PLEASEEE HELPPPP MEE WITH LINE PLOTSSS I DONT UNDERSTAND THISSSS PLEASEE HELPPPP!!!!!!!
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Answer:
Probability of at least 50 obese individuals in our sample is 0.92364 .
Step-by-step explanation:
We are given that a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8% .
Let X = Number of obese individuals
Firstly, X ~
For approximating binomial distribution into normal distribution, firstly we have to calculate and
.
Mean of Normal distribution, = n * p = 300 * 0.198 = 59.4
Variance of Normal distribution, = n * p * (1-p) = 300 *0.198 *0.802 = 47.64
So, now X ~ N()
The standard normal z score distribution is given by;
Z = ~ N(0,1)
So, probability of at least 50 obese individuals in our sample = P(X >= 50)
P(X >= 50) = P(X > 49.5) {using continuity correction}
P(X > 49.5) = P( >
) = P(Z > -1.43) = P(Z < 1.43) = 0.92364
Therefore, required probability is 0.92364 .
Answer:
If we compare the p value with a significance level for example we see that
so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.
Step-by-step explanation:
Data given and notation
represent the mean for the account balances of a credit company
represent the population standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
Calculate the P-value
First we need to calculate the degrees of freedom given by:
Since is a one-side lower test the p value would be:
In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"
Conclusion
If we compare the p value with a significance level for example we see that
so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.