Soft-drink bottles. A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor’s production process to verify the vendor’s claim. The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

a. Assuming the vendor’s claim to be true, what is the probability of obtaining a sample mean this far or farther below the process mean? What does your answer suggest about the validity of the vendor’s claim?

b. If the process standard deviation were 3 psi as claimed by the vendor, but the mean were 156 psi, would the observed sample result be more or less likely than in part a? What if the mean were 158 psi?

c. If the process mean were 157 psi as claimed, but the process standard deviation were 2 psi, would the sample result be more or less likely than in part a? What if instead the standard deviation were 6 psi?

A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi.

The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

a.

According to Central Limit Theorem, the sampling distribution of$\overline{x}$follows normal distribution with mean${\mu }_{\overline{x}}=\mu$and standard deviation${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Here, the mean is ${\mu }_{\overline{x}}=157$and the standard deviation is given below:

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{3}{\sqrt{40}}\\ & =& 0.474\\ & & \end{array}$

It is given that the mean is 1.3 psi below the process mean (157 psi).

Hence, the sample mean$\left(\overline{x}\right)$ is (157-1.3)=155.7.

Now,

$\begin{array}{rcl}P\left(\overline{x}⩽155.7\right)& =& P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{155.7-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& P\left(z⩽\frac{155.7-157}{0.474}\right)\\ & =& P\left(z⩽-2.74\right)\\ & & \end{array}$

From the table of Normal curve areas,

$$\begin{gathered} =0.5-0.4969 \\ =0.0031 \end{gathered}$$

The required probability of obtaining a sample mean below the process mean is 0.0031.

If the vendor’ claim is true, then the probability of obtaining a sample mean below the process mean is 0.0031 implies that there is very less chance of observing the sample mean 1.3 psi below 157 psi.

Hence, it is concluded that the population mean is not 157, and it would be less than 157.

b.Let,

From the table of Normal curve areas,

$$\begin{gathered} =0.5-0.2357 \\ =0.2643 \end{gathered}$$

The required probability of obtaining a sample mean below the process mean is 0.2643

If the vendor’s claim is true, then the probability of obtaining a sample mean below the process mean is 0.2643 implies that there is more chance of observing the sample mean 1.3 psi below 156 psi.

Now,

$\begin{array}{rcl}P\left(\overline{x}⩽155.7\right)& =& P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{155.7-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& P\left(z⩽\frac{155.7-158}{0.474}\right)\\ & =& P\left(z⩽-4.85\right)\\ & & \end{array}$

From the table of Normal curve areas,

$$\begin{gathered} =0.5-0.5 \\ =0 \end{gathered}$$

The required probability of obtaining a sample mean below the process mean is 0

If the vendor’s claim is true, then the probability of obtaining a sample mean below the process mean is 0 implies that there is no chance of observing the sample means 1.3 psi below 158 psi. That is, there is less chance of observing sample mean compared to the mean 157 psi.

c.

Obtaining $P\left(\overline{x}⩽155.7\right)$when the mean is 157 psi and the process standard deviation is 3 psi.

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{2}{\sqrt{40}}\\ & =& 0.316\\ & & \end{array}$

Let,

$\begin{array}{rcl}P\left(\overline{x}⩽155.7\right)& =& P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{155.7-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& P\left(z⩽\frac{155.7-157}{0.316}\right)\\ & =& P\left(z⩽-4.11\right)\\ & & \end{array}$

From the table of Normal curve areas,

$$\begin{gathered} =0.5-0.5\backslash \mathrm{hfill} \\ =0 \end{gathered}$$

The required probability of obtaining a sample mean below the process mean is 0. Hence, there is less chance of observing sample mean if $\sigma =2$compared to$\sigma =3$

Obtaining $P\left(\overline{x}⩽155.7\right)$when the mean is 157 psi and the process standard deviation is 6 psi.

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{6}{\sqrt{40}}\\ & =& 0.949\\ & & \end{array}$

Let,

$\begin{array}{rcl}P\left(\overline{x}⩽155.7\right)& =& P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{155.7-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& P\left(z⩽\frac{155.7-157}{0.949}\right)\\ & =& P\left(z⩽-1.37\right)\\ & & \end{array}$

From the table of Normal curve areas,

$$\begin{gathered} =0.5-0.4147 \\ =0.0853 \end{gathered}$$

The required probability of obtaining a sample mean below the process mean is 0.0853. Hence, there is more chance of observing sample mean if $\sigma =6$compared to$\sigma =3$