Hypothesis Tests One Sample Means

What are hypothesis tests? Calculations that tell us if a value, x, occurs by random chance or not – if it is statistically significant Is it . . . a random occurrence due to natural variation? a biased occurrence due to some other reason? Is it one of the sample means that are likely to occur? Is it one that isn’t likely to occur? Statistically significant means that it is NOT a random chance occurrence!

Steps: Identify Hypothesis statements, parameters & type of test Notice the steps are the same except we add hypothesis statements – which you will learn today Steps: Identify Hypothesis statements, parameters & type of test Conditions Calculations Conclusion, in context

Assumptions for z-test (t-test): YEA – These are the same assumptions as confidence intervals!! Have an SRS of context Sample size < 10% of population Distribution is (approximately) normal Given Large sample size Graph data

Writing Hypothesis statements: Null hypothesis – is the statement being tested; this is a statement of “no effect” or “no difference” Alternative hypothesis – is the statement that we suspect is true H0: Ha:

The form: Null hypothesis H0: parameter = hypothesized value Alternative hypothesis Ha: parameter > hypothesized value Ha: parameter < hypothesized value Ha: parameter = hypothesized value

In other words . . . is it far out in the tails of the distribution? P-values - Assuming H0 is true, the probability that the test statistic would have a value as extreme or more than what is actually observed In other words . . . is it far out in the tails of the distribution?

Statistically significant – The p-value is as small or smaller than the level of significance (a) If p > a, “fail to reject” the null hypothesis at the a level. If p < a, “reject” the null hypothesis at the a level.

Facts about p-values: ALWAYS make decision about the null hypothesis! Large p-values show support for the null hypothesis, but never that it is true! Small p-values show support that the null is not true. Double the p-value for two-tail (=) tests Never accept the null hypothesis!

Calculating p-values For z-test statistic – For t-test statistic – Use normalcdf(lb,ub) [using standard normal curve] For t-test statistic – Use tcdf(lb, ub, df)

Draw & shade a curve & calculate the p-value: 1) right-tail test t = 1.6; n = 20 2) left-tail test z = -2.4; n = 15 3) two-tail test t = 2.3; n = 25 P-value = .0630 P-value = .0082 P-value = (.0152)2 = .0304

Writing Conclusions: A statement of the decision being made (reject or fail to reject H0) & why (linkage) A statement of the results in context. (state in terms of Ha) AND

Be sure to write Ha in context (words)! “Since the p-value < (>) a, I reject (fail to reject) the H0. There is (is not) sufficient evidence to suggest that Ha.” Be sure to write Ha in context (words)!

Formulas: s known: m z =

Formulas: s unknown: m t =

Example 7: The Fritzi Cheese Company buys milk from several suppliers as the essential raw material for its cheese. Fritzi suspects that some producers are adding water to their milk to increase their profits. Excess water can be detected by determining the freezing point of milk. The freezing temperature of natural milk varies normally, with a mean of -0.545 degrees and a standard deviation of 0.008. Added water raises the freezing temperature toward 0 degrees, the freezing point of water (in Celsius). The laboratory manager measures the freezing temperature of five randomly selected lots of milk from one producer with a mean of -0.538 degrees. Is there sufficient evidence to suggest that this producer is adding water to his milk?

What are your hypothesis statements? Is there a key word? SRS? Assumptions: Normal? How do you know? I have an SRS of milk from one producer The freezing temperature of milk is a normal distribution. (given) Do you know s? s is known What are your hypothesis statements? Is there a key word? H0: m = -0.545 Ha: m > -0.545 where m is the true mean freezing temperature of milk Plug values into formula. p-value = normalcdf(1.9566,1E99)=.0252 Use normalcdf to calculate p-value. a = .05

Compare your p-value to a & make decision Conclusion: Since p-value < a, I reject the null hypothesis. There is sufficient evidence to suggest that the true mean freezing temperature is greater than -0.545. This suggests that the producer is adding water to the milk. Write conclusion in context in terms of Ha.

Example 8: The Degree of Reading Power (DRP) is a test of the reading ability of children. Here are DRP scores for a random sample of 44 third-grade students in a suburban district: (data on note page) At the a = .1, is there sufficient evidence to suggest that this district’s third graders reading ability is different than the national mean of 34?

