C HAPTER 4  Hypothesis Testing -Test for one and two means -Test for one and two proportions

WHY WE HAVE TO DO THE HYPOTHESIS? To make decisions about populations based on the sample information. Example :- we wish to know whether a medicine is really effective to cure a disease. So we use a sample of patients and take their data in effect of the medicine and make decisions. To reach the decisions, it is useful to make assumptions about the populations. Such assumptions maybe true or not and called the statistical hypothesis.

Definitions Hypothesis Test: It is a process of using sample data and statistical procedures to decide whether to reject or not to reject the hypothesis (statement) about a population parameter value (or about its distribution characteristics). Null Hypothesis, : Generally this is a statement that a population has a specific value. The null hypothesis is initially assumed to be true. Therefore, it is the hypothesis to be tested.

Alternative Hypothesis, : It is a statement about the same population parameter that is used in the null hypothesis and generally this is a statement that specifies that the population parameter has a value different in some way, from the value given in the null hypothesis. The rejection of the null hypothesis will imply the acceptance of this alternative hypothesis. Test Statistic: It is a function of the sample data on which the decision is to be based.

Critical/ Rejection region: It is a set of values of the test statistics for which the null hypothesis will be rejected. Critical point: It is the first (or boundary) value in the critical region. P -value: The probability calculated using the test statistic. The smaller the p -value is, the more contradictory is the data to.

Procedure for hypothesis testing 1. Define the question to be tested and formulate a hypothesis for a stating the problem. 2. Choose the appropriate test statistic and calculate the sample statistic value. The choice of test statistics is dependent upon the probability distribution of the random variable involved in the hypothesis. 3. Establish the test criterion by determining the critical value and critical region. 4. Draw conclusions, whether to accept or to reject the null hypothesis.

Example 4.7: A sample of 50 Internet shoppers were asked how much they spent per year on Internet. From this sample, mean expenses per year on Internet is and sample standard deviation is It is desired to test whether they spend in mean expenses is RM32500 per year or not. Test at.

Solution: The hypothesis tested are: Test Statistic: Critical Value : As two tailed (=), so alpha has to divide by two,becomes : Rejection Region: Conclusion: The Internet Shoppers spend RM32500 per year on the Internet.

Example 4.8: A random sample of 10 individuals who listen to radio was selected and the hours per week that each listens to radio was determined. The data are follows: Test a hypothesis if mean hours individuals listen to radio is less than 8 hours at.

Solutions: The hypothesis tested are: Test Statistic: n < 30 Critical Value: Rejection Region: Conclusion : Mean hours individuals listen to radio is greater or equal to 8 hours.

Exercise 4.1: 1. From the following data, test the null hypothesis that population mean is less than or equal to 100 at 5% significance level. (Ans: t = , Reject) A paint manufacturing company claims that the mean drying time for its paint is at most 45 minutes. A random sample of 35 trials tested. It is found that the sample mean drying time is minutes with standard deviation 3 minutes. Assume that the drying times follow a normal distribution. At 1% significance level, is there any sufficient evidence to support the company claim? (Ans: z = , Reject)

Hypothesis testing for the differences between two population mean, Test hypothesis Test statistics i) Variance and are known, and both and are samples of any sizes. ii) If the population variances, and are unknown, then the following tables shows the different formulas that may be used depending on the sample sizes and the assumption on the population variances.

Equality of variances, when are unknown Sample size

Example 4.9: The mean lifetime of 30 bulbs produced by company A is 50 hours and the mean lifetime of 35 bulbs produced by company B is 48 hours. If the standard deviation of all bulbs produced by company A is 3 hour and the standard deviation of all bulbs produced by company B is 3.5 hours, test at 1 % significance level that the mean lifetime of bulbs produced by Company A is better than that of company B. ( Variances are known)

Solutions: The hypothesis tested are: Critical Value: Rejection Region: Conclusion : The mean lifetime of bulbs produced by company A is better than that of company B.

