What is the variable of 6(x+5)= -36
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Answer:
a) - The population consists of all of Danielle's colleagues that could have been one of the randomly surveyed 50.
- The sample is Danielle's 50 colleagues that she ramdomly sampled.
b) From the statistical test performed, there is significant evidence to conclude that Danielle is truly overpaying for her auto insurance compared to her colleagues.
c) Check Explanation
Step-by-step explanation:
The full complete, correct question is attached to the solution of this question.
a) The population is normally the extended distribution where every selected random sample is extracted from. So, for this question, the population will be all of Danielle's colleagues.
The sample is the subset distribution obtained from the population. In the question, it is stated explicitly that Danielle randomly picked 50 of her colleagues to participate in the survey. Hence, the sample is Danielle's 50 colleagues that she ramdomly sampled.
b) The appropriate statistical inference for this question is to carry out the t-test hypothesis test.
For hypothesis testing, the first thing to define is 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, Danielle wants to prove that she is overpaying for her auto insurance compared to her colleagues.
So, the null hypothesis is that there is no significant evidence to conclude that Danielle is overpaying for her auto insurance compared to her colleagues.
That is, Danielle isn't overpaying for her auto insurance compared to her colleagues or better stated that her colleagues are paying more than or just about the same for auto insurance compared to her.
While, the alternative hypothesis is that there is significant evidence to conclude that Danielle is overpaying for her auto insurance compared to her colleagues.
Let μ be the mean Danielle's colleagues' auto insurance fees.
Mathematically,
The null hypothesis is represented as
H₀: μ ≥ 476
The alternative hypothesis is given as
Hₐ: μ < 476
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 = $447
μ₀ = Danielle's auto insurance bill that we're comparing the sample against = $476
σₓ = standard error = [σ/√n]
σ = Sample standard deviation = $75
n = Sample size = 30 (30 colleagues got back to Danielle)
σₓ = [75/√30] = $13.693
t = (447 - 476) ÷ 13.693
t = -2.117 = -2.12
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n - 1 = 30 - 1 = 29
Significance level = 0.05 (Most hypothesis tests are carried out at this level of significance)
The hypothesis test uses a one-tailed condition because we're testing only in one direction. (Checking whether Danielle is overpaying or that the mean is of her colleagues' auto insurance fees is less than Danielle's)
p-value (for t = -2.12, at 0.05 significance level, df = 29, with a one tailed condition) = 0.021342
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 = 0.021342
0.021342 < 0.10
Hence,
p-value < significance level
This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that Danielle is truly overpaying for her auto insurance compared to her colleagues.
c) The two type of errors associated with this test include the Type I and Type II errors.
In Hypothesis testing, A type I error involves rejecting the null hypothesis and accepting the alternative hypothesis when in reality, the null hypothesis is true.
A type II error involves failing to reject the null hypothesis when in reality it should have been rejected. It entails not rejecting the null hypothesis and making conclusions based on the null hypothesis, when in reality, the alternative hypothesis should have been accepted together with its conclusion.
For this question, a type I error entails obtaining from the statistical test that Danielle is overpaying when she isn't overpaying for her auto insurance compared to her colleagues in reality.
A type II error would be obtaining from the statistical test that Danielle isn't overpaying when she is truly overpaying for her auto insurance compared to her colleagues, in reality.
Hope this Helps!!!
