Presentation on theme: "1 One Tailed Tests Here we study the hypothesis test for the mean of a population when the alternative hypothesis is an inequality."— Presentation transcript:
1 One Tailed Tests Here we study the hypothesis test for the mean of a population when the alternative hypothesis is an inequality.
2 Let’s follow an example and then highlight some main points about hypothesis testing. At its drive-up windows McDonald’s has had a mean service time of 163.9 seconds. McDonald’s would like to improve on this. The null hypothesis in this case would be that the mean of the population of all service times would be greater than or equal to 163.9. We write the null as Ho: μ ≥ 163.9 The alternative, then, would be that the mean of the population of all service times are less than 163.9. We write H1: μ < 163.9. In this case the alternative hypothesis contains the improvement the company is attempting to make. The null hypothesis is a maintaining of the status quo. In a hypothesis test we will either 1) fail to reject the null hypothesis or 2) reject the null and go with the alternative.
3 Sample means service time 163.9 Here I show the distribution of sample means of service times. The center is at 163.9. Note some samples would have means less than 163.9 even when in reality the population has mean of 163.9.
4 Remember a Type I Error would have us reject the null when it is true. In this case a Type I Error would suggest the company is improving service time when it is fact not doing so. We want to be careful about this, so, we will only reject the null when the probability of an observed event is really low. What do I mean by the phrase, “probability of an observed event?” In a study like this we do not look at every drive-up case. This would be expensive. So a sample is taken. From the sample we calculate a statistic and figure a probability of this value or a more extreme value. This is the event to which I refer. On the next slide I show the distribution of sample means, assuming the mean is exactly 163.9 seconds. To have the standard error of this distribution we need to know the sample size and have either the standard deviation in the population or the sample standard deviation as an estimate of it. Here we say we do not know the population standard deviation, so we use the sample value and know we have to use the t distribution.
5 Sample means in seconds 163.9 In terms of a t = 0 Critical value Level of significance Reject null Accept null
6 Now, in the graph on the previous slide I put the mean of the distribution at 163.9. You will notice on the left side of the distribution I put another vertical line and wrote next to it the phrase level of significance. I also labeled the value on the horizontal axis as “critical value.” This is the lowest level the sample mean could be and still have the claim true. Here is the basic logic of the hypothesis test. Say in our one sample we get a value farther away from the center than the critical value. Although it could happen, the probability is small. This will lead us to reject the null.
7 Now, let’s refer back to the graph on slide 5. The area I called the level of significance is the “low” probability I referred to before. If our sample value falls in this area past the critical value we will reject the null. But we pick the level of significance to be low, like.05 or.01, and thus we make our chance of making a Type I Error low. Next I want to develop some equivalent ways of conducting the hypothesis test. They are all similar and related.
8 t distribution In terms of a t = 0 Critical value of t = Level of significance = Reject null Accept null Critical t method
9 Say that alpha, or the level of significance we want is.05 and also say we will use a sample of 25 stores to observe service time. Thus with df = 25 – 1 = 24 and all.05 in the lower tail we have a critical value of -1.7109. We put all of the alpha =.05 in the lower tail because are alternative is an inequality. Will you please put the alpha and critical t values in the graph on the previous slide? Up until this point I have not used a real sample mean value. I have talked in the hypothetical. The hypothetical allowed me to think of critical values. (In practical terms you would probably have a sample mean from data at this time, but we have not used it yet.)
10 To complete the test we have to take the sample mean and actually calculate the t value for it. This will be called the test statistic tstat. Say in our sample we get an X bar = 152.7 and a sample standard deviation of 20 seconds. Thus, tstat = (sample mean minus hypothesis value)/(sample standard dev/sqrt(n)) = (152.7 – 163.9)/(20/sqrt(25)) = -2.80. Since the test statistic is farther from a t of 0 than the critical value is we reject the null and go with the alternative in this example. McDonald’s has done something to speed up its service time (reduce the time to serve customers).
11 t distribution 0 Critical value of t = -1.7109 Level of significance =.05 Reject null Accept null p value method
12 p-value What is a p-value? Let’s develop a story. When we said the level of significance was picked to be.05 we could find the critical value of -1.7109 in terms of a t value. Then we had a sample mean of 152.7 and a corresponding tstat of minus 2.80. On the number line this tstat is farther from the center than the critical value. And for this reason we rejected the null. Now, if you go to the graph on the previous slide, and if you put your right pinky, palm up on the critical value, the area to the left of your pinky was chosen to be.05.
13 Now, move your pinky to the left to the value of the sample mean, or its corresponding t. The area under the curve to the left of your pinky is less than.05. This actual area is called the p- value. It is just the level of significance of the sample statistic we obtained. An equivalent way to test an hypothesis then is to reject the null if the p value is less than the level of significance established for the test. In the t table we see in row df = 24 the value 2.7969 under.005. With a symmetric distribution -2.7969 has a lower tail area of.005. Our sample value is even less than that so the p-value is less than.005. We reject the null here.
14 Remember, all this is a one tailed test on the low side of the mean. What if we had a one tailed test on the high side of the mean? Same logic. Generic example: Sample size is 50 so df = 49. Alpha is picked to be.01. The critical t is found in the table as 2.4049. 2.4049 t Do not reject null Reject null Say in a sample the sample mean is 6.034 and sample standard deviation is.02. SO our tstat = (6.034 – 6.03)/(.02/sqrt(50)) = 1.41 and since this is not as big as 2.4049 we do not reject the null. The p-value is somewhere between.05 and.10 because 1.41 is between 1.6766 and 1.2991.
15 Let’s summarize some things. When an alternative hypothesis H1 has a not equal sign we have a two tailed test. We take alpha and divide by 2 to get critical values. Reject the null if our tstat is more extreme than the critical values. If we go the p-value approach we have to take our tail area and multiply by 2. Reject the null if the p- value for the tstat is less than or equal to alpha. When an alternative hypothesis H1 has an inequality (either a less than or a greater than sign) sign we have a one tailed test. Alpha is all in one tail to get the critical value. Reject the null if our tstat is more extreme than the critical value. If we go the p-value approach the tail area with the tstat is the p-valule. Reject the null if the p-value for the tstat is less than or equal to alpha.
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