Question

Independent t-test (assuming unequal variances) for mass per pot at the 50 pot and 100 pot comparing radish to tomato. Is this asking to perform a t-test with tomato at 50 and radish at 50 and another t-test with tomato at 100 and radish at 100 or tomato at 50 and tomato at 100 and radish at 50 and raidsh at 100

Tomato at 50 pot
0.61
0.61
0.53
0.61
0.62
0.47
0.49
0.56
0.54
0.65
0.96
0.65
0.66
0.65
0.56
0.45
0.48

Tomato at 100 pot
0.66
0.83
0.95
0.95
0.99
0.9
0.45
1.01
0.99
0.98
0.81
0.9
0.98
0.94
1.14
0.99
1.05
0.89

Radish 100	Radish 50
1.35	1.07
1.23	1.1
1.3	0.89
1.16	1.05
1.22	1.05
1.24	1.01
1.24	0.96
1.27	1.06
1.31	1.08
1.33	1
1.37	1.03
1.45	1.13
1.3	1.1
1.22	1.03
1.13	1.06
1.23	0.97
1.14	0.92
0.83	0.95

I

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.02836
DF = 33
t = [ (x₁ - x₂) - d ] / SE

t = - 15.21

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 37 degrees of freedom is more extreme than -15.21; that is, less than -15.21 or greater than 15.21.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is significance difference between mass per pot at the 50 pot.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.04778
DF = 34
t = [ (x₁ - x₂) - d ] / SE

t = - 6.87

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 36 degrees of freedom is more extreme than - 6.87; that is, less than - 6.87 or greater than 6.87.

Thus, the P-value = less than 0.001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is significance difference between mass per pot at the 100 pot.