Answer:
The p-value obtained from the hypothesis test is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & say that there isn't enough evidence from the sample to conclude that the the calcium concentrations have changed since 2010.
Step-by-step explanation:
To perform this test, we need to first obtain the sample mean and sample standard deviation.
The data is 0.065 0.087 0.070 0.262 0.126 0.183 0.120 0.234 0.313 0.108
Mean = (sum of variables)/(sample size)
= (0.065+0.087+0.070+0.262+0.126+0.183+0.120+0.234+0.313+0.108)/10
= 0.1568 milligram per liter
Standard deviation = σ = √[Σ(x - xbar)²/N]
Σ(x - xbar)² = 0.00842724+0.00487204+0.00753424+0.01106704+0.00094864+0.00068644+0.00135424+0.00595984+0.02439844+0.00238144
= 0.0676296
σ = √[Σ(x - xbar)²/N] = √(0.0676296/10) = 0.08224 milligrams per liter
To do an hypothesis test, we need to first define the null and alternative hypothesis
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, the null hypothesis would be that there is no significant evidence to suggest that the the calcium concentrations have changed since 2010. That is, the calcium concentrations haven't changed since 2010.
The alternative hypothesis is that there is significant evidence to suggest that the the calcium concentrations have changed since 2010. That is, the calcium concentrations haven't changed since 2010.
Mathematically,
The null hypothesis is represented as
H₀: μ = 0.11 milligrams per liter
The alternative hypothesis is given as
Hₐ: μ ≠ 0.11 milligrams per liter
To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ₀)/σₓ
x = sample mean = 0.1568 milligram per liter
μ₀ = the mean calcium concentrations from 2010 = 0.11 milligrams per liter
σₓ = standard error = (σ/√n)
where n = Sample size = 10
σ = Sample standard deviation = 0.08224 milligrams per liter
σₓ = (σ/√n) = (0.08224/√10) = 0.026
t = (0.1568 - 0.11) ÷ 0.026
t = 1.80
checking the tables for the p-value of this t-statistic
Degree of freedom = df = 10 - 1 = 10 - 1 = 9
Significance level = 0.05
The hypothesis test uses a two-tailed condition because we're testing in two directions. (Greater than and less than)
p-value (for t = 1.80, at 0.05 significance level, df = 9, with a two tailed condition) = 0.105391
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.105391
0.105391 > 0.05
Hence,
p-value > significance level
This means that we fail to reject the null hypothesis & say that there isn't enough evidence to conclude that the the calcium concentrations have changed since 2010.
Hope this Helps!!!
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