Answer :
a. The critical t value is 2.451. b. The t-statistic is 3.4. c. Fertilizer A is significantly greater than Fertilizer B using an alpha of 0.025.
a) To find the critical t value for a one-tailed test with alpha = 0.025, we need to use the degrees of freedom (df) for the t-distribution. The degrees of freedom for a two-sample t-test with unequal variances is calculated as: [tex]\[ df = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{(s_1^2/n_1)^2/(n_1 - 1) + (s_2^2/n_2)^2/(n_2 - 1)} \][/tex]
where [tex]\( s_1^2 \) and \( s_2^2 \)[/tex] are the sample variances, and [tex]\( n_1 \) and \( n_2 \)[/tex] are the sample sizes.
For Fertilizer A: [tex]\( s_1 = 0.25 \) and \( n_1 = 15 \)[/tex]
For Fertilizer B: [tex]\( s_2 = 0.20 \) and \( n_2 = 13 \)[/tex]
First, we calculate the variances:
[tex]s_1^2 = (0.25)^2 = 0.0625[/tex]
[tex]\[ s_2^2 = (0.20)^2 = 0.04 \][/tex]
Now, we calculate the degrees of freedom:
[tex]\[ df = \frac{(0.0625/15 + 0.04/13)^2}{(0.0625/15)^2/(15 - 1) + (0.04/13)^2/(13 - 1)} \][/tex]
[tex]\[ df = \frac{(0.004167 + 0.003077)^2}{(0.004167)^2/14 + (0.003077)^2/12} \][/tex]
[tex]\[ df = \frac{0.0000525}{(0.00000889 + 0.00000637)} \][/tex]
[tex]\[ df = \frac{0.0000525}{0.00001526} \][/tex]
[tex]\[ df \approx 34.47 \][/tex]
Since degrees of freedom must be an integer, we round down to the nearest whole number: df = 34.
Using a t-distribution table or a calculator, we find the critical t value for a one-tailed test with df = 34 and alpha = 0.025. The critical t value is approximately 2.451.
b) Next, we calculate the t-statistic using the formula for a two-sample t-test with unequal variances: [tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
where [tex]\( \bar{x}_1 \) and \( \bar{x}_2 \)[/tex] are the sample means.
[tex]\[ t = \frac{12.92 - 12.63}{\sqrt{\frac{0.0625}{15} + \frac{0.04}{13}}} \][/tex]
[tex]\[ t = \frac{0.29}{\sqrt{0.004167 + 0.003077}} \][/tex]
[tex]\[ t = \frac{0.29}{\sqrt{0.007244}} \][/tex]
[tex]\[ t = \frac{0.29}{0.0851} \][/tex]
[tex]\[ t \approx 3.408 \][/tex]
Rounded off to one decimal place, the t-statistic is 3.4.
c) We compare the calculated t-statistic to the critical t-value. Since 3.408 > 2.451, we reject the null hypothesis in favour of the alternative hypothesis that Fertilizer A is significantly greater than Fertilizer B at the 0.025 level of significance.
Therefore,
a. The critical t value is 2.451.
b. The t-statistic is 3.4.
c. Fertilizer A is significantly greater than Fertilizer B using an alpha of 0.025.
The complete question is
Two equal groups of seedlings, and equal in height, were selected for an experiment. One group of seedlings was fed Fertilizer A, and the other group with Fertilizer B. The mean heights of the two groups of seedlings, their standard deviations, and sample sizes are listed below. Assume simple random sampling, independence, and normally distributed data. Alpha = 0.025.Sample Mean Height (inches, Std Deviation (inches), Sample Size Fertilizer A 12.92, 0.25, 15 Fertilizer B 12.63, 0.20, 13. What is the null hypothesis?
a. Using the data in the previous question, what is the critical t value using the simplified textbook method and Tables when testing the claim that Fertilizer A is significantly greater than Fertilizer B? Record the answer to three decimal places.
b. Using the data from the previous question, what is the t-statistic rounded off to one decimal place?
c. Using the data from above, Fertilizer A is significantly greater than Fertilizer B using an alpha of 0.025?
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
