Mr.martland decides the length of his square vegetable garden will be 17 ft. He calculates that the area of the garden is 34 ft^
1 answer:
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Using the formula of test statistic, the value of the test statistic is 2.6961.
The mean of the public accountant is
Mean x(p) =50.2+59.8+57.3+59.2+53.2+56.0+49.9+58.5+56.0+51.9/10
x(p) = 55.2
Now the standard deviation of public accountant is
SD(p) = √{∑(x-x(p))^2/n-1}
SD(p) = √(50.2-55.2)^2+(59.8-55.2)^2+..................+(51.9-55.2)^2/n-1
After solving;
SD(p) = 3.34
The mean of the financial planner is
Mean x(F) =48.0+49.2+53.1+55.9+51.9+53.6+49.7+53.9+52.8+48.9/10
x(F) = 51.6
Now the standard deviation of financial planner is
SD(F) = √{∑(x-x(p))^2/n-1}
SD(F) = √(48.0-51.6)^2+(49.2-51.6)^2+..................+(48.9-51.6)^2/n-1
After solving;
SD(F) = 2.57
Test Statistic (t) =
t =
After solving
t = 2.6961
Hence, the value of the test statistic is 2.6961.
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The right question is
public accountant 50.2 59.8 57.3 59.2 53.2 56.0 49.9 58.5 56.0 51.9
financial planner 48.0 49.2 53.1 55.9 51.9 53.6 49.7 53.9 52.8 48.9
Use a 0.05 level of significance and test the hypothesis that there is no difference between the starting annual salaries of public accountants and financial planner
Find the value of the test statistic.
Answer:
a)
And the 99% confidence interval would be given (0.630;0.730).
b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)
c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population
Step-by-step explanation:
Previous concepts
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".
represent the real population proportion of interest
represent the estimated proportion
n=575 is the sample size required (variable of interest)
represent the critical value for the margin of error
The population proportion have the following distribution
Part a
The confidence interval would be given by this formula
For the 99% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 99% confidence interval would be given (0.630;0.730).
Part b
We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)
Part c
For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population