Example 4.10: A mathematic placement test was given to two sample of classes of 45 and 55 student respectively. In the first class the sample mean grade was 75 with a standard deviation of 8, while in the second class the sample mean grade was 80 with a standard deviation of 7. Is there a significant difference between the mean performances of the two classes at 5% level of significance? Assume the population variances are equal.

Solutions : Hypothesis tested are: Test Statistic: Critical Value : Rejection Region : Conclusion : There is a significant difference between the performance of the two classes.

Exercise 4.2: A sample of 60 maids from country A earn an average of RM300 per week with a standard deviation of RM16, while a sample of 60 maids from country B earn an average of RM250 per week with a standard deviation of RM18. Test at 5% significance level that country A maids average earning exceed country B maids average earning more than RM40 per week. (Assume Variances are not equal) Answer : Z = , Reject

Example 4.11: When working properly, a machine that is used to make chips for calculators produce 4% defective chips. Whenever the machine produces more than 4% defective chips it needs an adjustment. To check if the machine is working properly, the quality control department at the company often takes sample of chips and inspects them to determine if they are good or defective. One such random sample of 200 chips taken recently from the production line contained 14 defective chips. Test at the 5% significance level whether or not the machine needs an adjustment.

Solutions: The hypothesis tested are: Test Statistic: Critical Value: Rejection Region : Conclusion : Machine needs adjustment

Exercise 4.3: A manufacturer of a detergent claimed that his detergent is least 95% effective is removing though stains. In a sample of 300 people who had used the Detergent and 279 people claimed that they were satisfied with the result. Determine whether the manufacturer’s claim is true at 1% significance level. Answer: Do not Reject

Example 4.12: A researcher wanted to estimate the difference between the percentages of two toothpaste users who will never switch to other toothpaste. In a sample of 500 users of toothpaste A taken by the researcher, 100 said that they will never switched to another toothpaste. In another sample of 400 users of toothpaste B taken by the same researcher, 68 said that they will never switched to other toothpaste. At the significance level 1%, can we conclude that the proportion of users of toothpaste A who will never switch to other toothpaste is higher than the proportion of users of toothpaste B who will never switch to other toothpaste?

Solutions: The hypothesis tested are: Test Statistic: Critical Value : Rejection Region : Conclusion : The proportion of user toothpaste A is less than and equal to Claim is not true.

Exercise 4.4: In a process to reduce the number of death due the dengue fever, two district, district A and district B each consists of 150 people who have developed symptoms of the fever were taken as samples. The people in district A is given a new medication in addition to the usual ones but the people in district B is given only the usual medication. It was found that, from district A and from district B, 120 and 90 people respectively recover from the fever. Test the hypothesis that the new medication better to cure the fever than the using the usual ones only using a level of significance of 5%. Answer: reject

EXERCISES

Exercise A paint manufacturing company claims that the mean drying time for its paint is at most 45 minutes. A random sample of 20 trials tested. It is found that the sample mean drying time is minutes with standard deviation 3 minutes. Assume that the drying times follow a normal distribution. (a) Construct a 99% confidence interval for the mean drying time of the paint. (b) At 5% significance level, is there any sufficient evidence to support the company claim? (c) Suppose that another manufacturing company wants to estimate the mean drying time for its paints at 95% confidence level. Given, what is the sample size of trials required in order to obtain an estimate that is within maximum error of 3 minutes?

2.A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no more than 4% of the electronic circuit boards fall outside the most rigid level of industry performance specifications. In a simple random sample of 300 electronic circuit boards from the shipment, 15 fall outside these specifications. (a) Construct the 95% confidence interval for the percentage of all boards in this shipment that fall outside the specifications. (b) Test whether the supplier’s claim would appear to be correct at 10% significance level.

3.Babies To Go Inc. is the manufacturer of car seats for children. They manufacture two distinct types of car seats. They want to test the car seats for durability. The results of the test are given in the following table. The mean and standard deviation of durability is in months. (Assuming variances are not equal) a) Construct a 95 % confidence interval for the mean difference in car seats durability. b) Can we conclude that the difference between the mean in car seats durability is more than 3.4 ( ). Type 1Type 2 Mean Life Sample Size1510 Standard Deviation1.20.8