Mrs. Morton has a special reward system for her class. When all her students behave well, she rewards them by putting 333 marble
1 answer:
Answer:
Step-by-step explanation:
Let r represent the number of additional times the class is rewarded by Mrs. Morton.
We have been given that when all students of Mrs. Morton's class behave well, she rewards them by putting 3 marbles into a marble jar. So marbles added in r additional times would be .
We are also told that right now, the jar has 24 marbles. So total marbles in the jar would be .
Since the students have a party, when the jar has 100 or more marbles, so for having a party for the students total number of marble should be greater than or equal to 100.
We can represent this information in an inequality as:
Therefore, our required inequality would be .
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Answer:
Check the explanation below.
Step-by-step explanation:
Hello!
The study variable is:
X: Arsenic level of well water (ppb)
If the mean concentration of arsenic is 8 ppb then the water is considered safe to use for agricultural use.
In texas a sample of 36 tests for arsenic was taken to see if the average level of arsenic of the water well is less than 8 ppb, symbolically μ < 8
The hypothesis is:
a)
H₀: μ ≥ 8
H₁: μ < 8
Level of significance is α: 0.05
b)
To study the population mean the variable needs to have at least a normal distribution. If the information of the sample was available we could make a normality test to check if it's distribution is normal or not. Since the Students t cannot be used unless the variable has a normal distribution, we cannot choose this statistic to make the hypothesis test, the same happens with the standard normal, if the condition of a variable with normal distribution is not met, the statistic cannot be used.
The most recommendable is to apply the Central Limit Theorem, this theorem states:
Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
X[bar]≈N(μ; σ²/n)
Then you can use the standard normal approximation:
Z= X[bar] - μ ≈N(0;1)
σ/√n
The critical region for this test is one-tailed (left tail)
If , you reject the null hypothesis.
If , you do not reject the null hypothesis.
c)
The statistic is:
Z= -1.75
The p-value is one-tailed with the same direction of the test, you calculate it as:
P(Z < -1.75)= 0.04006
To decide using the p-value you have to compare it against the significance level.
If it is greater than the level of significance, then you don't reject the null hypothesis.
If it is equal or less than the level of significance, then you reject the null hypothesis.
d) The calculated p-value (0.04006) is less than α (0.05), the decision is to reject the null hypothesis.
e) At a level of 5%, the decision is to reject the null hypothesis. This means that there is significant evidence to think that the average level of arsenic in the well water is less than 8 ppb.
I hope it helps!