Hypothesis Testing

Objective:

The objective of this assignment was to use hypothesis testing on sample and real world data to better understand z and t tests and when to use them. Also, this assignments goal was to learn how to calculated z and t values and to make a decision about the null and alternative hypothesis.

Part 1:

The first part of of this assignment was to finish a table that was provided. The table provides the interval type (one or two tailed), the confidence level, and the sample size. From this, the level of significance, whether a t or z test was required, and what the z or t value is. The table below (table 1) shows the table filled out with the missing information added. The interval type can be either a one

table 1. Finished statistics table

 or two tailed test. A one tail test means that only the positive or negative critical value is found. A two tailed test means that two values are found on both the positive and negative extremes and the level of significance is divided by 2. The confidence level is the amount of times out of 100 that sample data should be the same as population data. N is the sample size of the sample that is taken. Level of significance is set to determine the probability of a type 1 error occurring. To choose z or t is simple, if n is over 30, use a z test. If n is 30 or under, use t test. For the t or z values, a z or t value chart is needed and the level of significance is looked up in the chart to find the cut off for when the null hypothesis is rejected or not rejected.

The next part of this assignment looks at Kenya's agriculture. From a sample of 23 farmers, it needs to be determined if their yields for different types of crops are statically different from the county as a whole. The three crops are ground nuts, cassava, and beans. For all calculations, a confidence level of 95% is used with a two tailed test. This means that the test statistic for all three tests are +/- 2.074. The test done for these will be a t test because the sample size is under 30. Finding the t value for these follows the following equation: (sample mean - population mean)/(sample standard deviation/ square root of N).

Ground nuts:
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.
Equation: (0.52-0.57)/(.3/sqrt 23)= -0.799
Probability: 21.66%
-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.
Equation: (3.3-3.7)/(.75/sqrt 23) = -2.558
Probability: 1.07%
 -2.558 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.
Equation: (0.34-0.29)/(0.12/sqrt 23) = 1.998
Probability: 97.03%
1.998 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 meaning 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 only crop that rejected the null hypothesis. The beans was the only crop that had a sample mean larger than the population mean.

The last section of part one looks at the level pollution in a stream. There were 17 samples taken and with a sample mean pollution level of 6.4 mg/l and a standard deviation of 4.4. The allowable limit for stream pollution is 4.2 mg/l. The part uses hypothesis testing to determine if the stream is statically over the allowable limit.

Null hypothesis: There is no difference between the mean of the pollution samples and the allowable limit for the stream.
Alternative hypothesis: There is a difference between the mean of the pollution samples and the allowable limit for the stream.
Statistical test: The sample size is 17. 17 is under 30 so a t test is needed for this.
Level of significance: This was provided and is 95% of a one tailed test. For a level of significance of 95% for a one tailed t test is 1.746.
Equation: (6.4-4.2)/(4.4/sqrt 17) = 2.0616
Probability: 97.4%
2.0161 is over 1.746 so the null hypothesis will be rejected.
This means that there is a difference between the mean of the pollution samples and the allowable limit for the stream. The researcher was correct that the level of pollution in the stream is over the allowable limit. However, by how much it is over is unknown.

Part 2:

For the second part of the assignment, home values are compared between the block groups for the city of Eau Claire and the block groups for Eau Claire county as a whole to see if home values for the city are statistically different from that of the county. This can be done with hypothesis testing. The mean for the home values for Eau Claire county is 169438.13. The mean for the home values for the city of Eau Claire is 151876.509 with a standard deviation of 49706.919 with a n of 53.

Null hypothesis: There is no difference between the home values for the city of Eau Claire and the county of Eau Claire.
Alternative hypothesis: There is a difference between the home values for the city of Eau Claire and the county of Eau Claire.
Statistical test: The sample size is 53. 53 is larger than 30 so a z test will be used.
Level of significance: For this a level of significance will be 95% giving a critical value 1.64 but the sample mean is smaller than the population mean so the critical value needs to be multiplied by -1 giving a critical value of -1.64.  
Equation: (151876.509-169438.13)/(49706.919/sqrt 53) = -2.572
-2.572 is smaller than -1.64 so the null hypothesis will be rejected. This means that there is a statistical difference between the home values of the city of Eau Claire and the county of Eau Claire. The map below (map 1) shows the home values for the city of Eau Claire and the county of Eau

map 1. Home values by block group for Eau Claire county

Claire. This map helps show that home values in the city of Eau Claire are higher than the rest of Eau Claire county. The map also gives reference for where the city of Eau Claire is in Eau Claire county and where Eau Claire county is in Wisconsin.