An Exhibitor is selling decorative wreaths at an arts and craft show. The net profit P in dollars from the sales of the Wreaths
1 answer:
<u>Given</u>:
An Exhibitor is selling decorative wreaths at an arts and craft show.
The net profit P in dollars from the sales of the Wreaths is given by , where N is the number of wreaths sold.
We need to determine the number of wreaths sold to earn a net profit of $100.
<u>Number of wreaths sold:</u>
The number of wreaths sold to earn a profit of $100 can be determined by substituting P(n) = 100 in the equation , we get;
Thus, the number of wreaths sold is 200.
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Answer:
(1) The degrees of freedom for unequal variance test is (14, 11).
(2) The decision rule for the 0.01 significance level is;
- If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.
- If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.
(3) The value of the test statistic is 0.3796.
Step-by-step explanation:
We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.
The following is the amount ($000) earned per month by a sample of 15 Claiborne models;
$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .
The following is the amount ($000) earned by a sample of 12 Klein models;
$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.
(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;
Here, = sample of 15 Claiborne models
= sample of 12 Klein models
So, the degrees of freedom = () = (15 - 1, 12 - 1) = (14, 11)
(2) The decision rule for 0.01 significance level is given by;
- If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.
- If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.
(3) The test statistics that will be used here is F-test which is given by;
T.S. = ~
where, = sample variance of the Claiborne models data =
= 1.007
= sample variance of the Klein models data =
= 2.653
So, the test statistics = ~
= 0.3796
Hence, the value of the test statistic is 0.3796.
Answer:
The 98% confidence interval would be given by (0.326;0.354)
Step-by-step explanation:
1) Notation and definitions
number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states
random sample taken
estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states
true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
2) Confidence interval
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by and
. And the critical value would be given by:
The confidence interval for the mean is given by the following formula:
If we replace the values obtained we got:
The 98% confidence interval would be given by (0.326;0.354)
Answer:
The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly
This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So
X = 1900
By the Central Limit Theorem
has a pvalue of 0.9535.
X = 1300
has a pvalue of 0.0465.
0.9535 - 0.0465 = 0.907.
The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.