Chapter 8, Problem 57SE

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of 3.05 mm and a sample standard deviation of .34 mm. The desired true average thickness of such lenses is 3.20 nun. Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using α = .05

To determine

Test whether the true average thickness of such lenses is other than 3.20 or not.

Answer to Problem 57SE

There is sufficient evidence to suggest that the true average thickness of such lenses is something other than desire.

Explanation of Solution

Given info:

For checking the thickness of the lenses 50 samples were taken. The mean thickness of the sample is 3.05 mm. The standard deviation is 0.34. The desired true average thickness of the lenses is 3.2 mm. $\alpha =0.05$.

Calculation:

Step 1: parameter of interest

Let $\mu$ be the average thickness of the lenses.

Step 2: Null hypothesis:

$H_0:\mu =3.20$

That is, the thickness of the lenses is 3.2 mm.

Step 3: Alternative hypothesis:

$H_a:\mu \ne 3.20$

That is, the thickness of the lenses differs from 3.2 mm.

Step 4: Test statistic:

Software procedure:

Step by step procedure to obtain the test-statistic value using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Summarized data, enter the sample size as 50 and mean as 3.05
• In Standard deviation, enter 0.34
• In Perform hypothesis test, enter the test mean as 3.2
• Check Options; enter Confidence level as 95%.
• Choose not equal in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the test statistic value is –3.12.

Step 5: P-value:

From the output of step 3, the P-value is given by 0.0018.

Step 6: Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.0018\right)<\alpha \left(0.05\right)$.

By rejection rule, reject null hypothesis.

Step 7: Interpretation:

Thus, it can be concluded that there is sufficient evidence to support that the thickness of the lenses differs from 3.2 mm at 5% level of significance.