91Ó°ÊÓ

n $=10$

c $=95\%=0.95$

The formula used:$E=t_{a/2}×\frac{s}{\sqrt{n}}$

Conditions:

Random and independent are the three requirements.

Normal/Large sample ($10$percent condition).

Because the sample is a random sample, I'm satisfied.

Because the sample of $20$tables represents less than $10\%$of the total number of tables, independent: fulfilled.

Because the pattern in the normal quantile plot is generally linear, the normal/large sample is satisfied.

Because all of the prerequisites have been met, it is appropriate to calculate the population's confidence interval.

The mean is

$\overline{x}=\frac{230.328}{20}$

$\overline{x}=11.5164$

The variance is

$s=0.0950$

In the table of the Students T distribution, look in the row starting with degrees of freedom $df=n−1=20−1=19$and in the column with $c=95$percent to find the $t-$value:

$t_{a/2}=2.093$

The margin of error is

$$\begin{gathered} E=t_{a/2}×\frac{s}{\sqrt{n}} \\ =2.093×\frac{0.0950}{\sqrt{20}} \\ =0.0445 \end{gathered}$$

The boundaries of the confidence interval are

$$\begin{gathered} \overline{x}−E=11.5164−0.0445=11.4719 \\ \overline{x}+E=11.5164+0.0445=11.5609 \end{gathered}$$

There are $95\%$confident that the true hardness of the tablets in this batch is between $11.4719$and $11.5609$

From the Result part (a):

$(11.4719,11.5609)$

It is noted that the confidence interval contains the value $11.5$indicating that the mean hardness is likely to be $11.5$As a result, the claim that the mean is $11.5$is not rejected. There is insufficient evidence to support the claim that the genuine hardness of the tablets in this batch is lower than $11.5$

We then realized that we had arrived at the same conclusion as we had in the prior experiment.