Question

The data set contains the compressive strength, in thousands of pounds per square inch (psi), of 30 samples of concrete taken two and seven days after pouring.

Sample	Two Days	Seven Days
1	2.830	3.505
2	3.295	3.430
3	2.710	3.670
4	2.855	3.355
5	2.980	3.985
6	3.065	3.630
7	3.765	4.570
8	3.265	3.700
9	3.170	3.660
10	2.895	3.250
11	2.630	2.850
12	2.830	3.340
13	2.935	3.630
14	3.115	3.675
15	2.985	3.475
16	3.135	3.605
17	2.750	3.250
18	3.205	3.540
19	3.000	4.005
20	3.035	3.595
21	1.635	2.275
22	2.270	3.910
23	2.895	2.915
24	2.845	4.530
25	2.205	2.280
26	3.590	3.915
27	3.080	3.140
28	3.335	3.580
29	3.800	4.070
30	2.680	3.805

(a) At the 0.10 level of significance, is there evidence of a difference in the mean strengths at two days and at seven days?

(b) Find the p-value in (a) and interpret its meaning.

(c) At the 0.10 level of significance, is there evidence that the mean strength is lower at two days than at seven days?

(d) Find the p-value in (c) and interpret its meaning.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.12276
DF = 58
t = [ (x₁ - x₂) - d ] / SE

t = - 4.71

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

b)

Since we have a two-tailed test, the P-value is the probability that a t statistic having 58 degrees of freedom is more extreme than -4.71; that is, less than -4.71 or greater than 4.71.

Thus, the P-value = less than 0.001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is evidence of a difference in the mean strengths at two days and at seven days.

c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁> u ₂
Alternative hypothesis: u₁ < u ₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.12276
DF = 58
t = [ (x₁ - x₂) - d ] / SE

t = - 4.71

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

d)

Since we have a two-tailed test, the P-value is the probability that a t statistic having 58 degrees of freedom is less than -4.71.

Thus, the P-value = less than 0.001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the mean strength is lower at two days than at seven days.