The purpose of the following assignment is to understand "z" and "t" tests and how they're different. Also, learning how to calculate a "z" and "t" test, using steps of hypothesis testing, making decisions about the null and alternative hypotheses, as well as using real-world data to connect statistics and geography.
Steps of hypothesis testing are below:
- State the null hypothesis
- State the alternative hypothesis
- Choose a statistical test
- Choose the significance level (α)
- Calculate test statistic
- Make a decision about the null & alternative hypothesis
Null Hypothesis
A null hypothesis states there is no difference between the sample mean (derived from personal data) and the mean of the entire population which you are comparing the data against.
Alternative Hypothesis
A alternative hypothesis states there is difference between the sample mean (derived from personal data) and the mean of the entire population which you are comparing the data against.
The null null or alternative hypothesis only tells you if there is a difference, but not how much the difference is.
We either reject, or fail to reject the null hypothesis and never accept the null hypothesis.
Question 1
We were given a chart that gave the interval type, confidence interval or level, and the number of test samples. Then I determined which test type was appropriate and the Significance Level (α). Two-tailed tests had 2 z or t values.
Question 2
Dr. Weichelt provided the following question and directions:
1. A
Department of Agriculture and Live Stock Development organization in Kenya
estimate that yields in a certain district should approach the following
amounts in metric tons (averages based on data from the whole country) per
hectare: groundnuts. 0.57; cassava, 3.7; and beans, 0.29. A survey of 23 farmers had the following
results:
μ σ
Ground
Nuts 0.52 0.3
Cassava 3.3 .75
Beans 0.34 0.12
a. Test
the hypothesis for each of these products.
Assume that each are 2 tailed with a Confidence Level of 95% *Use the
appropriate test
b. Be
sure to present the null and alternative hypotheses for each as well as
conclusions
c. What
are the probabilities values for each crop?
d. What
are the similarities and differences in the results
I then followed the appropriate steps for hypothesis testing to answer the questions.
1) State the null hypothesis for each of the three crops. Ground Nuts, Cassava, and Beans.
2) State the alternative hypothesis for all three crops.
3) Choose a statistical test based on (n).
4) Choose the significant level (α)
5) Calculate the test statistic
6) Make a decision about the null and alternative hypothesis based on where they fall.
Null hypothesis- There is no difference between the yield of ground nuts between the sample farmers and the county as a whole.
Alternative hypothesis: There is a difference between the yield of ground nuts between the sample farmers and the county as a whole.
Test statistic: (0.52-0.57)/(.3/sqrt 23)= .05/.063= -.794
Probability: .78344
-0.779 falls between -2.074 and 2.074 so for ground nuts, the null hypothesis will fail to be rejected.
Cassava:
Null hypothesis: There is no difference between the yield of cassava between the sample farmers and the county as a whole.
Alternative hypothesis: There is a difference between the yield of cassava between the sample farmers and the county as a whole.
Test statistic: (3.3-3.7)/(.75/sqrt 23) = .4/.156= -2.56
Probability: .5938
-2.56 falls outside of -2.074 and 2.074 so for cassavas, the null hypothesis will be rejected.
Beans:
Null hypothesis: There is no difference between the yield of beans between the sample farmers and the county as a whole.
Alternative hypothesis: There is a difference between the yield of beans between the sample farmers and the county as a whole.
Test statistic: (0.34-0.29)/(0.12/sqrt 23) = .05/.025= 2
Probability: .96037
2 falls between -2.074 and 2.074 so for beans, the null hypothesis will fail to be rejected.
Similarities:
The beans and ground nuts both failed to reject the null hypothesis. This means that they both are not statistically different from the population mean. Another similarity is between ground nuts and cassava. Both of sample means for these two crops were lower than the population mean.
Differences:
The cassava was the one crop that rejected the null hypothesis. Additionally, beans were the only crop that had a sample mean larger than the population mean.
Question 3
Dr. Weichelt provided the following question:
1. A
researcher suspects that the level of a particular stream’s pollutant is higher
than the allowable limit of 4.2 mg/l. A
sample of n= 17 reveals a mean pollutant level of 6.4 mg/l, with a standard
deviation of 4.4. What are your
conclusions? (one tailed test, 95%
Significance Level) Please follow the hypothesis testing steps. What is the corresponding probability value of
your calculated answer
5) t=(6.4-4.2)/(4.4/sqrt(17)) = 2.2/1.067 = 2.062
6) Since 2.062 is over 1.746, we reject the null hypothesis. This meant there is no difference between the standard of the pollution samples and the allowable time limit for the stream.
Part 2
For part two home values are compared between between block groups for the city of Ea Claire and blog groups for the county of Eau claire as a whole to see if home values for the city are different from the county. Hypothesis testing was again used to determine the answer.
1) Null hypothesis: there is no difference between the home values for the city of Eau Claire and the county of Eau Claire.
2) Alternative: There is a difference between the home values for the city of Eau Claire and the county of Eau Claire.
3) (n) is 53, therefore we have to use a z-test.
4) The confidence level was 95%, and I used a one-tailed test, so the significance level (α) is .05.
5) Test statistic: z = ((151876.51-169438.13)/(49706.92/sqrt(53))=-17561.62/6827.77=-2.57
6) Since -2.57 falls out of -1.64, I rejected the null hypothesis.
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