If the rejection zone is on the right-hand side above, the test is known as a right trailing test. One finding is that the average maintenance cost is not more than Rs. 2) The quality control department of the processing company determines that the average net weight per package of its products must be 20 g. Is this sufficient evidence that the actual average weight of packages has decreased (use 5% significance levels).
The Z test used earlier is based on the assumption that the sampling distribution of the mean is a normal distribution. In light of this data, there is a suggestion that the average height in the population is 66. Can the average weight of college students be considered to be 120 pounds.
Test at 5% level of significance that children from the locality are on average less than 65 cm. The average weekly sales of the chocolate bar in a candy store was 146.3 bars per store. Test whether the weights of the bags are below the average expected value of 3,500 g or not.
Is it reasonable to believe that the average weight of the machine part is greater than 640 gm.
TEST FOR EQUALITY OF TWO MEANS β PAIRED t TEST
This test is used when, for the same sample, values ββare given that indicate an increase or decrease in the variable. Before and after values ββof the variable are usually given, or the differences can be specified directly.
TEST FOR EQUALITY OF PROPORTION
Solution: The null hypothesis is that the proportions of defects before and after overhaul are equal.
CHI SQUARE DISTRIBUTION (X 2 DISTRIBUTION)
Can we say that the variance of the distribution of the weights of all the students from whom the above sample was drawn is equal to 20 square kg. Specifications for the manufacture of the type of jewelry parameters state that the variance in weight should not exceed 0.015 mgm square. The conclusion is that the data are consistent with the hypothesis of an unbiased death.
Eg.(2) A Personal Manager is interested in trying to determine whether absence is greater on one day of the week than on another. Solution: We set up H0: Number of absences is uniformly distributed over the week versus H1: Number of absences is not uniformly distributed over the week. Since calculated value of x2 = 7.5 is less than its critical value, the null hypothesis is accepted.
Frequencies related to two attributes can be tested to find out whether a relationship exists between the attributes or not for the two-way table between the attributes, the expected frequencies are calculated for each cell of the observed frequencies. Rejection rule: If the calculated value of Chi-Square is greater than or equal to the tabulated value of (r-1)(c-1) d.o.f. Note: for 2Γ2 tables, the Chi-square formula for testing independence of attributes is simplified and can be used directly as
Rejection rule: The null hypothesis is rejected when the Chi-square value equals or exceeds the tabulated value for the d.o.f. To get a better result of x2 it is necessary to make a correction to the above formula and corrected formula is given by.
TESTS OF HYPOTHESIS ABOUT THE VARIANCE OF TWO POPULATIONS While comparing the variance of two populations, data are collected from two independent
The critical value m on the left side of the F distribution is found in the table of critical values ββon the right side of the F distribution.
ANALYSIS OF VARIANCE-A TEST FOR HOMOGENEITY OF MEAN,
The subscript / indicates the row or observation of the sample, while the subscript / indicates the column or population from which the observation came. Its use is there in the case of inference, when two or more samples are involved and the test is made on the basis of variances. The variances of random samples drawn from a normal universe have a distribution that is positively skewed; therefore, we need to have a separate distribution called the F-distribution to test for significance based on variances.
Its application is normally in the case of data in the form of a table with specified rows and columns. If there is only one attribute, then it is called a one-way classification and has as many columns as there are sample sets of a given size. If, however, there is more than one characteristic, then there is a table with a large number of rows and columns.
The various sums of squares involved in this are as follows: First consider the data classified in one way. If the obtained value is more than the table value, the difference is significant and the null hypothesis is rejected. If the obtained value is less than the table value, the difference is not significant and the null hypothesis is accepted.
ANALYSIS OF VARIANCE IN MANIFOLD CLASSIFICATION
He controls his manufacturing process so that the proportion of defective products is supposed to be only 5%. In the light of the above data, discuss the proposition that the average weight in the universe is 65 g. Set up ANOVA table for the following yield per hectare for three varieties of wheat, each grown on four plots, and perform the test for the varieties.
The following table shows the number of plane accidents that occur on different days of the week. Using the Γ· 2 test, the hypothesis that the digits were evenly distributed in the table is assessed. Examine whether the data are compatible with the assumption that this particular is visited by birds belonging to these six categories in the proportion.
From the table below, check whether the son's eye color is related to his father's eyes. Check whether the nature of the area is related to the voting preference in this election. Of the vaccinated persons, 15 persons fell ill, and 25 persons in the second group.
Test whether any distinction is made in the appointment based on female or male criteria. Test whether there is any significant difference in the petrol consumption of these two types of cars. Answer: There is significant difference in petrol consumption of the two types of new cars.
It is likely that the mean of the first population is less than that of the second. A manufacturer of golf balls introduces a new type of ball (brand 2) that it claims is superior to the previous brand (brand 1) in terms of the distance it will carry under identical playing conditions. To test this claim, a randomly selected golfer hits 35 shots using the old brand (brand 1), while another randomly selected golfer hits 40 shots using a new brand (brand 2). The distance in meters for each brand is recorded and summarized as follows (the raw data is shown in the computer solution below);.
Is it likely that the mean of population I is less than that of population II. Can we say that the sd of the distribution of grades of all the students from which the above sample was drawn is equal to 4.47.
