Answer to Question #153100 in Statistics and Probability for rehan tahir

Question

From the area planted in one variety of guayule, 25 plants were selected at random. Of these plants, 13 were “ Off types” and 12 were “Aberrant” The rubber
percentages of these plants were:

Off types 4.47 5.88 6.21 5.55 6.09 5.70 5.82 4.84 5.59 5.59 5.22 4.45 6.76
Aberrant 6.48 6.36 4.28 7.71 6.40 7.06 5.51 8.93 7.71 7.20 7.37 5.91

Compute a 90% confidence interval for the difference of two population means. Also interpret your results. Also test the hypothesis that two types of plants have equal average rubber production.

Answer 1

Off types:

$n_1 = 13$

$\bar x_1=\frac{\sum x_i}{n}\frac{4.47+5.88+6.21+5.55+6.09+5.70+5.82+4.84+5.59+5.59+5.22+4.45+6.76}{13}=5.55$

$s_1=\sqrt{\frac{\sum(\bar x_1 - x_i)^2}{n-1}}=0.67$

Aberrant:

$n_2 = 12$

$\bar x_2=\frac{6.48+6.36+4.28+7.71+6.40+7.06+5.51+8.93+7.71+7.20+7.37+5.91}{12}=6.74$

$s_2=\sqrt{\frac{\sum(\bar x_2-x_i)^2}{n-1}}=1.2$

Since samples are less than 30 in this problem we are dealing with t-distirbution with n₁ + n₂ – 2 = 13 + 12 – 2 = 23 degrees of freedom.

The table t-value for a 90% confidence interval with 23 df is t_{0.05, 23} = 1.714

The formula for a 90% confidence interval for the difference of two population means:

$$(\bar x_1-\bar x_2)\pm t_{0.05, 23}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

where

$$s_p=\sqrt{\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}}$$

$s_p=\sqrt{\frac{0.67^2\cdot12+1.2^2\cdot11}{12+13-2}}=0.96$

Confidence interval:

$(5.55-6.74)\pm1.714\cdot0.96\sqrt{\frac{1}{12}+\frac{1}{13}}=-1.19\pm0.66$

We are 90% confident that the difference in the two population means is between –1.85 and –0.53. Zero is not in this interval so there is a significant difference in the average rubber production between “off types” and “aberrant” plants.

$H_0:\mu_1=\mu_2$

$H_a:\mu_1\ne\mu_2$

The hypothesis test is two-tailed. Since samples are less than 30 and the population standard deviations are unknown this is t-test.

The critical value for significance level $\alpha=0.1$ and df = 23 is t_{0.05, 23} = $\pm$ 1.714

The critical region is (-∞, -1.714] ∪ [1.714, ∞)

Test statistic:

$$t=\frac{(\bar x_1-\bar x_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{5.55-6.74-0}{0.96\sqrt{\frac{1}{12}+\frac{1}{13}}}=-3.1$$

Since –3.1 < –1.714 thus t falls in the rejection region, we reject the null hypothesis.

At the 10% significance level the data do provide sufficient evidence to not support the claim. We are 90% confident to conclude that two types of plants have not equal average rubber production.