T-Distribution – Examples

### Question 1 It is known that the number of oranges on the trees in Orchard A is normally distributed with a mean of 100 (standard deviation unknown). Following the use of new fertilizer, the claim was made that the number of oranges on the trees increased. To test the claim, a sample of nine trees was conducted. The number of oranges on the sample trees was as follows: 110, 108, 115, 109, 105, 117, 111, 110, 108. The claim must be tested at a significance level of 5%. #### Solution steps 1. Define the null hypothesis – the fertilizer is ineffective (the mean is 100).
2. Defining the alternative hypothesis – the garbage is effective (expectancy greater than 100).
3. The significance level is 5%, the number of degrees of freedom is 8. The corresponding value from the t-table is 1.86.
4. The sample mean is 110 oranges.
5. The standard error of the baseline variable, ( T_1 ), is 3.5 oranges.
6. The standard error of the mean variable, ( T_9 ), is 1.167 oranges ([3 .5=””] ).
7. The red line will be set at a value of 102.17 oranges ((102.17=1.167times1.86+100)).
8. The statistical conclusion is that the counter-hypothesis is accepted (the garbage is effective), because the mean is in the abnormal range. — ### Question 2 From previous studies it is known that the number of kernels in a hotel is normally distributed with a mean of 200 (standard deviation unknown). A researcher claims that due to global warming, this number has changed and is no longer 200. To do this, he conducted a sample of 36 melons and counted the number of kernels in each hotel. He calculated the mean and found that it was 210. After finding the mean, he calculated the estimated standard deviation (which is the standard error) of the base variable and found that it was 54 kernels. According to the sample results, is the researcher’s claim correct at a significance level of 5%? #### Solution steps 1. Define the null hypothesis – warming does not affect the number of kernels (the mean is 200).
2. Defining the counter-hypothesis – warming affects the number of nuclei (expectancy different from 200).
3. The significance level is 5%, the number of degrees of freedom is 35.

The value corresponding to 30 degrees of freedom is 2.042.
4. The sample average is 210 kernels.
5. The standard error of the base variable, ( T_1 ), is 54 kernels.
6. Standard error of the mean variable, ( T_{36} ), is 9 nuclei ([54/sqrt{36}=9] ).
7. The red lines will be set at the following values:
– ( 218.38=2.042 times 9+200 )
– ( 181.62=2.042times9-200 )
8. The statistical conclusion is that the null hypothesis is accepted (the number of nuclei did not change due to warming), because the average is within the reasonable range. — ### Question 3 It is known that the temperature in Tel Aviv on a summer day (from the beginning of June to the end of September) is normally distributed. A meteorologist measured the temperature on 9 days in the summer and obtained the following sample: 26, 30, 32, 27, 30, 28, 35, 29, 33. Find the confidence interval for the temperature expectation at a confidence level of 95%. #### Solution steps 1. The sample mean is 30 degrees.
2. The confidence level is 95%, the number of degrees of freedom is 8. The corresponding value from the t-table is 2.306.
3. The standard error of the base variable, ( T_1 ), is 2.915 degrees.
4. The standard error of the mean variable, ( T_9 ), is 0.972 degrees.
5. The upper limit of the confidence interval is (32.24=0.972times2.306+30).
6. The lower bound of the confidence interval is ( 27.76=0.972times2.306-30 ). — ### Question 4 Loulan wants to find the mean of the weight of the eggs laid by the hens in his coop. It is known that their weight is normally distributed. Since Loulan knows that he will not be able to find the exact mean, he is satisfied with finding a confidence interval for the mean at a confidence level of 90%. To do this, Loulan conducted a sample in which 16 eggs were weighed. The results (in grams) were as follows: 120, 150, 132, 139, 155, 141, 132, 127, 142, 140, 151, 136, 136, 144, 138, 141. #### Steps to the solution 1. The sample mean is 139 grams.
2. The confidence level is 90%, the number of degrees of freedom is 15.

The value corresponding to 13 degrees of freedom is 1.771.
3. The standard error of the base variable, ( B_1 ), is 8.892 grams.
4. Standard error of the mean variable, ( B_{16} ), is 2.223 grams.
5. The upper limit of the confidence interval is (142.9=2.223times1.771+139).
6. The lower limit of the confidence interval is (135.1=2.223times1.771-139).