Last year Beth's annual salary was $38,350. This year she received a promotion and now earns $46,462 annually. She is paid biwee
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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] ± * (S/√n)
20.36 ± * (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= 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!