Question

A seed company is developing many strains of tomatoes by selective breeding. Trials of two similar but not identical strains with favorable qualities were done in two fields under similar conditions. The company would like to know if the population average weight of tomato for strain 2 (u2) is statistically significantly larger than the population average weight for stain 1 (u1). Consequently they picked at random 15 tomatoes from the field with strain 1 and 14 from the field with strain 2 and weighed them.

Weight grams:

strain 1

132, 68, 74, 93, 61, 81, 62, 68, 103, 72, 64, 104, 62, 86, 95.

strain 2

40, 88, 112, 127, 114, 124, 95, 125, 989, 86, 142, 130, 70, 81.

Does the strain 2 have significantly greater mean weight than for strain 1? Test this hypothesis at the alpha = 0.01 and 0.05 levels, using the two samples. Make the assumption that the weight is distributed normally in both populations with equal variances. You will be testing u1 vs u2.

1) Which diagram shows reject/ fail to reject regions for this problem?

2) what is the test statistic is use for question?

3) compute the sample means and standard deviations that you will need for 5 &6.

4) what is the test statistic value computed from the data in question 1 &3?

5) if the level of significance alpha is .01 state the critical values which you would use relevant to questions 1&4.

6) if the level of significance alpha is 0.05 state the critical value which you would use relevant to questions 1 and4.

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means.

Null hypothesis: H₀: the strain 2 has same mean weight as it is for strain 1.

Alternative hypothesis: H_a: the strain 2 has significantly greater mean weight than for strain 1.

H₀: µ₁ = µ₂ versus H_a: µ₁ < µ₂

µ₁ = Mean for strain 1

µ₂ = Mean for strain 2

This is a lower tailed test.

We are given

X1bar = 81.66666667

X2bar = 165.9285714

S1 = 20.44388371

S2 = 238.5257491

n1 = 15

n2 = 14

df = n1 + n2 – 2 = 15 + 14 – 2 = 27

α = 0.01 and 0.05

Critical value for α = 0.01 = -2.4727

Critical value for α = 0.05 = -1.7033

(by using t-table)

1) Which diagram shows reject/ fail to reject regions for this problem?

The t curve diagram shows the required regions for this problem. The required diagram is given as below:

2) what is the test statistic is use for question?

The test statistic used for this question is t test statistic.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

3) compute the sample means and standard deviations that you will need for 5 &6.

From given data, we have

X1bar = 81.66666667

X2bar = 165.9285714

S1 = 20.44388371

S2 = 238.5257491

n1 = 15

n2 = 14

4) what is the test statistic value computed from the data in question 1 &3?

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(15 – 1)* 20.44388371^2 + (14 – 1)* 238.5257491^2]/(15 + 14 – 2)

S_p² = 27610.3801

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (81.66666667 – 165.9285714) / sqrt[27610.3801*((1/15)+(1/14))]

t = -84.2619 / 61.7484

t = -1.3646

Test statistic value = -1.3646

5) if the level of significance alpha is .01 state the critical values which you would use relevant to questions 1&4.

Critical value for α = 0.01 = -2.4727

Test statistic value = -1.3646

Test statistic value > critical value

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the strain 2 has significantly greater mean weight than for strain 1.

6) if the level of significance alpha is 0.05 state the critical value which you would use relevant to questions 1 and4.

Critical value for α = 0.05 = -1.7033

Test statistic value = -1.3646

Test statistic value > critical value

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the strain 2 has significantly greater mean weight than for strain 1.