Answer to Question #154495 in Statistics and Probability for Mano

Question

From the area pianted in one variety of guayule, 54 plants were selected at random. Of these plants. 15 were "Offrypes" and 12 were "Aberrants. The rubber per centages of these plants were:

Offtypes4.47, 5.88, 6.21, 5.55, 6.09, 5.70, 5.82, 4.84, 5.59, 5.59,5.22. 4.45, 6.74. 6.06, 6.04· Aberrants 6.48, 6.36, 4.2S, 7.71, 6.40, 7.06, 5.02, &.93. 7.72, 7.20,7.37, 5.91

Compute a 90% confidence interval for the difierence of two population means.

Answer 1

Off types:

$n_1 = 15$

$\bar x_1=\frac{\sum x_i}{n}$

$\bar x_1=\frac{4.47+5.88+6.21+5.55+6.09+5.70+5.82+4.84+5.59+5.59+5.22+4.45+6.74+6.06+6.04}{15}=5.62$

$s_1=\sqrt{\frac{\sum(\bar x_1 - x_i)^2}{n-1}}=0.64$

Aberrant:

$n_2 = 12$

$\bar x_2=\frac{6.48+6.36+4.20+7.71+6.40+7.06+5.02+6.93+7.72+7.20+7.37+5.91}{12}=6.53$

$s_2=\sqrt{\frac{\sum(\bar x_2-x_i)^2}{n-1}}=1.07$

Since samples are less than 30 in this problem we are dealing with t-distirbution with n₁ + n₂ – 2 = 15 + 12 – 2 = 25 degrees of freedom.

The table t-value for a 90% confidence interval with 25 df is t_{0.05, 25} = 1.708

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

$$(\bar x_1-\bar x_2)\pm t_{0.05, 25}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

where

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

$s_p=\sqrt{\frac{0.64^2\cdot14+1.07^2\cdot11}{15+12-2}}=0.86$

Confidence interval:

$(5.52-6.53)\pm1.708\cdot0.86\sqrt{\frac{1}{15}+\frac{1}{12}}=-1.01\pm0.57$

We are 90% confident that the difference in the two population means is between –1.58 and –0.44. Zero is not in this interval so there is a significant difference in the average rubber production between “off types” and “aberrant” plants.