Top Special Offer! Check discount

Get 13% off your first order - useTopStart13discount code now!

81 views
2 pages ~ 534 words

Get a Custom Essay Writer Just For You!

**Experts** in this subject field are ready to write an original essay following your instructions to the dot!

In statistics, a null hypothesis is that hypothesis, which is assumed true before the research is conducted and is assumed that the observed sample result exactly from chance. H0 is the denotation for null hypothesis. The alternative hypothesis on the other hand denoted by H1. This hypothesis states that some nonrandom cause influences the sample observations.

The mean of the data (84.4) here, has been compared by the actual figures in the documented statistics (720, 60, 42). According to the analysis, the researcher is able to test hypothesis using the chi-square goodness of fit, which is non-parametric and tests whether the value observed in a given sample shows a significant difference from the value expected (Bryant and Albert 378). In this case, the observed values from the documented statistics are; 720, 60, and 42 while the expected value is the mean of the data (84.4).

Hypothesis

The null hypothesis is that observed mean and the expected mean shows no significant difference whereas the alternative hypothesis is that the observed mean and the expected mean will show a significant difference.

This hypothesis is appropriate since the research seeks to compare the values in the documented statistics with the average value obtained from the data set as displayed in the table below.

Observed values

Expected values

3 to 5

720

84.4

6 to 17

60

84.4

18 to 75

42

84.4

To compute the test statistic, (chi-square) χ2calc= {(O - E)/ E} 2 where;

O=Observed value

E=Expected value

χ2calc= Chi-square goodness of fit

{(720 – 84.4)/ 84.4}2 + {(60 – 84.4)/ 84.4}2 + {(42 – 84.4)/ 84.4}2

(7.53)2

+ (-0.29)2 + (-0.5)2

56.7009 + 0.0841 + 0.25

Χ2

calc = 57.035

We find the degrees of freedom (df) as follows:

(c-1)×(r-1) where;

c=no. of columns

r=no. of rows

In this case, the df = (2-1) × (3-1) =2

The degrees of freedom are used with a given significance level for example, 0.05 to find the tabulated value (χ20.05, df) from the chi-square table. The researcher then compares this value with the calculated value above to help him/her arrive at a conclusion. For example, a greater calculated value as opposed to the tabulated value would result into rejecting the null hypothesis (H0). If the null hypothesis is rejected, it means that the sample from which the data was picked is biased and does not give a fair representation of the population.

On the other hand, do not reject the null hypothesis in the event the value calculated is less than the value in the table. As a result, deduce that significant difference lacks between the expected value and the value observed, which therefore means the sample used represents the population well.

The analysis can also use the p – value approach. We reject the null hypothesis when the p –value is less than 0.05. When the p- value is greater than 0.05, we do not have enough evidence to say that there is a significant statistical difference between the expected values and the observed values.

One can also carry out a One Sample Two –Tailed t-test to conduct a hypothesis to compare the sample mean and the population mean, that is:

H0: µ = M

H1: µ ≠ M

Where;

µ = sample mean of the data

M = population mean

Work Cited

Bryant, Fred B., and Albert Satorra. "Principles and practice of scaled difference chi-square testing." Structural Equation Modeling: A Multidisciplinary Journal 19.3 (2012): 372-398.

September 25, 2023

Eliminate the stress of Research and Writing!

Hire a Pro
Hire one of our **experts** to create a completely original paper even in **3 hours**!