Step 1

The given 95% confidence interval for the true average amount of warpage of laminate sheets under specified conditions is $\left( {1.81,1.95} \right)$.

Confidence level, C = 0.95.

Sample size, n = 15.

The amount of warpage is normally distributed.

The null hypothesis is ${H_0}:\mu = 2$.

The alternate hypothesis is ${H_1}:\mu \ne 2$.

 
Step 2

a)

Significance level is calculated below.

\[\begin{array}{c} \alpha = 1 - C\\ = 1 - 0.95\\ = 0.05 \end{array}\]

The 95% confidence interval for the true average amount of warpage of laminate sheets under specified conditions is $\left( {1.81,1.95} \right)$. It means that there is 95% confidence that the true average amount of warpage of laminate sheets under specified conditions lies between 1.18 and 1.95.

 
Step 3

b)

Significance level, $\alpha = 0.01$.

Use below formula to find confidence level.

\[\begin{array}{c} C = 1 - \alpha \\ = 1 - 0.01\\ = 0.99 \end{array}\]

Confidence level is 99%.

\[ME = 1.88 - 1.81\]

Degrees of freedom for t distribution are calculated below.

\[\begin{array}{c} df = n - 1\\ = 15 - 1\\ = 14 \end{array}\]

In T-table, find the value with row headed 14 and column headed 0.025 for one tailed area.

\[{t_{0.025}} = 2.145\]

In T-table, find the value with row headed 14 and column headed 0.005 for one tailed area.

\[{t_{0.005}} = 2.977\]

Formula to find confidence interval for true average is given below.

\[\begin{array}{c} CI = \bar x \pm {t_{0.005}} \times se\\ = 1.88 \pm 2.977 \times 0.0326\\ = 1.88 \pm 0.09705\\ = \left( {1.783,1.977} \right) \end{array}\]

The interval does not contain hypothesized mean 2.

Hence, there is sufficient evidence that the true average amount of warpage of laminate sheets under specified conditions differs from 2.