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Nadusha1986 [10]
3 years ago
13

Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In 2010, the concentrati

on of calcium in precipitation in New York State was 0.11 milligrams per liter. A random sample of 10 precipitation dates in 2017 results in the following data (in milligrams per liter). Does the sample evidence suggest that the calcium concentrations have changed since 2010? Perform the test at the.05 level. Assume the measured calcium levels follow a normal distribution.
0.065 0.087 0.070 0.262 0.126 0.183 0.120 0.234 0.313 0.108
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

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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Step-by-step explanation:
If Abe rode for 77mins
Let ta represent Amil time and tb represent Abe time.
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