I have an SRS of third-graders Normal? How do you know? Since the sample size is large, the sampling distribution is approximately normally distributed OR Since the histogram is unimodal with no outliers, the sampling distribution is approximately normally distributed Do you know s? What are your hypothesis statements? Is there a key word? s is unknown H0: m = 34 where m is the true mean reading Ha: m = 34 ability of the district’s third-graders Plug values into formula. p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212 Use tcdf to calculate p-value. a = .1

Compare your p-value to a & make decision Conclusion: Since p-value > a, I fail to reject the null hypothesis. There is not sufficient evidence to suggest that the true mean reading ability of the district’s third-graders is different than the national mean of 34. Write conclusion in context in terms of Ha.

Example 9: The Wall Street Journal (January 27, 1994) reported that based on sales in a chain of Midwestern grocery stores, President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Does this indicate that the sales of the cookies is different from the earlier figure?

Assume: Have an SRS of weeks Distribution of sales is approximately normal due to large sample size s unknown H0: m = 1323 where m is the true mean cookie sales Ha: m ≠ 1323 per week Since p-value < a of 0.05, I reject the null hypothesis. There is sufficient to suggest that the sales of cookies are different from the earlier figure.

Example 9: President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Compute a 95% confidence interval for the mean weekly sales rate. CI = ($1105.30, $1310.70) Based on this interval, is the mean weekly sales rate statistically different from the reported $1323?

Why are we getting different answers? What do you notice about the decision from the confidence interval & the hypothesis test? Remember your, p-value = .01475 At a = .02, we would reject H0. What decision would you make on Example 10 if a = .01? What confidence level would be correct to use? Does that confidence interval provide the same decision? If Ha: m < 1323, what decision would the hypothesis test give at a = .02? Now, what confidence level is appropriate for this alternative hypothesis? A 96% CI = ($1100, $1316). Since $1323 is not in the interval, we would reject H0. You would fail to reject H0 since the p-value > a. In a one-sided test, all of a (2%) goes into that tail (lower tail). You should use a 99% confidence level for a two-sided hypothesis test at a = .01. The 98% CI = ($1084.40, $1331.60) - Since $1323 is in the interval, we would fail to reject H0. Why are we getting different answers? Tail probabilities between the significant level (a) and the confidence level MUST match!) In a CI, the tails have equal area – so there should also be 2% in the upper tail CI = ($1068.6 , $1346.40) - Since $1323 is in this interval we would fail to reject H0. a = .02 .96 .02 That leaves 96% in the middle & that should be your confidence level

A special type of t-inference Matched Pairs Test A special type of t-inference

Matched Pairs – two forms Pair individuals by certain characteristics Randomly select treatment for individual A Individual B is assigned to other treatment Assignment of B is dependent on assignment of A Individual persons or items receive both treatments Order of treatments are randomly assigned or before & after measurements are taken The two measures are dependent on the individual

Is this an example of matched pairs? 1)A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employment No, there is no pairing of individuals, you have two independent samples

Is this an example of matched pairs? 2) In a taste test, a researcher asks people in a random sample to taste a certain brand of spring water and rate it. Another random sample of people is asked to taste a different brand of water and rate it. The researcher wants to compare these samples No, there is no pairing of individuals, you have two independent samples – If you would have the same people taste both brands in random order, then it would bean example of matched pairs.

Is this an example of matched pairs? 3) A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again. Yes, you have two measurements that are dependent on each individual.

A whale-watching company noticed that many customers wanted to know whether it was better to book an excursion in the morning or the afternoon. To test this question, the company collected the following data on 15 randomly selected days over the past month. (Note: days were not consecutive.) You may subtract either way – just be careful when writing Ha Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Morning After-noon Since you have two values for each day, they are dependent on the day – making this data matched pairs First, you must find the differences for each day.

-1 -2 I subtracted: Morning – afternoon Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Morning After-noon Differences -1 -2 I subtracted: Morning – afternoon You could subtract the other way! Assumptions: Have an SRS of days for whale-watching s unknown Since the normal probability plot is approximately linear, the distribution of difference is approximately normal. You need to state assumptions using the differences! Notice the granularity in this plot, it is still displays a nice linear relationship!

Differences -1 -2 1 2 Is there sufficient evidence that more whales are sighted in the afternoon? Be careful writing your Ha! Think about how you subtracted: M-A If afternoon is more should the differences be + or -? Don’t look at numbers!!!! If you subtract afternoon – morning; then Ha: mD>0 H0: mD = 0 Ha: mD < 0 Where mD is the true mean difference in whale sightings from morning minus afternoon Notice we used mD for differences & it equals 0 since the null should be that there is NO difference.

finishing the hypothesis test: Differences -1 -2 1 2 finishing the hypothesis test: Since p-value > a, I fail to reject H0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning. In your calculator, perform a t-test using the differences (L3) Notice that if you subtracted A-M, then your test statistic t = + .945, but p-value would be the same