Question

A process that produces bottles of shampoo, when operating correctly, produces bottles whose contents weight, on average, 20 ounces. A random sample of nine bottles from a single production run yielded the following content weights (in ounces):
21.4 19.7 19.7 20.6 20.8 20.1 19.7 20.3 20.9

Assume that the population distribution is normal.

a. Enter the data in Excel and get the descriptive statistics.
b. Conduct a 95% confidence interval for the population mean. Show your work.
c. Test at the 5% level against a two-sided alternative that the null hypothesis that the process is operating correctly.
d. Use PHStat to get the test results for part c. (PHStat is an Add-in from your textbook. You need to install this Add-in before you can use PHStat. See page 53 of the course packet.)
e. The PHStat printout in part d gives you the p-value. Use this p-value to conduct the test for part c.

Answer 1

Answer:

a. Descriptive statistics for the sample.

n= 9

mean X[bar]= 20.36

median Me= 20.30

min: 19.70

max: 21.40

standad deviation S= 0.61

b. 95% Confidence interval [19.88; 20.83]

c. Decision: Support H₀

Step-by-step explanation:

Hello!

I don't have excel so I cannot install PHstat, I've used a statistic software called Infostat for the calculations.

There was a random sample of 9 bottles of shampoo taken to test if the process is operating correctly. If it is, the bottles should weight on average 20 ounces (this would be our study parameter)

The study variable is X: the weight of a bottle of shampoo.

X~N(μ;σ²)

a.

Descriptive statistics for the sample.

n= 9

mean X[bar]= 20.36

median Me= 20.30

min: 19.70

max: 21.40

standad deviation S= 0.61

b. 95% Confidence interval.

Since the variable has a normal distribution and the population variance is unknown, the statistic to use to construct this interval is the Students t:

X[bar] ± $t_{n-1; 1- \alpha /2}$ * (S/√n)

20.36 ± $t_{8; 0.975}$ * (0.61/√9)

[19.88; 20.83]

c. You have to test the hypothesis that the production is operating correctly, if it is so, then:

H₀: μ = 20

H₁: μ ≠ 20

α: 0.05

The statistic value is t$_{H0}$= 1.74

p-value: 0.1198

When you use the p-value to decide on a statistical hypothesis, you should always contrast it with the level of significance using the following rule:

If p-value ≤ α, then you reject the null hypothesis.

If p-value > α, then you do not reject the null hypothesis.

Since the p-value is greater than the significance level, you do not reject the null hypothesis. This means that the average weight of the shampoo bottles is 20 ounces, i.e. the process is operating correctly.

I hope it helps!