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples consisted of paired data, and the mean differences were normally distributed. In addition, we used the approximation formula to compute the standard error, since the sample size was small relative to the population size.

Set Null hypothesis Alternative hypothesis Number of tails 1 1 - 2 = d 1 - 2 d 2 2 1 - 2> d 1 - 2 < d 1 3 1 - 2< d 1 - 2 > d 1

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