Question

A simple random sample of 5 (young) spruce trees from a nursery weighed 132, 145, 162, 166, and 175 grams. A simple random sample of 4 fir trees from the same nursery weighed 137, 147, 158, and 170 grams.

(a) Suppose each species’ population of weights is normally distributed. Decide whether the population mean weights of spruce and fir in this nursery are different. Using a 2 sample t-test.

• Hypotheses:

• Assumptions:

• Test statistic:

• P-value:

• Conclusion:

(b) Suppose each species’ population of weights is normally distributed. Find a 98% confidence interval for μspruce − μfir. (Hint: The problem says “98%,” not “95%.”)

Do by hand.

Answer 1

a)

H0 : mu1 = mu2
Ha: mu1 not equals to mu2

Assumptions:

1.The data are continuous
2. The data follow the normal probability distribution.
3. The variances of the two populations are equal.
4. The two samples are independent.

test statistics;

x1 =156, x2 = 153
s1 = 17.2772, s2 = 14.2127
n1 = 5 , n2 = 4

t = (x1 - x2)/sqrt(s1^2/n1+s2^2/n2)
= ( 156 - 153)/sqrt(17.2772^2/5 + 14.2127^2/4)
= 0.2858

p value = .7936

conclusion : Fail to reject H0

b)
t value at 98% = 4.5407

CI = (x1 - x2) +/- t * sqrt(s1^2/n1+s2^2/n2)
= ( 156 - 153)+/- 4.5407 * sqrt(17.2772^2/5 + 14.2127^2/4)
= (-44.6667,50.6667)

The 98% confidence interval for μspruce − μfir. is (-44.6667,50.6667)