Section 8 - Breazeal
Section 10.3 – Hypothesis Tests for a Population Mean
Objectives
1. Test hypotheses about a mean
2. Understand the difference between statistical significance and practical significance
Notation:
n = sample size µ = population mean
[pic] = sample mean s = sample standard deviation
df = degrees of freedom = n - 1
Objective 1 – Test hypothesis about a mean
Why use the Student’s t-distribution?
We learned in Chapter 8 that [pic]follows a standard normal distribution. However, what if we don’t know (? The best option is to replace ( with s, the sample standard deviation. But when ( is replaced with s, then [pic]follows the Student’s t-distribution with n-1 degrees of freedom, not a normal distribution.
Review the properties of the Student’s t-distribution.
Verify before the hypothesis test
The requirements that must be fulfilled before we can do hypothesis testing on the sample mean with ( unknown are
1. The sample is obtained using simple random sampling or is the result of a randomized experiment
2. The sampled values are independent of each other
3. The sample has no outliers (check with a Boxplot) and at least one of the following hold
a. the population from which the sample is drawn is normally distributed, or
b. the sample size is large (n ( 30)
Steps for Hypothesis Testing for a Sample Mean using the P-value Approach
1. State the null and alternative hypothesis mathematically
2. State the level of significance, ( based on the seriousness of making a Type I error
3. Calculate the test statistic that follows the Student’s t-distribution with n-1 degrees of freedom (steps 3 and 4 can be done at the same time on the calculator). Draw a graph.
[pic]
4. Compute the P-value, which is an area under the normal curve related to the test statistic. Use either Table VI with n-1 degrees of freedom or a calculator. Calculator is preferred.
5. Compare P-value to (
If P–value < α, Reject H0
If P–value > α, Do Not Reject H0
6. State conclusion
Note, test statistic and P-value from steps 3 and 4 above can be found in one step on the calculator.
The following graphic shows the meaning of the P-value.
[pic]
Calculator Instructions
[pic]
Example
Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05. At the ( = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?
Example
According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data:
5.70 5.67 5.73 5.61 5.70 5.67
5.65 5.62 5.73 5.65 5.79 5.73
5.77 5.71 5.70 5.76 5.73 5.72
At the ( = 0.05 level of significance, is there evidence to conclude that the state quarters have a weight different from 5.67 grams?
Example
Consumer Reports indicated that the mean acceleration time (0–60 mph) for the Dodge
Intrepid was 10.2 seconds. In most tests of this type, regular unleaded gasoline is used. Suppose that 41 such tests were made using premium unleaded gasoline, and the sample mean acceleration time was 9.7 seconds with a standard deviation of 2.1 seconds. Does this indicate that premium gasoline tends to reduce average acceleration time? Use α = 0.05.
Example
The mean waiting time at the drive-through of a fast food restaurant from the time an order is placed to the time the order is received is 84.9 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at the restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided below
101.2 66.4 57.3 76.6 64.8
79.7 94.5 86.3 71.8 81.4
At the ( = 0.01 level of significance is the new system effective?
Objective 2 - Understand the difference between statistical significance and practical significance
When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance
[pic]
Beware of studies with large sample sizes that claim statistical significance because the differences may not have any practical meaning.
[pic]
Example
According to the American Community Survey, the mean travel time to work in Collin County, TX, in 2008 was 27.3 minutes. The Department of Transportation reprogrammed all traffic lights in Collin County in an attempt to reduce travel time. To determine if there is evidence that travel time has decreased as a result of the reprogramming, the Department of Transportation obtains a random sample of 2500 commuters, records their travel time to work, and finds a sample mean of 27.0 minutes with a standard deviation of 8.5 minutes. At the ( = 0.05 level of significance, does the sample suggest that travel time has decreased?
The following calculator instructions are useful if you perform all steps manually and not use the calculator’s T-Test.
You can duplicate the critical values from the t-chart table (Table VI, pg A-13) by using the instructions below. The calculator instructions in blue below are not in the textbook!
[pic]
[pic]
-----------------------
TI-83/84 Plus
1. If necessary, enter raw data into L1
2. Press STAT, highlight TESTS, and select 2:T-Test
3. Choose DATA or STATS
a. If the data are raw, highlight DATA; make sure list is set to L1 and Freq is set to 1
b. If summary statistics are known, highlight STATS and enter summary statistics. For the value of (0, enter the value of the mean stated in the null hypothesis
4. Select the direction of the alternative hypothesis
5. Highlight Calculate or Draw and press ENTER. The TI-83/84 gives the P-value
TI-84 Procedure
The TI-84 offers direct access to inverse t — that’s one of its few differences from the TI-83. The format is invT(ltail,df). Suppose you want to find critical t for 27 degrees of freedom and a 0.025 one-tailed test. Here s the procedure:
1. Press [2nd VARS makes DISTR] [4] for invT(.
2. Enter the area of the left-hand tail. Since 0.025 is al t for 27 degrees of freedom and a 0.025 one-tailed test. Here’s the procedure:
Press [2nd VARS makes DISTR] [4] for invT(.
Enter the area of the left-hand tail. Since 0.025 is the right-hand tail, 1−0.025 is the left-hand tail.
3. Press the comma and enter the degrees of freedom, [pic]
4. Press [)] [ENTER].
Answer: t(27,0.025) = invT(1−.025,27) = 2.0518
TI-83
The TI-83 doesn’t have a specific function for t(df, α/2) but you can get at it by a “back door”. Here’s how you would find critical t for 27 degrees of freedom in a 0.025 one-tailed test:
1. Select TInterval using the keystrokes you already know, [STAT] [◄] [8].
2. Pick “Stats”.
[pic]
3. Set [pic]=0.
4. For Sx, you want the square root of the sample size. Press [2nd x² makes √] and then enter the sample size (which is df+1).
5. For n, again enter the sample size df+1.
6. For C-Level, enter the area in the middle (the non-rejection region). You can enter it as a calculation, 1−2( (right-hand tail). For instance, in computing t(27, 0.025), you can enter 1−2×0.025 or just enter .95 directly.
7. Select CALCULATE and read the upper bound of the interval. That is the inverse t you’re looking for. In this case the answer is 2.0518.
Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important.
Large sample sizes can lead to results that are statistically significant, while the difference between the statistic and parameter in the null hypothesis is not enough to be considered practically significant.
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