2. A popcorn manufacturer claims that the bags of popcorn the factory produces have a mean weight of 6.5 ounces. The population standard deviation is known to be 0.22 ounces. A simple random sample of 40 popcorn bags has a mean of 6.7 ounces. Assume a confidence level of [tex]$ \alpha = 0.10 $[/tex].

Find the value of the test statistic [tex]$z$[/tex] using

[tex]
z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
[/tex]

A. -5.75
B. 4.00
C. 2.70
D. 5.75



Answer :

To determine the value of the test statistic [tex]\( z \)[/tex] using the provided information, follow these steps:

1. Identify the given information:
- Sample mean ([tex]\(\overline{x}\)[/tex]) = 6.7 ounces
- Population mean ([tex]\(\mu\)[/tex]) = 6.5 ounces
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 0.22 ounces
- Sample size ([tex]\(n\)[/tex]) = 40
- Significance level ([tex]\(\alpha\)[/tex]) = 0.10

2. Use the formula for calculating the test statistic [tex]\( z \)[/tex]:

[tex]\[ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

3. Substitute the known values into the formula:

[tex]\[ z = \frac{6.7 - 6.5}{\frac{0.22}{\sqrt{40}}} \][/tex]

4. Simplify the expression:
- Calculate the denominator: [tex]\(\frac{0.22}{\sqrt{40}}\)[/tex]

Let's first find [tex]\(\sqrt{40}\)[/tex]:

[tex]\[ \sqrt{40} \approx 6.3246 \][/tex]

Now, divide the population standard deviation by this value:

[tex]\[ \frac{0.22}{6.3246} \approx 0.03478 \][/tex]

So the expression becomes:

[tex]\[ z = \frac{6.7 - 6.5}{0.03478} \][/tex]

5. Calculate the numerator:

[tex]\[ 6.7 - 6.5 = 0.2 \][/tex]

6. Divide the numerator by the denominator:

[tex]\[ z = \frac{0.2}{0.03478} \approx 5.75 \][/tex]

Hence, the value of the test statistic [tex]\( z \)[/tex] is approximately [tex]\( 5.75 \)[/tex].

Therefore, the correct answer from the given options is:

[tex]\[ 5.75 \][/tex]