ExamplE 1 - WPMU DEV

ExamplE 1

To test H0: ? = 100 versus H1: ? 100, is a simple random sample size of n = 20 is obtained from a population that is known to be normally distributed. If = 104.7 and s = 8., compute the test statistic.

t = 2.416 (rounded to three places)

If the researcher decides to test this hypothesis at the = 0.01 level of significance, determine the critical values. = 0.01 So, /2 = 0.005 Since n = 20, we have n ? 1 = 19 degrees of freedom

The critical values are -2.861 and 2.861 (rounded to three places) The area to the left is 0.995 because 1 ? 0.005 = 0.995. However, they will always be opposites, so after finding -2.861 you know the other one will be +2.861. Choose the t-distribution that depicts the critical region(s).

The critical regions are in both tails because this is a "not equal to" test. It could be either more or less than the critical values. Will the researcher reject the null hypothesis? There is NOT sufficient evidence for the researcher to reject the null hypothesis since the test statistic is between the critical values. (2.416 is between -2.861 and 2.861 rather than being less than -2.861 or greater than 2.861.)

ExamplE 2

Several years ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older results in a mean height of 64.3 inches.

State the null and alternative hypotheses. To test H0: ? = 63.7 in. versus H1: ? > 63.7 in. (We use `greater than' because we are testing to see if women are taller ? therefore their height would be greater.)

Suppose the P-value for this test is 0.17. Explain what the value represents. This means there is a 0.17 probability of obtaining a sample mean height of 64.3 inches or more (taller) from a population whose mean height is 63.7 inches.

Write a conclusion for this hypothesis test assuming an = 0.05 level of significance. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today. (Because the P-value is GREATER than .)

ExamplE 3

Researchers wanted to measure the effect of alcohol on the development of the hippocampal region in adolescents. The researchers randomly selected 13 adolescents with alcohol use disorders. They wanted to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with = 8.40 and s = 0.9. Conduct an appropriate test at the = 0.01 level of significance. First we must state our hypotheses. H0: ? = 9.02 cm3 versus H1: ? < 9.02 cm3 (Because the researchers are testing to see if the hippocampal volumes are less in these adolescents with the alcohol use disorders.)

t = -2.484 (rounded to three places)

Since we are using the = 0.01 level of significance, determine the critical value(s). = 0.01 So, /2 = 0.005 Since n = 13, we have n ? 1 = 12 degrees of freedom

The critical value is -3.055 (rounded to three places) The area to the left is 0.005 because we only want the "less than" or critical value on the left. Will the researcher reject the null hypothesis? There is NOT sufficient evidence for the researcher to reject the null hypothesis since the test statistic is greater than the critical value, not less than it. (-2.484 is greater than -3.055.) So we do not reject H0 since the test statistic is not less than the critical value.

ExamplE 4

Several years ago, the reported mean age of an inmate in death row was 44.4 years. A sociologist wondered whether the mean age of a death row inmate has changed since then. She randomly selects 35 death-row inmates and finds that their mean age is 46.2, with a standard deviation of 8.1. Construct a 90% confidence interval about the mean age of the death row inmates. What does the interval imply? H0: ? = 44.4 versus H1: ? 44.4 (We use ? because it's the "mean age" and because it said the researcher wants to know if it has "changed" not whether it is greater or less than.)

Confidence interval: (43.89, 48.52) (rounded to two places)

The interval implies that since the given mean age is IN the interval (44.4 is in the interval 43.89 to 48.52) we do not reject the null hypothesis. In other words there is not sufficient evidence that we have a mean greater than or less than 44.4 because it's inside the interval, rather than being less than 43.89 or greater than 48.52. So we can't say with 90% confidence that the mean age is different than 44.4 years old.

ExamplE 5

We have all been told for years that the mean temperature of humans is 98.6? F. However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6? F. They measured the temperatures of 148 healthy adults 1 to 4 times daily for 3 days, obtaining 600 measurements. The sample data resulted in a sample mean of 98.2? F and a sample standard deviation of 0.8? F.

Use the classical approach to judge whether the mean temperature of humans is less than 98.6? F at the = 0.01 level of significance.

H0: ? = 98.6 versus H1: ? < 98.6 (We use ? because it's the "mean temperature" and < because it said the researchers want to know if the average temp is less than 98.6.)

t = -2.33

The test statistic is t0 = -12.247 599

-12.247 < -2.33 Therefore we reject H0. Approximate the P-value.

We will write the P-value as 0.0000 since it says to round to four places.

ExamplE 6

We are testing to see if the mean (mu) is equal or not equal to 0.15. Because we have a small sample size (n = 10) we are using a t-test and will have the calculator give us a "TInterval". Enter the data into L1 in your calculator.

Then go to

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